Thanks to Harvey and ION for bringing this experiment to light (again). [ION, this was the "treat" I was going to offer the other day].
Video demonstration: Part 1 (http://www.youtube.com/watch?v=eqjl-qRy71w) and Part 2 (http://www.youtube.com/watch?v=1bUWcy8HwpM&feature=related)
We've discussed this experiment and concluded that Professor Lewin is incorrect in his assessment of his observations made regarding this experiment. The fields are indeed fully conserved, and it is more of a measurement error combined with an assumption which allowed him to come to this conclusion that the voltages measured are different depending on the direction taken.
Well, this is true, but Lewin has incorrectly assumed that he is measuring the same exact points in the circuit for each resistor. He is not.
I enjoy the challenge of being able to simulate various things using SPICE. This was no exception. I believe I have clearly shown that this experiment can be simulated, and that Professor Lewin was incorrect in his assessment. I also feel that Lewin may owe those other professors he talked about in his video-taped class regarding this experiment, an apology. They were correct in their disbelief, but they should have known why and where Lewin was in error.
Lewin_NCF_01.png shows the circuit and the resulting voltage measurements. Each probe is positive at the top. L3 is the solenoid, and L1, L2 represent the wiring between the resistors, and comprise part of the induction loop. Clearly Lewin overlooked the fact that the wiring represents an inductance in the loop and in a dynamic sense, points A, A' and D, D' are not the same at all. This is why the voltage is measured to be quite different.
Lewin must have had the scope probes set across each resistor as shown.
Lewin_NCF_02.png illustrates the probe locations had he actually placed them across the exact center of the wiring, as he has "implied".
Can anyone have a go at what the actual voltage measurement would have been, had he placed the probes in the center at position A and D (purple probes)?
Note that in this case, BOTH scope probes would have measured exactly the same voltage.
I'll post the solution after you've all had a chance to contemplate it a little and offer your answers.
.99
No comments or answers? :-X ?
hmm...
Quote from: poynt99 on 2011.02.20, 01:33:21
No comments or answers? :-X ?
hmm...
I've been very busy with other things but will post a reply as time permits.
I drew a diagram to understand your point. In my picture, I show that wires represent an exaggeration of coils. The two points of measuring vary accordingly. The difference is the inductance in betweens. Am I close?
Quote from: GibbsHelmholtz on 2011.02.20, 15:25:28
I drew a diagram to understand your point. In my picture, I show that wires represent an exaggeration of coils. The two points of measuring vary accordingly. The difference is the inductance in betweens. Am I close?
Your drawing appears to be a rearrangement of mine, so I think it is ok. I'm not sure I understand your question though.
.99
Quote from: poynt99 on 2011.02.20, 15:37:25
Your drawing appears to be a rearrangement of mine, so I think it is ok. I'm not sure I understand your question though.
.99
Question? What question? ;D Just want to understand your thinking. No question... atleast not yet. >:-)
Quote from: GibbsHelmholtz on 2011.02.20, 15:25:28
The difference is the inductance in betweens. Am I close?
This is what I am not sure about what you mean exactly.
The inductance of the wiring will be very close to the same in both sections between the resistors. Just as I have drawn it.
.99
poynt99,
Your analysis is correct but I think you may wish to dig a bit deeper into the physics. Lewin is famous for dropping bombs to make you think
The big problem with determining if the good prof is wrong is you must take a stand on whether magnetic fields are truly conservative, in this sense.
By that, I mean conservative in equal distribution of energy on all points of the loop.
Since one side has a resistor 9x the other side, are both sides producing the same amount of waste heat? If they aren't then both sides are not equal in the initial eddy current or the current due to Ampere/Turns.
I'm not disagreeing with your ideas. There are some unknowns about his experiment causing me to wonder if he really did provide a true example of a non-conservative field.
Conservative fields store energy without loss.
The net work done by a non-conservative field taken around a closed loop is non-zero. The net work is invariably negative. Or, a non-conservative field dissipates energy.
Magnetic fields cannot gain or loose energy. Magnetic fields, or magnetic flux density in motion, will actually change motion (direction, density, etc.) because this is true.
So, what I'm wondering is this a special case where the uneven loss of energy from the uneven resistances is cause for also uneven eddy currents & CEMF? If so, the minor inductance you show on your circuit may be the means for the separation between the loop sides and Lewin made use of those.
There doesn't seem to be any close-ups of the exact setup. So, saying he is wrong or right are only words, unless you are convinced there is no true example of non-conservative induction. :)
In any case, those other professors mentioned don't deserve an apology. They should have corrected Lewin or agreed with his experiment. They did neither.
WW, Thanks for responding, and your points have been considered.
However, I stand by the analysis as being accurate. I also stand by the notion that professor Lewin measured each resistor directly across the resistor terminals.
If we eliminate the possibility of induction in the measurement probe cables, two probes placed literally across precisely the same two points anywhere in the circuit, will measure precisely the same voltage on each.
The amount of power dissipated in each resistor is dependent on Ohm's law, and varies according to R value and circuit current. The current in the circuit is always equal for both resistors, therefore calculating the power is a simple task.
If the two resistors are increased to 1k and 9k respectively, what parameters change?
The eddy current idea is interesting, but I think the effective dissipation resulting from any eddy currents in those resistors would be magnitudes lower than that from I2R dissipation.
Maybe Lewin is fully aware of this illusion, which is why there are no close-up views of the experiment? Illusion: 2. The condition of being deceived by a false perception or belief.
Lewin seems to have a very strong personality, and I doubt many, if any question him. In one of his lectures he makes an obvious error with simple Ohm's law, and there was not a single peep from his class about it. Those other professors are probably "afraid" of Lewin, or out or respect, don't question him, just as his students don't.
.99
Lewin's experiment didn't show the rise time on his scope display. I suspect this is due to the settings for trigger & capture.
Is there a reason for the same not shown on your simulation?
Just curious.
Quote from: WaveWatcher on 2011.02.23, 11:15:09
Lewin's experiment didn't show the rise time on his scope display. I suspect this is due to the settings for trigger & capture.
Is there a reason for the same not shown on your simulation?
Just curious.
With reference to the schematic diagram, I would be happy to post any wave form you wish.
In terms of rise time, you can see the rise and fall time of the generator driving the solenoid coil, and you can see the inductance and DCR values of the solenoid coil. Is it the solenoid current that you want to see? No problem.
I posted the wave forms I thought were relevant, and I've shown much more than Lewin showed in terms of the actual measurement points and "hidden" components that actually make up the circuit. I am not trying to obscure anything, if that is what you are getting at. ???
.99
I had no thoughts of you hiding anything.
Now that I'm on something besides my phone, I can see the red and green trace going from zero. So the rise time (on the loop, not the solenoid) may be a couple of ns?
With Lewin's lectures is is easy to spot his frequent mistakes. It isn't so easy when he is trying to emphasize something with his usual grandstand methods. There always seems to be something, aside from the experiment, he tries to sneak in. Normally you have to go through what he says in great detail.
One such detail is his emphasis on the solenoid being in the center of the loop. Visually, you can see the magnetic coupling is rather loose. The oddity of this setup is he is using the equivalent of a loose coupled transformer with a single winding secondary. The odd thing about the secondary is one half of that single turn has nine times the resistance of the other half. This is not something you would find in the real world.
In this case, references to non-conservative action has nothing to do with free energy. My interest is due to the odd transformer layout.
When I get some time I'll perform the experiment to satisfy my curiosity.
I can't watch the videos here at work, but I'm wondering if this is not the old tricky experiment shown to physics students at most universities.
From what I remember, the voltage measurements at the same point are different depending on what side the probe wires lie. It has to do with induction and loops, since the circuit is subjected to magnetic fields, or is sitting over a coil fed AC voltage.
EM
Here are two plots:
sol_loop_current01.png shows both the solenoid and loop currents.
R_power01.png shows the instantaneous power in each resistor.
.99
Another showing rise time.
@EM, I don't think this experiment involves induction into the measurement probes. As you can see, the results work out perfectly at the most fundamental level with no mumbo-jumbo required.
WW, I'm glad you are going to perform the experiment; it saves me from having to do it to prove it to you guys. O0
Is anyone going to take a stab at the now two or three un-answered questions?
.99
Quote from: WaveWatcher on 2011.02.23, 22:22:10
The odd thing about the secondary is one half of that single turn has nine times the resistance of the other half. This is not something you would find in the real world.
It only appears strange. The fact that there are two "load" resistors rather than one is the only "oddity". With a single 1k resistor placed across the ends of the loop, the resulting induced circuit current would be identical, and the total power dissipation would be as well.
.99
Is this professor saying something heretical about this experiment? Is he claiming that it can't be explained? just currious.
Quote from: EMdevices on 2011.02.24, 02:06:16
Is this professor saying something heretical about this experiment? Is he claiming that it can't be explained? just currious.
I'm not sure if he knows what he is claiming. Have a look at his lecture supplement (attachment). His first diagram and assumption that all the currents are clockwise can not be correct.
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=3967)
.99
Quote from: poynt99 on 2011.02.24, 02:38:02
I'm not sure if he knows what he is claiming. Have a look at his lecture supplement (attachment). His first diagram and assumption that all the currents are clockwise can not be correct.
.99
;D
Ah! I never think about reading the supplements.
This is why I wish to perform the experiment.
When he made the statement about Kirchhoff not applying but Faraday always applies, I had an idea of what may be going on.
If I understand him correctly, KVL has limitations. This experiment is supposed to be an example.
What I want to know.... Is it a true example? I'll have to find out :)
Notice that Professor Lewin has shown his volt meters connected to the resistors, not to points A and D. Again I'll ask: What voltage would he measure if the meters were connected to points A and D?
I really fail to "get" his point of this experiment. All seems well and good in EM Theory land. There will be -0.1 and 0.9 volts across those resistors regardless if there is a volt meter (and it's associated leads) connected or not (assuming the meters are very high impedance as he noted).
.99
I doubt KVL has any limitations, and I think it holds in this case as well.
Can anyone tell me how Professor Lewin has skewed this experiment and thus claimed KVL does not work or apply?
.99
I finally watched the video, and I assumed correctly, it's that experiment they love to show, to confuse those young minds.
KVL does not hold, because adding all the voltage drops around a loop should give zero, but here it does not. The reason is that we have magnetic induction and KVL assumes no induction. That's all he's trying to say. And yes you measure different voltages from different sides, I saw that years ago. It's really and important experiment to ponder.
Hope that helps.
EM
@Poynt99
QuoteCan anyone tell me how Professor Lewin has skewed this experiment and thus claimed KVL does not work or apply?
First Professor Lewin did not skew the measurement and second KVL never has applied, the professor has simply provided proof that there is an exception to the rule as there usually is if we look hard enough. The law does not apply to open systems such as this whereby the energy has a "choice" as to the path it takes, that is we have not forced it around a defined path in which the results must always be the same. I should also add that this has nothing to do with the conservation of energy and the issue is in the effects which change the measurement and the fact it does not work out as the law stipulates in this case. I also like this because it follows another truely fundamental law --- If we do things differently sometimes we get different results.
Regards
AC
AC, EM.
I respect both you guys and your opinions, but these responses are a little disappointing. :-\
EM, sorry but you're wrong.
AC, you're the one always lecturing us on how we should be looking beyond what we think we see, and how our minds should be open to "other" possibilities; well free your minds guys, you're not seeing this for what it truly is.
Put on your thinking caps and try again. :D
.99
Start by reviewing KVL: ;)
Kirchhoff's Voltage Law (KVL)
The voltage law says that the sum of voltages around every closed loop in the circuit
must equal zero. A closed loop has the obvious definition: Starting at a node, trace
a path through the circuit that returns you to the origin node. KVL expresses the fact
that electric fields are conservative: The total work performed in moving a test charge
around a closed path is zero. The KVL equation for our circuit is
v1+v2−v=0
(4)
(http://cnx.org/content/m0015/latest/circuit4a.png)
In writing KVL equations, we follow the convention that an element's voltage enters
with a plus sign if traversing the closed path, we go from the positive to the
negative of the voltage's definition.
KVL does not hold for circuits where there's magnetic induction into the circuit loops itself, becuase this creates an EMF that is not accounted for on the circuit diagram as an explicit voltage source, and so when you sum up voltage drops around the circuit it does not add up to zero.
Assume the current flow around the loop is 1mA, and the loop has two resistors, 100 ohms and 900 ohms. The voltage drop due to the current is .1mV and .9mV, and so as you sum up the voltage drops ( - 0.1 - 0.9 = - 1V) you do not obtain zero.
This is not some mysterious circuit, just an example to get the student to realize there are ASSUMPTIONS to every equation they encounter in their books.
EM
P.S. The professor says it's a non-conservative because this example violates the definition of conservative fields, meaning that as you add up voltage drops along any path that returns to the starting point, or it's closed, you should sum up to zero, and for this example it's obviously not the case. But I would not say the magnetic field is non-conservative, just the voltage around the loop in this specific example.
This example is a bit more involved then the professor illustrates, and I have seen the full explanations. for example it depends on the loop the probe wires form and what angle they are respective to the plane of the paper, because we have flux changes here so we need to think 3D. for example if you place the probes plane verticaly in the center, the flux from the solenoid will not flow through your probe plane, so what voltage will it measure? 0.1V or 0.9 V? This always baffles my mind, but there is an answer, can you figure it out? :)
@Poynt99
QuoteThe voltage law says that the sum of voltages around every closed loop in the circuit must equal zero. A closed loop has the obvious definition: Starting at a node, trace a path through the circuit that returns you to the origin node. KVL expresses the fact that electric fields are conservative: The total work performed in moving a test charge around a closed path is zero. The KVL equation for our circuit is
I would have to disagree on several points, first the circuit you have depicted does not resemble the circuit in question in any way as you have incorrectly placed a single fixed voltage source in the circuit. Why would you think you can completely change a circuit to justify the action of another that is nothing like it?, that's crazy.
QuoteIn writing KVL equations, we follow the convention that an element's voltage enters with a plus sign if traversing the closed path, we go from the positive to the negative of the voltage's definition.
I would disagree again, I have used electron flow notation exclusively for the last 15 years ever since I learned that conventional flow notation (+ to -) is in fact incorrect. To this day I still find it bizarre that so many talk of proof, truth and doing things correctly yet still use an antiquated and incorrect form of flow notation.
@Emdevices
I would agree this is too easy --- maybe that is the problem. ;D
Regards
AC
I would say that KVL applies. There is no hocus-pocus going on with some kind of exception. I haven't watched the Walter Lewin clip but I will just state a simple example based on what EMdevices said.
QuoteAssume the current flow around the loop is 1mA, and the loop has two resistors, 100 ohms and 900 ohms. The voltage drop due to the current is .1mV and .9mV, and so as you sum up the voltage drops ( - 0.1 - 0.9 = - 1V) you do not obtain zero.
The net voltage drop around the loop is zero because the interconnect wires are not just dumb equipotential nodes anymore, they become voltage sources. If there is a total voltage drop of 1 volt due to the two resistors then there has to be a total voltage increase of 1 volt between the two interconnect wires which are acting as EMF sources.
For example, use drops for the resistors and increases for the EMF sources:
-0.1 + 0.1 - 0.9 + 0.9 = 0
or
-0.1 + 0.5 - 0.9 + 0.5 = 0
or
-0.1 + 0.6 - 0.9 + 0.4 = 0
As long as EMF1 + EMF2 = 1 volt then 1 mA of current will flow through the loop. The two EMFs only exist if there is changing flux going through the loop.
MileHigh
MH, I would suggest watching the videos first. C.C
Folks,
First I'll say, 'please forgive me, in advance'.
Throughout history we've lived with incorrect rules. When we learn more or use better judgment the rules are corrected or we discover new ones.
What happens when we misapply the rules or just stop teaching the details? Even worse, what happens when we teach these details by providing false examples?
I know KVL has exceptions. This was part of my education. It was in the course material and worked on the bench. (On what might be a funny note: I was also taught that magnetic fields are neither conservative(non-dissipative) nor non-conservative(dissipative). ??? )
I'm just not so sure Lewin's example was truly an example of this case. If it is then the KVL shouldn't be applied and the example is of how laws can be misapplied.
From the little bit of detail we have.... .99 could be completely correct by considering the loop inductance to cover a more common solution.
... more thoughts on other similar issues but maybe another day and thread :-X
EM, AC, KVL does indeed hold.
First, lets look at a few definitions of what KVL is:
1) Electrical Engineering Dictionary - CRC Press, 2000
Kirchoff's voltage law (KVL) a fundamental
law of electricity that states that the
sum of the voltage drops and rises in a closed
loop must equal 0.
2) Fundamentals of Electric Circuits - 3ed. - p.37
Kirchhoff's voltage law (KVL) states that the algebraic sum of all voltages
around a closed path(or loop) is zero.
or alternatively;
Sum of voltage drops = Sum of voltage rises
3) Modern Dictionary of Electronics_7ed - p.408
Kirchhoff's laws- No.2:
The algebraic sum of the voltage drops in any closed path in a circuit is equal to the algebraic sum
of the electromotive forces in that path.
4) The Resource Handbook of Electronics - CRC Press, 2001 - sect. 6.2
Kirchoff's voltage law (KVL). The algebraic sum of instantaneous voltages
around a closed loop is zero. **
** In my opinion, this is the best one of the four offered here.
The first observation one should make by examining what professor Lewin has presented along with his objection, is that his explanation of KVL and his use of the law, is incorrect. Applying KVL involves using voltage drops or losses across the circuit elements, but always using the same direction or polarity. Lewin has arbitrarily flipped the voltage meter across one resistor and this accounts for his opposite polarity problem. In fact, the voltage, if assumed to be correctly measured across the resistors, will sum to either +1V or -1V, not -0.1V and +0.9V. That is his error number 1. How could a prestigious college professor mess up something as basic as KVL?
MH, you are on the ball my friend. :) O0
The second observation one should make, and this is how Lewin skewed the experiment, is that there are two (or four, depending on the position of points A and D) voltage drops that are not accounted for. The interconnecting wire between the resistors must in fact be included in this KVL computation, as per the above definitions, but they were not, and no one except MH has recognized this.
The sum of the instantaneous voltages around this closed loop do indeed equal zero volts!
But you won't get this result if you do not measure all the voltage drops!
See the attached scope shots of all four voltages in the first diagram of my simulation. The sum is 0V.
KVL holds regardless if the emf is induced by induction, or if impressed by a source or emf such as a battery.
@EM, this experiment has nothing to do with induction in the measurement probe cabling or wires. That is a misnomer I'm afraid. @AC, the conventional vs. electron current debate is irrelevant in regards to the results of this experiment. The results are equal either way.
.99
I'll bank some Brownie Points in case I make seven years worth of mistakes! lol
@poytn99
I would disagree, first you are not using a real world test only a simulator, second you have incorrectly used a coil acting as a point source--why not justify your point by just simulating a battery as the source and be done with it?. Changing all the parameters to get the results you like is not what I would call good science and when you do a real experiment with the correct parameters then we can talk about facts.
Quote@AC, the conventional vs. electron current debate is irrelevant in regards to the results of this experiment. The results are equal either way.
I didn't imply that it applied here specifically I just wanted to clarify the fact that most everyone is still doing something incorrectly by repeating a mistake that was corrected decades ago, electron flow notation is correct. So why is most everyone still repeating this mistake?, because people are creatures of habit and they fear change almost as much as death. To be honest I find it kind of comical that so many are ready to defend their normalacy to the bitter end, to defend the past and present no matter how silly or pointless it is because we consider it normal. Then we speak of change but do nothing to change ourselves, we demand change but shun anyone who acts differently, who speaks differently or promotes technology which is different. I think anyone who is brave enough to take that one big step backwards and take a look at the bigger picture may find as I have that there is nothing normal about what we call normal.
There is a good book to justify this, Gustav LeBon--The Crowd: A Study of the Popular Mind, I think old Gustav had it right that if you convince people (or sheep) that something no matter how @#$%up or bizarre is "normal" they will not only do it but also defend it to the bitter end. It is said that Adolf Hitler kept a copy of this book on his nightstand, you see the problem with normal is that you have to step outside the box in order to recognize it, normal people don't have a god damn clue what "normal" is.
Regards
AC
Quote from: poynt99 on 2011.02.24, 02:38:02
I'm not sure if he knows what he is claiming. Have a look at his lecture supplement (attachment). His first diagram and assumption that all the currents are clockwise can not be correct.
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=3967)
.99
Hi Poynt99,
The solenoid has an expanding field during the time frame of the test. That expanding field and its associated flux moves outward from the center of each winding radially in such a way so as to induce a specific voltage in the circuit everywhere simultaneously. It is the change in magnetic flux that does this. The field is round and radial, the circuit is squared, so there are points where the flux does not intersect the conductors at a true 90° and there are places where it does. But because the coil is centered in the middle of the circuit, it induces a current flow in the same direction everywhere, and in this case the orientation of the flux as it passes through the conductors the current flow (conventional opposite electron direction) is clockwise everywhere.
The voltage around a given looped conductor is the the closed loop integral of E·
dl. I'm sure the conductor has some resistance, I think the professor used something like a 4AWG wire at about 4" long between the resistors IIRC (please check this to be sure), so you could factor that in and using this calculator (http://www.consultrsr.com/resources/eis/induct5.htm) you could even get some measure of inductance to plug in there too as tiny as that would be. But the end result is simply that the coil puts energy into the circuit everywhere at the same time, and it all flows in the same direction.
A changing magnetic field is not conservative. Thank goodness that is the case because all of our motors and generators depend on that fact. If they were conservative we would never be able to get any transformations from them. A uniform static magnetic field is conservative though. That's why static magnetic fields cannot be used in transformers. It has to change in order to get energy to transform.
Assuming no voltage is being induced in the resistors themselves (even though in reality we know that some is in this experiment) we can treat the two conductors as voltage sources. So there are two voltage sources that have equal voltage induced in them because they are identical, and the fields react with them identically. It just so happens that each represents 1/2 Volt end to end. So there is 1/2V across each resistor. Relative to point A, one end of the wire is -0.25V and the other end is +0.25V, same with D -0.25V and +0.25V. And that is what the professor was trying to impress on the students, is that near zero impedance can still exhibit a voltage drop across it when it is acting as a power source - it is a path dependent exercise. And it is very non intuitive because it causes you to think there is an inversion in your measurements, like "somebody must have made a mistake". But there is no mistake, it just is what it is - two instantaneous power sources each feeding two resistors of different values in a loop.
So the total series voltage in the circuit then is 1V. Each resistor has +0.25V and -0.25V across it, or 0.5V. So using this measurement method we can now say that the voltage drop from A to D is 0V. Therefore, I can use either A or D to be my reference to any of the four nodes at the resistors and I will get the same reading for each.
How is it possible then, that we measure -0.1V and +0.9V ? If KVL holds, then KCL breaks and if KCL holds the KVL breaks. Or if we somehow fix those two to hold, then we must throw out our laws of induction because in this case the only way we can get the circuit to agree, is if we do the math with path dependence, aka Faraday's law which allows that to happen. Or are we to fool ourselves into thinking that somehow the wires make an adjustment to be nonlinear?
You have only probably watched the two videos that whet your appetite for the explanation. But the whole lesson is nearly an hour long (51 minutes) and fully explains it mathematically by building up to how the current and voltage are induced. http://videolectures.net/mit802s02_lewin_lec16/ (http://videolectures.net/mit802s02_lewin_lec16/) Step by Step derivation PDF http://videolectures.net/site/normal_dl/tag=28248/non-conservative_fields-do_not_trust_your_intuition.pdf (http://videolectures.net/site/normal_dl/tag=28248/non-conservative_fields-do_not_trust_your_intuition.pdf)
I love the part where he says "Their brains couldn't handle it"
Essentially we have an EMF of 1V in the loop and the total loop has 1000 Ohms so the current in the loop is 1/1000A. 1mA * 900 Ohms is 0.9V, and -1mA * 100 Ohms is -0.1V. What is the true voltage across each voltage source (wires) and why is it path dependent?
:)
Quote from: Harvey on 2011.02.25, 11:11:51
Assuming no voltage is being induced in the resistors themselves (even though in reality we know that some is in this experiment) we can treat the two conductors as voltage sources. So there are two voltage sources that have equal voltage induced in them because they are identical, and the fields react with them identically. It just so happens that each represents 1/2 Volt end to end. So there is 1/2V across each resistor. Relative to point A, one end of the wire is -0.25V and the other end is +0.25V, same with D -0.25V and +0.25V. And that is what the professor was trying to impress on the students, is that near zero impedance can still exhibit a voltage drop across it when it is acting as a power source - it is a path dependent exercise. And it is very non intuitive because it causes you to think there is an inversion in your measurements, like "somebody must have made a mistake". But there is no mistake, it just is what it is - two instantaneous power sources each feeding two resistors of different values in a loop.
You should rethink this one out Harvey because you are over-complicating things. You are implying that the centers of the wires are at a kind of ground reference for the loop when that's not the case. You are implying that there is a constraint where the voltage across each resistor is 0.5 volts and that's not the case. The entire loop is floating relative to the outside world and it is a reference unto itself.
Just go back to basics and do the walk around closed loop integral of E·dl.
Look at my one of my examples:
Quote-0.1 + 0.5 - 0.9 + 0.5 = 0
You start your walk and you drop 0.1 volts. At this point along the walk, it's a free-for-all, the EMF can go anywhere. So you continue and find yourself up 0.5 volts and now you are sitting 0.4 volts above from where you started. You continue the walk around the loop and you drop 0.9 volts. Finally, you go up 0.5 volts and you connect the dots. It's really very simple.
MileHigh
Quote from: allcanadian on 2011.02.25, 07:54:41
@poytn99
I would disagree, first you are not using a real world test only a simulator, second you have incorrectly used a coil acting as a point source--why not justify your point by just simulating a battery as the source and be done with it?. Changing all the parameters to get the results you like is not what I would call good science and when you do a real experiment with the correct parameters then we can talk about facts.
I either don't get your points, or they are not valid.
First, a simulator can be successfully applied to many situations, this one included.
Second, why are you wanting to change the experiment by using a voltage source? I am simulating the experiment in such a way that it models what is actually happening, minus the negligible inductance that might occur in the measurement wires. But this can be included as well. However, professor Lewin said himself that the currents in the measurement wires is negligible compared to that through the inner loop where the resistors and interconnecting wires are.
The experiment involves the production of an emf within a loop, and this emf is produced via induction from a solenoid coil located inside that loop. It would seem that proposing to use a battery instead of what the true scenario involves, is unscientific.
The KVL holds.
.99
Harvey,
I'm not really certain what your final point might be, but my posts stand.
I've shown that:
1) professor Lewin is using KVL incorrectly to sum his voltages; the total using ONLY the resistors should be 1V total, not 0.8V.
2) the professor has omitted two of the voltage measurements that clearly if accounted for, show that KVL holds for this case.
3) the two "missing" voltages cancel out the resistor voltages for a net loop voltage of 0V, as per KVL.
It really isn't any more complicated than this.
.99
Harvey, AC, EM,
Your challenge, should you choose to accept it, is to successfully argue against the 3 points in the above post #36.
.99
Hi Darren,
You need to place your source coil inside the loop, not outside - you have inverted one of the sources there. This is why you keep making the mistake with voltage sums thinking that the Professor made a mistake when it is you that have made the mistake. The current all flows in a single direction in the center loop, it is induced by an expanding magnetic field and has a specific direction in the wire. Using a single reference point (D for example) the polarities are inverse for each resistor as current flows into D from one, and out from D through the other.
The Professor shows the math for both Kirchhoff and Faraday. Kirchhoff's method results in 1V for loop (not 0.8 as you mistakenly claim). Faraday's method clearly shows 0V as it should be.
I am confident that once you get your polarities sorted out correctly in your mind and the induced values properly modeled in your simulator that you will gain a much greater appreciation for this demonstration and the professor's methods of teaching.
:)
It would also be good consider this point found on Wikipedia for KVL: http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Limitations (http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Limitations)
Harvey:
Let's assume that the current flow is clockwise and you travel clockwise around the loop. Every time you encounter a resistance you observe a voltage drop. So in this case you observe two voltage drops and they add together to give you a larger voltage drop.
The polarities are not inverse for each resistor as long as you continue in the same direction along the path (or loop). I will try to watch the clip again. I watched it several months ago and seem to recall being uncomfortable with what he was saying for this specific example. Of course in general his series is excellent.
MileHigh
Quote from: MileHigh on 2011.02.27, 02:25:45
Harvey:
Let's assume that the current flow is clockwise and you travel clockwise around the loop. Every time you encounter a resistance you observe a voltage drop. So in this case you observe two voltage drops and they add together to give you a larger voltage drop.
The polarities are not inverse for each resistor as long as you continue in the same direction along the path (or loop). I will try to watch the clip again. I watched it several months ago and seem to recall being uncomfortable with what he was saying for this specific example. Of course in general his series is excellent.
MileHigh
You are correct. Thus the closed loop integral according to KVL results in 1 rather than 0. This is why it fails. In order to apply KVL you must introduce the non-existent power source(s). As the Professor states, that 1V is exactly the induced EMF.
You could theoretically use a flat piece of copper with two rectangular slots in it, and solder the resistors across those slots and get the same effect. As long as the change in flux induces an eddy current circular through the path of those two resistors of 1mA there will be those voltage drops. Of course the actual demonstration was reduced by a factor of 10, so that would permissible in any real world experiment. This matter can be very counter intuitive because in that situation the resistor is shorted out by copper all around it. But the magnetic field if properly centered would induce those currents circularly, and independent of the material on either side of the path.
At the end of the PDF by Lewin, he has a section called "Test Yourself" that was added to impress on the students what is happening. If you try to use Kirchhoff's math on that problem and compare those results to Faraday's method, you will see a large difference. The only way to make Kirchhoff's method work is to add inductors as Darren has, and include them in the equation and at a very fundamental level this is why the simulators perform as well as they do, because they do break things down to those levels.
But it is not necessary to do that if you use Faraday's law to calculate the EMF :)
Harvey,
In SPICE, locations of the inductors is immaterial. All the inductors are coupled (like a transformer) with a chosen coupling factor K. I used 0.5 for K, but this is not critical.
It may appear that one inductor is inverted, but this is not the case. The two inductors representing the wire are connected in series-adding. See my new diagram (attached). The loop is a series loop, and therefore not only must the inductors be placed in series adding, but when you go around with your voltage meter, you must pay careful attention to the polarity of the leads. Professor Lewin has inverted the leads measuring R1, and that is why his voltage measurement is -0.1V rather than the correct +0.1V. Mine is in fact correct. Do you agree now?
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=3991)
It is not I that has implied the loop voltage for the resistors ONLY is 0.8V, it is in fact professor Lewin that has done so. Carefully read my post again. I tried to state that the correct loop voltage is 1V, not 0.8V (+0.9V - 0.1V = +0.8V) as Professor Lewin has implied with his stated measurements.
It can not be denied that professor Lewin failed to mention and measure the two other voltages present in the circuit. By definition all voltage gains and losses must be measured in the loop and summed to formulate the result. This is what is shown in the simulation, and this is what would be measured in real life as well. Not executing the law properly by omitting measurements, does not constitute proper grounds to invalidate that law.
It can also not be denied that professor Lewin has made an error with his KVL loop calculation by inverting one voltage meter.
KVL holds. O0
.99
Darren,
Where exactly in your diagram above do you place the common ground reference for your scope probes?
Professor Lewin was very precise in showing that the reference remained unchanged. It is you who are moving your references.
In addition, you are adding imaginary power sources to the circuit, these do not exist in his diagram or his demonstration and that is why they are excluded from the KVL calculations. As I stated in my last post, it is necessary to add them in to pretend that KVL has not failed when the math clearly shows that it has. Without the power sources (reality) you get 1V around the loop, not 0V. Faraday does not need those imaginary inductors that you have included falsely. Did you even think to measure the real inductance of those wires and model that? :-\
Pretending that the good professor has made some mistake because you don't fully comprehend the magnitude of the demonstration is bad form. It is well known that KVL cannot be used where a changing magnetic field is concerned. Look again at the beginning of the first video when the battery is in the circuit. The professor clearly shows how KVL holds in this case because the battery is included in the calculations that net to zero. By removing the battery, KVL fails both in the math and in real life.
I have tried to help you learn this as tactfully as I can, but your I guess your "brain just cant handle it" and that tells us something about you ;)
EDIT to add:
You may wish to fully avail yourself of the discussion already recorded here which takes this to the most accurate levels:
http://www.physicsforums.com/showthread.php?t=405700 (http://www.physicsforums.com/showthread.php?t=405700)
Darren:
You have the voltage polarities for the inductors in your diagram backwards.
Harvey:
QuoteThe only way to make Kirchhoff's method work is to add inductors as Darren has, and include them in the equation and at a very fundamental level this is why the simulators perform as well as they do, because they do break things down to those levels.
You made reference to the fact that the wires themselves have the property of inductance. That's why Poynt added them to his circuit, it's just to model the physical reality on paper and make it a tangible entity that a layperson can recognize and PSpice can crunch on. The wires
are a power source because of the changing magnetic flux going through the loop that they form.
Anyway, it looks pretty straightforward to me. Let's look at a few cases:
1. Ordinary wire loop with the two series resistors: Here you have current flow and the voltage drops across the resistors in balance with the EMFs from the wires for net zero voltage around the loop. If you stick ideal voltage probes in two spots along a given wire you will measure a voltage.
2. Ordinary wire loop with no resistors: Here you have current flow where the linear EMF in the wire and the distributed resistance in the wire cancel each other out. I'm pretty sure if you stick ideal voltage probes at any two points along the loop of wire you will read zero volts.
3. A superconducting loop of wire with no resistors: Here you are in pure current loop territory. The net flux through the loop is always zero, and your ideal voltage probes will measure zero volts anywhere along the wire.
Anyway, I hope to really look at the clip again soon.
MileHigh
Quote from: MileHigh on 2011.02.27, 08:23:24
Darren:
You have the voltage polarities for the inductors in your diagram backwards.
Actually they are correct. On the voltage meters , I have indicated the polarity of the meter,
NOT the polarity of the voltage that will be measured. ;)
Harvey:
Either one is attempting to sum the voltage drops and gains in a loop
per KVL, or they are not. Is professor Lewin attempting to compute the KVL loop or not? If he IS, and it appears to me that he is, then he is clearly going about it incorrectly. If he is not, then he is attempting to perform a different experiment, and he is mixing apples and oranges. What I have shown is the correct way to compute and use KVL.
We may disagree about how KVL applies, and about what the professor was or was not trying to demonstrate, but
there is absolutely nothing mysterious or unexplainable in the demonstration professor Lewin gave, and that really is the bottom line. And the electric fields are conserved. ;)
.99
<sigh> I really, really don't have time for this (apologies ahead of time for not being available to respond)... but...
If I were Walter giving that lecture and demonstration the thing I would be hoping people would get is loop integrals in changing magnetic fields. Don't focus on just the changing magnetic field in the central circuit -- consider also the changing field in the loops of the measurement circuits.
If you wanted to accurately measure the potential across R1 and R2 in practice you would have to take extreme measures to reduce the loop area for the measuring circuit -- to as close to zero as possible. If there is any area in the measuring loop then you have to include the induced EMF for that loop in any calculation of measured potentials. Even if you wanted to measure the potential between two points A and D accurately you would have to ensure the the measuring loops were as close to zero area as possible no matter where you located the measuring device.
Also -- don't forget the potential induced as you traverse the line integral that the circuit follows. There is an induced EMF per distance (flux cut actually) that is independent of any circuit resistances -- that's what induction is... the sections of wire joining R1 and R2 are EMF sources. To see what's happening don't model this in Spice - use FEM package like Maxwell (or any other competent FEM package) that will capture the induced EMF's in the actual topology.
Thanks for your input Mark, and thank you for affirming the emf sources in the wiring as I have shown and been arguing for.
First, SPICE does an excellent job showing us how this circuit behaves, but yes it would be great to confirm with some magnetic modeling software indeed.
Second, I have run simulations taking into account induction in the measurement wiring, and the contribution to the net current through each resistor is inconsequential. There would have to be hundreds of turns with the measurement wire before its induced emf had a significant impact on the opposing current in each resistor.
Again, professor Lewin is mixing apples and oranges here by introducing this dynamic to the experiment, but even so, it can be accounted for and shown that it has little effect on the measured voltages.
.99
Here are some numbers with a simulation run taking into account the measurement wiring on the left hand voltage meter across R1:
The wiring inductance was increased over that of the inner loop inductance by 1000 fold. So rather than 400nH in each leg, we have 400uH in each leg of the voltage meter. Also, the coupling factor is the same 0.5 from the solenoid to the voltage meter wiring. Meter impedance is 1M.
The voltage measured across R1 goes from the previous 100mV, down to about 98mV. Conversely, if the meter leads are swapped, the voltage measured is about 102mV. With the meter leads at a nominal inductance of 1uH or so, there is no detectable difference in the voltage across R1. Moving the scope probes across the meter's internal 1M resistor produces the same measurement as that directly across R1.
So it is obvious that the meter leads in this experiment have no appreciable effect on the voltage that the meter is measuring. This reinforces the fact that this experiment really is about the total induced inner-loop emf, and how it is distributed among the 4 inner loop components to produce a sum-zero net loop voltage, as prescribed by KVL.
.99
Quote from: Harvey on 2011.02.27, 06:58:47
I have tried to help you learn this as tactfully as I can....
.....You may wish to fully avail yourself of the discussion already recorded here which takes this to the most accurate levels:
http://www.physicsforums.com/showthread.php?t=405700 (http://www.physicsforums.com/showthread.php?t=405700)
Harvey,
There have been a few other discussions about this. I was part of one. The problem is... applying the modified/generalized KVL still yields the correct answers. This practice is also widely accepted. I don't understand how it can be. When this happens either the laws of Kirchhoff or Faraday are misapplied.
I don't see anyone denying the inductance of the wires. What does appear to me is the blurring of the definitions for a field and a force. It is undeniable that the summation will result in zero but only if you apply the correct law, or some generalized form. If the circuit elements include inductance Faraday must be used vs. Kirchhoff's KVL. I'm sure any circuit simulator uses a generalized form otherwise there would be no electric field difference across an inductor.
My memory says terms like 'EMF' & 'voltage drop' and the word 'zero' weren't part of the Kirchhoff's voltage law. I've been wasting a lot of time trying to find the original German text to prove that. I see there is no point. The only time someone will run into a problem with the misuse are discussions like this or the rare case where they should be using a network analyzer.
Maybe this argument is just another way of weeding out the young from the old?
Or the old from the young ;D
WW,
Sorry that you disagree with me also. I guess MH, ION and I stand against AC, EM, Harvey, and WW on this one.
It's not about young or old btw; age is irrelevant when it comes to actual observations.
I think what is important and what should be taken away from this discussion, is that observation is what counts, either on the bench, or in a simulator. Definitions and derivatives of one law to form another, is secondary. Also of importance is stating and understanding the goal of what is being demonstrated.
I gave 4 definitions of KVL, and not one mentioned anything about sources or static conditions etc. The experiment as presented by me in the simulation strictly obeys KVL as stated in those definitions, but there may be other definitions that DO state that induction does not apply. I would disagree, and the results I've shown prove this through.
If all this dissonance is simply the result of semantics, then this has perhaps all been a waste of time.
I stand by my results however, and the fact that professor Lewin has made one error with the voltage meter, and only revealed half of the actual story. What professor Lewin has proven imo, is that he is human, and that even professors make mistakes from time to time.
.99
sorry -- perhaps I was not clear enough...
A more realistic way to Spice this is to model the circuit as 4 equal voltage sources with differing impedances - one for each side of the receiving loop: R1, link, R2, link back to R1. That is as close to reality as you will get with Spice... it correctly models the induced EMF that is independent or resistance, but rather proportional to length of loop (percentage of total flux cut by the loop).
Hi Mark.
I probably don't fully understand what you mean exactly, so perhaps you could SPICE it up and show us?
But in general, I think I must disagree.
The best SPICE can do imo is to emulate what the experiment actually involves. This is what I have done by coupling the solenoid induction coil to the loop wiring. How much closer can you get than that? Isn't the experiment simply a loosely-coupled air-core transformer? Where are my results in error to suggest that this method is not correct or not the most accurate?
.99
Quote from: poynt99 on 2011.02.27, 15:17:14
WW,
Sorry that you disagree with me also. I guess MH, ION and I stand against AC, EM, Harvey, and WW on this one.
It's not about young or old btw; age is irrelevant when it comes to actual observations.
I think what is important and what should be taken away from this discussion, is that observation is what counts, either on the bench, or in a simulator. Definitions and derivatives of one law to form another, is secondary. Also of importance is stating and understanding the goal of what is being demonstrated.
...
It is rare that we disagree. Having different perspectives is an asset for any group. Lets just look at it that way ;)
Observation is not the only thing that counts. Even more important is what is actually happening. You are convinced a meter connection was swapped. I am convinced it wasn't because the negative swing should have appeared. I'm afraid observations won't silence this disagreement. It hasn't before.
In the end I think we'll learn Lewin's lesson was not about physics or electricity.
The important thing to note is we should all agree 1V was applied and 1V was returned.
Quote from: WaveWatcher on 2011.02.28, 03:08:03
It is rare that we disagree. Having different perspectives is an asset for any group. Lets just look at it that way ;)
Agreed.
Quote
Observation is not the only thing that counts. Even more important is what is actually happening.
I had hoped that by using the word "observation", I had implied actual processes.
Quote
You are convinced a meter connection was swapped. I am convinced it wasn't because the negative swing should have appeared.
It is a matter of perspective and goals. I would be interested to hear your take on both, in terms of what this experiment was supposed to be about.
From the perspective of traveling along the loop starting at point D,
and having the voltage meters oriented as they are,
and ignoring the emf produced in the wiring,
and traveling in opposing directions from D, yes I would agree the voltage meters should read as Lewin stated and showed.
However, is there something unusual that happened there? :-\
My question to you and anyone for that matter is this; what was the purpose of this experiment, and what did it prove?
Quote
The important thing to note is we should all agree 1V was applied and 1V was returned.
Can you clarify John what you mean by that statement?
Thanks. :)
.99
btw WW,
Have you now performed the experiment?
.99
I see the students are still second guessing the professor.
Perhaps this will help.
http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/lecture-notes/lecsup41.pdf (http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/lecture-notes/lecsup41.pdf)
It specifically addresses the reasons why many textbooks are wrong in their attempts to modify Kirchhoff's Laws.
We must recognize here that KVL only applies to an electrostatic field. The reason for this is because it is impossible to begin your integral and end your integral at the exact same instant in time. Therefore, if the field changes before you get back to the starting point the loop will not be zero.
This fact is quite evident both in Darren's Scope shots as well as Professor Lewin's scope shots, it is clearly a time changing field i.e. non-conservative. You can tell readily that in both cases (Darren and Lewin) the ground reference for the scope probes is common for both R1 and R2 (on the A side of the resistors).
Rewriting the laws to say "instantaneous" or adding unrealistic inductance in order to simulate a voltage source does not solve the problem. In fact, it perpetuates the misunderstanding and obfuscates the core principle of why KVL should be avoided when measuring and calculating nonconservative fields.
It is precisely these types of misunderstandings that lead to erroneous claims of "COP INFINITY" in systems dowsed with nonconservative field activity.
It can be argued that with fast enough equipment and small enough time slices that we can bury the variances of the loop under the error margins and make everything 'ok' but in reality what we are saying is "I'm too lazy to learn how to apply Faraday's Law, so let me take the easy way out and use KVL - it's close enough"
You may be very interested in knowing how Spice arrives at these results. Search this document for the word "Faraday":
http://ewh.ieee.org/soc/emcs/acstrial/newsletters/summer09/HowSpiceWorks.pdf (http://ewh.ieee.org/soc/emcs/acstrial/newsletters/summer09/HowSpiceWorks.pdf)
grrrr... don't have time -- really don't...
OK -- here is schematic and output -- didn't have time to mess with values but you can see that the result is essentially as Walter shows experimentally. What he gets is what you expect -- just ignore his explanation.
There is no mystery here.
Quote from: marksnoswell on 2011.02.28, 12:27:33
grrrr... don't have time -- really don't...
OK -- here is schematic and output -- didn't have time to mess with values but you can see that the result is essentially as Walter shows experimentally. What he gets is what you expect -- just ignore his explanation.
There is no mystery here.
Thanks Mark.
Looks to me that you've just modified my circuit slightly by incorporating inductance and resistance in two of the components. Your results are similar, but I don't see how this proves anything different than what my results are?
Anyway, I will look at the lecture supplement that Harvey uploaded. Just skimming it so far has revealed some interesting things. ;) This ain't over. ;) LOL.
.99
I have a question for all here, and in particular to those in disagreement with the simulation results and conclusions:
Are points A and D on Lewin's diagram the same as the two points directly across each respective resistor R1 and R2?
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=3967)
.99
@All
I think the most interesting part of this circuit concerns the fact that here we have very many knowledgable people who cannot seem to agree on what occurs in a circuit comprised of two pieces of wire joining two resistors in a loop, lol. Can you imagine such a thing, we assume we know so much and yet the dabate over such a simple circuit extends into areas where the supposed experts including professor's who are teaching our future generations cannot even agree ---- about two pieces of wire joining two resistors. To be honest this does not instill a great deal of confidence in me that anyone knows what their doing, that maybe we believe many things not because they are a fact or because we have verified them as fact but because someone simply told us they are facts and we believed them. What I find even more troubling is the fact that we seem to have no shortage of opinions here but nobody can seem to be bothered to actually ,in reality, join two resistors together with two pieces of wire and prove the matter for themselves, lol, priceless.
Friday I built a 500KV Van De Graaff generator as well as a variation of a Belli machine with a multiplying function for my daughters science project so I am quite confident I can manage joining two resistors together with two pieces of wire and providing some actual facts for the people here.
Regards
AC
GO AC! O0
If I understand this correctly, then here is why you have so many questions. It is because you are only observing one set of wires/resistors placed around one plane of that solenoid.
If you used more wires layered one on top of the other with that solenoid placed inside then you will have more perspective on the effect and this will give you a better idea on what's going on. I had made this diagram shown below a few days ago but forgot to post it. By using more wire/resistor sets and seeing if there is or not any differences between the sets, this will give you a better picture of what's going on.
First of all, according to my own understanding, the solenoid when energized, (well forget it because what I really have to say will just be taken again as mass drudgery so I will play the standard game), each end of the coil will produce a magnetic polarity. The north and south polarity are each turning. Some will say the poles turn in opposite directions, but I do not agree with that. I think they turn in the same direction but it is only the perspective that changes. Like if a tornadoes tail was passing through the solenoid always turning one way but you see two directions if looked down from the north or from the south.
Actually, if you used many coils wound in the identical way and stacked them with the solenoid in the center, then measured output off the coils that are each connected to a separate volt meter all in the same manner, you will see it right away if the reading is going from + or - or if they are all the same polarity or if they are all random. lol
wattsup
this simple circuit experiment has a lot more to teach then KVL limitations and assumptions.
The interesting phenomena here is that the voltmeters will register different voltages depending on their location or angle relative to the circuit plane.
These kinds of problems are not for the Circuits 101 student, but for the electrodynamic crowds that understand magnetic induction and can integrate.
EM
Quote from: poynt99 on 2011.02.28, 13:41:08
I have a question for all here, and in particular to those in disagreement with the simulation results and conclusions:
Are points A and D on Lewin's diagram the same as the two points directly across each respective resistor R1 and R2?
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=3967)
.99
No. Essentially you have +1/2V from A to D with respects to R1 and +1/2V from D to A with respects to R2. Therefore, you have +1V from A to A around the closed loop, or +1V from D to D around the closed loop. This is the nature of the nonconservative field in transition inducing an EMF in a loop of conductor.
Consider that I have drawn attention to the inductance of the wire a couple of times which would be (SWAG) 5.12 nH for a #4AWG at 4" long according to the calculator I have posted a link to here before. So that is part of the clue as to how EMF is being produced in the loop, but it is only a fraction of the equation. Lewin was quite specific with regards to the area of the solenoid and even took pains to shade it in in the drawing. There is good reason for that, because that too is part of the equation.
Let me illustrate the problem like this: You get in your car and drive 100 miles in a convoluted loop of hills and valleys ranging from 100 feet below sea level to 14,000 feet above sea level. You record your mileage, compass heading and elevation at every 10 minute mark. Upon arriving back home you tally up each thing. First you add up all the compass headings and realize that it totals to zero. Then you add up all the elevation changes, these too add up to zero. Then you add up your mileage, being careful mind the polarities of N and E = plus while S and W = minus and again, the result is zero. KVL HOLDS! Fantastic.
Now the next day, you do the same thing. But little did you know that half way around the loop an Earthquake occurred and your home elevation was lifted to exactly 10 feet higher than it was before. The closed loop integral of the compass readings is still zero, the mileage is still zero, but the elevation is not zero. KVL failed. Bummer. So you dig into the matter and find out why it failed. Then you say "no problem, all I have to do is change what Gustav stated in his Law originally and introduce a new factor to account for the difference. (That's what we do in Spice, we introduce a specific variable in the inductors that allow us to fool the KVL routines)
KVL is based on an electrostatic field.
Faraday's Law is based on a dynamic field because it is rooted in the change of flux over the change in time (http://upload.wikimedia.org/math/9/7/a/97acc856d0b23b7fd64872968abef047.png) (Borrowed from http://en.wikipedia.org/wiki/Faraday%27s_law_of_induction (http://en.wikipedia.org/wiki/Faraday%27s_law_of_induction))
Consider the attached image. I have purposely excluded the polarities so that you don't know which way the current is nor the direction of the induced EMF. It is a flat sheet of copper with holes punched out and resistors soldered in place. A dotted line shows a solenoid in the back ground. What conditions are necessary so as to produce a 0.1V drop across the left side resistor (call it R1) and a 0.9V drop in the right side resistor (call it R2) ? Which law is best suited to solving this problem, KVL or Faraday's law?
Harvey
Thanks Harvey for your responses and your time and effort in trying to help my stubborn mind understand. There are several comments and questions I have from your responses, however I am holding those back for now while I hopefully contemplate posted results of the actual experiment. AC has expressed interest in doing the test, and I've been advised by WW that he too will be doing it and posting his results.
I will however respond to one point in your recent post. The loop inductance calculator I found and used gave me a result that corresponded very closely with what my chosen values are for the interconnecting wiring.
http://www.qsl.net/in3otd/ind1calc.html
Judging from the MIT video, I estimate the solenoid diameter to be between 4 to 6 inches. I also estimate the loop to have a circumference of about 0.75m (it does not appear to be closely-wound on the solenoid).
Try 0.75m for the "perimeter", and 1mm wire diameter. The result is 833nH, and I used a total of 800nH in the first schematic.
.99
EDA:
I should also mention that the actual inductance values present in the loop wiring is of little consequence to the results of the experiment anyway. The induced emf will vary according to relative inductance and K factor between the solenoid and the loop, but the ratio of voltages measured across the two resistors, is a constant, regardless of the total induced emf. Therefore over all, the actual inductance of the loop wiring in relation to the solenoid coil is largely inconsequential to the resulting different voltages and their polarities.
EDA2:
The inductance result with your calculator and my parameters comes to 1088nH. Note that the calculator you provided is for a straight piece of wire, mine is for a circular loop, which of course is what professor Lewin would have placed around the solenoid.
Quote from: Harvey on 2011.03.01, 21:29:22
No. Essentially you have +1/2V from A to D with respects to R1 and +1/2V from D to A with respects to R2. Therefore, you have +1V from A to A around the closed loop, or +1V from D to D around the closed loop. This is the nature of the nonconservative field in transition inducing an EMF in a loop of conductor.
In regards to the 1/2V drops, is this in line with what professor Lewin is saying? I'm confused. See "Lewin_path01.png" below.
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=4069)
Quote
Consider the attached image. I have purposely excluded the polarities so that you don't know which way the current is nor the direction of the induced EMF. It is a flat sheet of copper with holes punched out and resistors soldered in place. A dotted line shows a solenoid in the back ground. What conditions are necessary so as to produce a 0.1V drop across the left side resistor (call it R1) and a 0.9V drop in the right side resistor (call it R2) ? Which law is best suited to solving this problem, KVL or Faraday's law?
Harvey
First, you need a homogeneous field, and you will not attain one with a solenoid placed as you described. Your version would require a Helmholtz coil with your copper board situated in the middle.
Having said that, there needs to be changing flux from the coil to induce an emf in the copper foil. This is essentially the same scenario as Lewin's setup, but with his, the current is confined to the wire loop. In yours, the current will flow along the path of least impedance.
In regards to solving the "problem", please re-state what the "problem" is from your perspective
Thanks,
Darren
Quote
Hi Darren,
First, the Helmholtz coil in unnecessary, all that is needed is for a change in flux to occur in the copper, and this will as the solenoid is energized.
A uniform magnetic field is what is prescribed. You don't get that at the end of a solenoid.
Quote
The proposal by Kirchhoff is that the sum of those two line integrals from A to D and D to A, or alternatively as shown D to A via R1 and D to A via R2 should equal zero. But for that to work E must be stable, and when E is stable there is no induced EMF and if there is no induced EMF then the EMF everywhere is zero and KVL would hold.
KVL holds in both dynamic and static conditions. It is the measurement method that is in error, not the application of KVL.
Quote
Why is it a bad idea to try and force KVL to work in changing magnetic fields by introducing changing power sources to match the drops?
It is perfectly valid to use KVL when the circuit and measurements are fully understood.
.99
Harvey,
What professor Lewin is discussing at that moment of the video (7:10, part1), is that if we go from D to A (or A to D), the answer depends on the path. Going through the left-hand path, we get -0.1V. Going through the right-hand path we get +0.9V.
You said this:
QuoteEssentially you have +1/2V from A to D with respects to R1 and +1/2V from D to A with respects to R2.
...which seems to contradict what the professor is describing and illustrating on the board.
So, I am confused by your statement.
Darren
Hi Darren,
http://www.youtube.com/watch?v=eqjl-qRy71w#t=7m40s (http://www.youtube.com/watch?v=eqjl-qRy71w#t=7m40s)
Perhaps you didn't watch it to the end? Just after the Professor tells us NOT to use Kirchhoff's Law with that demonstration you are alluding to at 7:10, he discusses going from D to D for around the whole loop as I have alluded to in my prior posts.
The whole loop is 1V as he clearly shows, Kirchhoff demands the loop to be zero. Remember my illustration regarding the earthquake earlier? That was to help you grasp the non-conservative action of a changing magnetic field.
As regards the 1/2V references, I went into some detail on that in my prior post. I'm sorry if you are still confused on that matter, but I'm running out of ways to explain it to you, and quite frankly I know you are much smarter than you are letting on there - in fact, I would venture a guess that you are not confused at all, but that you are trying to find something to argue about and this is an apparent discrepancy and you think you are exposing a flaw of some sort. You are mistaken.
Since the Professor and I both agree that there is 1V from D to D around the closed loop, then it stands to reason that there is 1/2V from D to A or from A to D because the circuit is in parallel when evaluating only half of the measurement. In this case I am offering an alternate 'outside the box' perspective for viewing the events in place by giving an "essentially" equivalent scenario that includes the sources within the network. The professor however, is addressing the issue that the sources do not exist and the EMF is introduced dynamically via induction. We are both adding EMF to the circuit, but in different ways.
In my case I draw attention to the fact that using the "length" in the calculation offers us the ability to actually measure the size of the wire and the size of the resistors. This is a very important thing to consider when answering your question regarding whether or not A and D are the 'same' as measuring right at the resistors. It is clearly not the same, because there is a connecting wire between them. Therefore, an EMF must exist between D and R2 and it must be an arbitrary value that was not measured, but we know it exists because there is "length" between D and R2 and an EMF is present in that length. So if I measure 0.5V from D to A, and 0.9V across R2 then I know that somewhere between D and R2 there is an EMF and somewhere between R2 and A there is an EMF and that combined EMF must be equal to 0.4V. If it were completely linear, and I don't expect it is, then we could say that we have a +0.2V from D to R2 and a -0.2V from R2 to A which results in the 0.9V across R2. A similar thing could be present for R1 But with greater magnitudes. The only reason I picked 1/2V for A to D and D to A was because these points are drawn equidistant and I arbitrarily chose to split the 1V down the middle to illustrate the importance of NOT assuming D or A was positioned directly on the resistors.
Perhaps the entire exercise would have been less confusing for you if we removed the wires from the circuit as I mentioned in my prior post. Imagine a carbon ring that has an asymmetrical resistance. From exactly 0° to 180° it measures (non parallel) 900 ohms to the right and 100 ohms to the left. Now we can place A at 0° and D at 180° and your question regarding the position of A and D would be true. In that case, no wires would exist other than the conductivity value of the carbon ring. Then we can use all the same parameters as before and everything Lewin stated would be true, your sim would fail and my 1/2V insertions would vaporize. I think that would finally put an end to your confusion. O0
I hope that helps ;)
Harvey
@All
Perhaps a simple analogy is in order, I have a toy train set on a circular track now if the beginning of the train is connected to the end of the train then the train is continuous and if I apply an external point of force anywhere on the train the whole train must move equally. However it should be painfully obvious that the train cars ahead of the applied force are pushed together and the train cars behind the applied force are pulled apart as well if part of the train is pushed and another part is pulled then on the opposite side of the track where push meets pull there must be a singular point where the train car is neither pushed nor pulled, it is both and neither and the forces sum to zero.
Now if on the left side of the track I have a resistive hump 1cm high and on the right side of the track I have a resistive hump 9cm high and every train car has an equal applied force then something interest happens. The greatest measurable force pushing the train cars together is obviously just "before" the 9cm high hump on the right hand side and the greatest force pulling the train cars apart is obviously just "after" the 9cm high hump on the right hand side. At the very peak of the 9cm hump there must be a singular point where a train car is both pushed from behind and pulled from the front and between these forces of push and pull there is a singular point that has both and neither forces applied or sums to zero.
Now let's consider the left hand side of the track where we find the rather small 1cm high resistive hump in the track, obviously the train cars before the 1cm hump are pushed together and the train cars after the 1cm hump are pulled apart but we have one little problem here. Can you see it? If not then draw a sketch of the system.
The problem here is that the greater forces pushing the train cars "together" before the larger 9cm high resistive hump on the right have extended all the way back through the train cars and onto the 1cm high hump on the left, the train cars falling down the 1cm hump on the left are not being pulled apart they are being pushed together by all the other cars before the 9cm hump on the right. As well we find that all the train cars falling "after" the larger 9cm hump on the right are being pulled apart but at one singular point the train cars cease being pulled apart and start being pushed together very near the smaller 1cm hump on the left.
If we examine the forces in this system we find that the greatest difference between the push and pull forces is directly across the peak of the larger 9cm hump on the right, we also find that the smallest difference in forces occurs very near the peak of the smaller 1cm hump on the left. That is the difference in forces across the larger 9cm hump is a large number and the differences in forces across the smaller 1cm hump is a small number near zero. It should be obvious when we look at the differences in the forces involved that there is no way in hell the numbers on the right and left could possibly sum to zero, that is absurd.
Maybe I am simple minded because I never cease to be amazed even when considering things as simple as a child's toy and all the dynamics involved. If you want to understand the dynamics of the Prof Lewis circuit simply replace the train cars with free electrons, the "humps" or hills with a resistance and consider that each train car or free electron has a singular external force applied to it.
Regards
AC
Harvey,
I'm sorry that you take me as being disingenuous. I'm mainly interested in technical discussions and understanding the point of view of others regarding this experiment. I believe I have a good grasp of this experiment and I do not yet buy into the assertion that professor Lewin et al are promoting, but I am also trying to understand why others see something different than I do.
I have difficulty understanding what people write sometimes, which is why I have been asking you for clarification, that is all. It seems to be upsetting or frustrating you though, so I'll simply refrain from discussions on this topic with you, until perhaps a later date.
As it seems no one else is really interested in this discussion, I guess it may be quiet here for a while.
Cheers,
Darren
Quote from: poynt99 on 2011.03.08, 23:57:00
I'm mainly interested in technical discussions and understanding the point of view of others regarding this experiment. I believe I have a good grasp of this experiment and I do not yet buy into the assertion that professor Lewin et al are promoting, but I am also trying to understand why others see something different than I do.
To that end....
One of the points about voltage drop sticks in my mind. I'll try to explain it in my current foggy mind state (it has been a long day).
Using the generalized KVL to produce the result of zero requires the addition of voltage drops across inductors. It is easy and common to think this is realistic but in reality...
It doesn't make sense. Mind you... I find Lewin's explanation of the measured voltage drop interesting but don't quite think it is completely correct.(The thing where the actual voltage drop being measured is of the meter input circuit not the drop across the inductor.)
Using our favorite spring analogy, voltage drop across an inductor makes even less sense.
If you apply a compression force to the free end of a spring, the other end secured in position, the spring compresses and stores the energy being applied.
At any point in time during the compression the force applied to the free end is equal to the force applied by the spring to the securing surface at the other end. (not talking about breaking the sound barrier when compressing the spring ;) )
In other words, the voltage difference between the ends of an inductor is zero. There is no voltage drop across an inductor. The force isn't delayed from one end of the inductor to the other.
Of course, this isn't always true but I'm talking about this situation. 'Single Pulse and not considering wire resistance'
So, KVL being what it is - a law about electric fields - does not apply to an inductor. There is no electric field across an inductor.
So what did the wires do? They were voltage sources during the pulse induction. The simulation would be more correct if you replaced the wires with batteries (for that couple of milliseconds). The problem with that is they would each be required to produce 1/2V.
The reason I say that is the total voltage dropped across the resistors is 1.0V.
The reason I say that is KVL is an electric field only law. It cannot be applied in it's original form due to the wires becoming transformer secondaries - due to Faraday's laws.
So what law could be used without mangling it for ease of obtaining results while using completely incorrect methods and not espousing obfuscation nor eschewing elucidation of what is really happening? (Cough, Cough, gag!)
Faraday's.
The problem is some will still be stuck on the odd looking results. I'm referring to the -.1 + .9 volts. or the voltage drop across the resistors. That adds up to .8V :D
Using Faraday the wires become voltage sources. This means their sine changes. Granted, since 'time' is now part of the equation due to induction things become unclear for many.
Just like KVL has become.
You can simplify the problem by rearranging the circuit. You can do that. This is a series circuit. Put the two voltage sources together. They add up to 1V. Put the resistors in series together. They add up to 1V..... Yes, they do.
Another fine point no longer clarified... simple measured potential is the measured difference between 2 points. The difference between -.1V and +.9V is 1 Volt. The voltage sources add up to 1 Volt. Since sources and loads are algebraically added with opposite sine... The result is zero Volts. I don't really care because Kirchhoff didn't use the word 'zero', or the German equivalent.
It all still works but is harder to obtain. (Even when doing it verbally and not exactly correct according to Faraday) At least the hard way means better understanding of what really is going on. This understanding includes things like:
...Zero potential is BS. It is only a reference point to hang your brain on. Like kissin' cousins say, "Its all relative" C.C
...When an unaccounted for variable jumps in it is time to apply a law that includes that variable, not time to 'adjust' the one you understand. Just because the result is correct doesn't mean the method was.
...a current loop can take the same path it would through a loop of wire - even if you replace that loop of wire with the negative equivalent - a sheet of copper with slots for the resistors.
...Rise and fall times do matter :o
One point is...
Applying some mangled form of a perfectly good law may get the correct results but it doesn't help folks understand what is actually happening. One result might be somebody using the reliable and correct spring analogy one moment and then.... contradicting the same idea the next.
Not picking on anyone in particular. I'm guilty of exactly the same thing.
I'll admit to not being precise but the point was to get the point across and give .99 some reference for my opinions on this subject.
Thank you WW for expressing your POV. :)
.99
@Wavewatcher
QuoteSo what law could be used without mangling it for ease of obtaining results while using completely incorrect methods and not espousing obfuscation nor eschewing elucidation of what is really happening? (Cough, Cough, gag!)
Well, we could try the laws of nature instead of distorting the truth to fit the laws dictated by any one person.
QuoteYou can simplify the problem by rearranging the circuit. You can do that. This is a series circuit. Put the two voltage sources together. They add up to 1V. Put the resistors in series together. They add up to 1V..... Yes, they do.
Does it sound reasonable that we should have to completely change the circuit thus the effects so that it agrees with any given law?
QuoteAnother fine point no longer clarified... simple measured potential is the measured difference between 2 points. The difference between -.1V and +.9V is 1 Volt. The voltage sources add up to 1 Volt. Since sources and loads are algebraically added with opposite sine... The result is zero Volts. I don't really care because Kirchhoff didn't use the word 'zero', or the German equivalent.
Does it sound reasonable that we should have to start changing all the figures and facts so that they agree with any given law?
You see in science most would find it unreasonable that when someone does not like the results of an experiment they completely change the experiment and the results then call it the same thing as the previous experiment. It just does not sit well with the scientifically minded because it is a complete distortion of the facts and science is about facts not laws.
Quote...Zero potential is BS. It is only a reference point to hang your brain on
Here is something to consider, we are all intelligent and reasonable people but what if you examined everything with a magnifying glass and all your facts were based on this. Then someone came along with new facts that contradicted your own, it is natural to assume they may be wrong, but what if this other person was using a microscope magnitudes more powerful that your magnifying glass?.
Let's put this in perspective, while you may be using an off the shelf multimeter and oscilloscope I use an instrument I invented which can detect the change in potential when my dog walks across the carpet on the other side of my house. While you are bound to the limited measurement of potential between two points I measure the instantaneous absolute potential at one singular point not relative to "one" other point but "every" other point because my reference is not a point it is a field which as you know are spherical in nature. I understand this is completely beyond your level of understanding and it makes no sense whatsoever but this is how science works, when we develop better instruments we get better facts.
Regards
AC
Ok.
Hmmm.... I guess it is my turn.
I'm not distorting the truth. I offered my opinion.
I didn't change the circuit to change the effects or the truth. Sometimes you can look at things from a different perspective in order to gain understanding. That is another scientific tool.
Figures and facts changing? - Never mind.
>>Edit...
Sorry AC. Your post caught me at a bad time. No reason for me to share the bad times here.
Quote from: poynt99 on 2011.03.08, 23:57:00
Harvey,
I'm sorry that you take me as being disingenuous. I'm mainly interested in technical discussions and understanding the point of view of others regarding this experiment. I believe I have a good grasp of this experiment and I do not yet buy into the assertion that professor Lewin et al are promoting, but I am also trying to understand why others see something different than I do.
I have difficulty understanding what people write sometimes, which is why I have been asking you for clarification, that is all. It seems to be upsetting or frustrating you though, so I'll simply refrain from discussions on this topic with you, until perhaps a later date.
As it seems no one else is really interested in this discussion, I guess it may be quiet here for a while.
Cheers,
Darren
Here is the thing Darren, you're smart and your aptitude is evident in your posts - so when you claim to be confused after repeated diverse explanations it comes across quite disingenuous or even insulting. Not that I'm offended in any way, its only that such tedious process of regurgitating the same things over and over in different textures gets wearisome and nonproductive - especially when they are focused on contextual semantics. I think the word I see thrown around all the time on the forums is 'pedantic' :P. What comes across to me, and other readers is that you just want to be right about this on some level even if its to argue that one divided by two cannot possibly be one half because the professor shows it to be different. Apart from that, I rather like you - you have skills with the Sims that puts many of us to shame. And you have a root fundamental understanding of things that is beneficial in workable ways.
Here is the crux of it: KCL and KVL cannot be applied in a changing scenario. No matter how hard you try, you must extrapolate in between your time slices because each time slice must equal zero for the loop. That is not reality.
So, as is stated in many places, Kirchhoff's Law is only a special case of Faraday's law where the closed loop integral equals zero. The only place this happens is in a conservative field aka static. Therefore, Kirchhoff's Law should not be used or manipulated into use where non-conservative fields are in play.
This matter really isn't of much importance in simulations where we have time to manipulate things to get the desired results. But it becomes very important as we get deeper into the physics of how substrates and conductors interact with changing fields at near quantum levels. The professor recognized that it was unfair to teach our students that it was a certain way when reality is different because it was just setting them up for the big "Why did that happen?" when they first experience it in the real world.
The most valuable thing I took away from that lecture was that changing magnetic fields are non-conservative - ergo, energy can be extracted from them. IMHO, that is way more important than "the wires are inductors". The second most important thing is that non-conservative fields are path dependent and time variant - so it matters both 'when' and 'where' you take you measurements regardless of how the energy got in the circuit.
Kirchhoff's Law is all about path and time independence and that is why it is generally a bad idea to try and use it to evaluate the experiment outlined in this thread. But hey, I've used a butter knife as a screwdriver in a pinch so I'm sure we can get it to work if we tweak it just right. ;)
Cheers,
Harvey
Harvey,
I'm starting to think about something Lewin said. Something about 'you'll never see this again'.
I think what he meant is you will see this constantly throughout your career but not as well detailed or clearly placed in front of you. I think it is a warning to avoid the pitfalls of having to rely on an adjusted rule vs. a real understanding of what happens.
My work is filled with this. I'm supposed to identify a root cause because the previous root cause found proved incorrect and the wrong solution was applied. There was nothing wrong with the previous result. The problem was always the method or the understanding of what transpired. So the wrong fix is applied.
Some of the arguments I get about how wrong I am are amazing. In the end the correct solution is always clear when the answer is correct. Not so with this example, for many.
@wavewatcher
I was going to edit my post as well just after I wrote it, We all know where that kind of critical debate leads, usually nowhere, and I apologise if I was out of line, which I was -- again :D. That was my somewhat twisted way of actually agreeing with you that I do not think KVL can apply here and I believed you were implying we should not have to mangle the law to make it fit where it cannot which I agree with. I would also agree that I went a little overboard when responding to your statement that zero potential is BS. I feel very strongly about this issue because I have spent a great deal of time in research and experiments to get to the point that I can nail down where and why this phenomena occurs. I will try to explain my point of view a little more rationally next time.
QuoteI think what he meant is you will see this constantly throughout your career but not as well detailed or clearly placed in front of you. I think it is a warning to avoid the pitfalls of having to rely on an adjusted rule vs. a real understanding of what happens.
My work is filled with this. I'm supposed to identify a root cause because the previous root cause found proved incorrect and the wrong solution was applied. There was nothing wrong with the previous result. The problem was always the method or the understanding of what transpired. So the wrong fix is applied.
I would agree completely, I am in the process of finishing a new measuring instrument which will show exactly what is happening in the Prof Lewis circuit, where it is happening and why it is happening to hopefully clear this up once and for all. I think we need to clarify as many loose ends in as much detail as we can if we are ever going to progress and from what I have seen so far in my testing of this circuit leads me to believe that the effect misleading many people applies to a great deal of other circuits which do not do what we expect.
Regards
AC
AC,
I can't say that zero potential doesn't exist at all. That question might be a little too deep for me right now with other 'life' going on.
What I do believe is it would be difficult to find a point or level that appears as zero from all perspectives. What I was referring to was zero on a scope or meter in even the more strange setups.
The zero I speak of would better be described as a reference point. 'Bird on a wire' kinda stuff. The ground we walk on is considered zero but it has a charge with respect to everything else in our normal life. While I normally avoid mentioning subatomic particles and don't care to know much more about them, 'neutral' is the closest thing to zero charge I can think of.
'Neutral' is a better word. Basically meaning it is no charge with respect to something else.
I'll admit my attempt to open other minds to the understanding of problems possible with massaging a perfectly good law for the purpose of getting the 'right' answer was stretching it. It is important that folks understand this stuff.
Just believing Lewin's example requires understanding that time and path are core components AND the need for these components to make things like induction, phase relationships and so much else work and understandable.
It torks me off when I speak to folks from our IT department about implementing some well known function. The poor people are highly trained and very smart. The problem is... they have no idea how a computer really works. When you mention 'assembly' to get something done they think you are talking about what must be done after Santa arrives and before the kids wake up.
I had one argue about a related issue until I finally told him he was right. Then I had time to do it myself. I use it on a local server but IT refuses to implement it enterprise wide. They can't understand the source code and are afraid of what might be in it :(
Every discipline needs to avoid short-cutting themselves to the point - where they think what was published a relatively short time ago - is gospel or completely correct. I don't care what the rule is. There is a good chance it isn't complete, is not one-size-fits-all, has been translated to death or just plain wrong.
When you look at occasions where Faraday trumps Kirchhoff, Maxwell contradicts relativity, Maxwell can't work without throwing in a little Lorentz, Faraday looses and Lorentz takes over, Lenz misses one, physic's use of 'covariance' and 'imaginary' (I could go on).... Is there any wonder I don't trust any law unless it passes on my bench?
I still want to perform Lewin's example. Not because I doubt it. I don't. Because it needs to be done in a way it can be tried by others so we can avoid discussions like this in the future. Hell, maybe we'll come up with a better explanation than 'the meter circuit induction'.
I hope so. ;D
The thread is locked for now so that I may respond to a few posts.
Quote from: WaveWatcher on 2011.02.21, 04:54:22
poynt99,
Your analysis is correct but I think you may wish to dig a bit deeper into the physics. Lewin is famous for dropping bombs to make you think
If it was correct, then your follow-up resistance to it is puzzling. ???
Quote
There doesn't seem to be any close-ups of the exact setup. So, saying he is wrong or right are only words, unless you are convinced there is no true example of non-conservative induction. :)
He is wrong and right, depending on your perspective. Magnetic fields are non-conservative, but the emfs induced by them are conservative.
.99
Quote from: allcanadian on 2011.02.24, 15:51:21
@Poynt99
I would have to disagree on several points, first the circuit you have depicted does not resemble the circuit in question in any way as you have incorrectly placed a single fixed voltage source in the circuit. Why would you think you can completely change a circuit to justify the action of another that is nothing like it?, that's crazy.
It is not my circuit btw. This was taken from a web page or book, I don't recall. The purpose of posting it was to get you guys thinking about how KVL is properly applied...in
any circuit.
Quote
@Emdevices
I would agree this is too easy --- maybe that is the problem. ;D
Regards
AC
If it was that easy, you perhaps would have figured it out, but that is not the case.
.99
Quote from: WaveWatcher on 2011.02.24, 23:40:29
I know KVL has exceptions. This was part of my education. It was in the course material and worked on the bench. (On what might be a funny note: I was also taught that magnetic fields are neither conservative(non-dissipative) nor non-conservative(dissipative). ??? )
I'm just not so sure Lewin's example was truly an example of this case. If it is then the KVL shouldn't be applied and the example is of how laws can be misapplied.
It is not KVL that is being misapplied here.
Quote
From the little bit of detail we have.... .99 could be completely correct by considering the loop inductance to cover a more common solution.
This is on the right track.
.99
Quote from: allcanadian on 2011.02.25, 07:54:41
@poytn99
I would disagree, first you are not using a real world test only a simulator, second you have incorrectly used a coil acting as a point source--why not justify your point by just simulating a battery as the source and be done with it?. Changing all the parameters to get the results you like is not what I would call good science and when you do a real experiment with the correct parameters then we can talk about facts.
AC
Again, this is pure nonsense AC. You did not respond to my rebuttal on this (http://www.overunityresearch.com/index.php?topic=739.msg11317#msg11317), so it is worth stating again.
I've not changed any parameters, but those proposing to insert DC sources, would. Seems you are contradicting yourself here. ???
.99
Quote from: Harvey on 2011.02.25, 11:11:51
A changing magnetic field is not conservative.
Agreed.
Quote
Assuming no voltage is being induced in the resistors themselves (even though in reality we know that some is in this experiment) we can treat the two conductors as voltage sources. So there are two voltage sources that have equal voltage induced in them because they are identical, and the fields react with them identically. It just so happens that each represents 1/2 Volt end to end. So there is 1/2V across each resistor.
Incorrect. R1 has -0.1V and R2 has 0.9V with the probe polarity as shown.
Quote
So the total series voltage in the circuit then is 1V. Each resistor has +0.25V and -0.25V across it, or 0.5V. So using this measurement method we can now say that the voltage drop from A to D is 0V. Therefore, I can use either A or D to be my reference to any of the four nodes at the resistors and I will get the same reading for each.
Incorrect. R1 has -0.1V and R2 has 0.9V with the probe polarity as shown.
Quote
How is it possible then, that we measure -0.1V and +0.9V ? If KVL holds, then KCL breaks and if KCL holds the KVL breaks. Or if we somehow fix those two to hold, then we must throw out our laws of induction because in this case the only way we can get the circuit to agree, is if we do the math with path dependence, aka Faraday's law which allows that to happen. Or are we to fool ourselves into thinking that somehow the wires make an adjustment to be nonlinear?
It is quite possible and quite explainable. There is something wrong here, but it is not the application of KVL.
Quote
I love the part where he says "Their brains couldn't handle it"
Me too. ;)
.99
Quote from: Harvey on 2011.02.28, 10:29:03
I see the students are still second guessing the professor.
Perhaps this will help.
http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/lecture-notes/lecsup41.pdf (http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/lecture-notes/lecsup41.pdf)
It specifically addresses the reasons why many textbooks are wrong in their attempts to modify Kirchhoff's Laws.
There is nothing wrong with KVL, and no mod is required.
Quote
We must recognize here that KVL only applies to an electrostatic field. The reason for this is because it is impossible to begin your integral and end your integral at the exact same instant in time. Therefore, if the field changes before you get back to the starting point the loop will not be zero.
This fact is quite evident both in Darren's Scope shots as well as Professor Lewin's scope shots, it is clearly a time changing field i.e. non-conservative. You can tell readily that in both cases (Darren and Lewin) the ground reference for the scope probes is common for both R1 and R2 (on the A side of the resistors).
I can't agree with that. Your concern with "time" in this case is misplaced. It is not an issue. At any instant in time, KVL holds.
Quote
Rewriting the laws to say "instantaneous" or adding unrealistic inductance in order to simulate a voltage source does not solve the problem. In fact, it perpetuates the misunderstanding and obfuscates the core principle of why KVL should be avoided when measuring and calculating nonconservative fields.
This is perfectly correct and realistic. KVL holds, because although the magnetic field is non-conservative, the induced emf
is conservative.
.99
Quote from: Harvey on 2011.03.01, 21:29:22
No. Essentially you have +1/2V from A to D with respects to R1 and +1/2V from D to A with respects to R2. Therefore, you have +1V from A to A around the closed loop, or +1V from D to D around the closed loop. This is the nature of the nonconservative field in transition inducing an EMF in a loop of conductor.
Incorrect. R1 has -0.1V and R2 has 0.9V with the probe polarity as shown.
.99
Quote from: Harvey on 2011.03.04, 08:32:39
Hi Darren,
First, the Helmholtz coil in unnecessary, all that is needed is for a change in flux to occur in the copper, and this will as the solenoid is energized.
A uniform magnetic field is what is prescribed. You don't get that at the end of a solenoid.
Quote
The proposal by Kirchhoff is that the sum of those two line integrals from A to D and D to A, or alternatively as shown D to A via R1 and D to A via R2 should equal zero. But for that to work E must be stable, and when E is stable there is no induced EMF and if there is no induced EMF then the EMF everywhere is zero and KVL would hold.
KVL holds in both dynamic and static conditions. It is the measurement method that is in error, not the application of KVL.
Quote
Why is it a bad idea to try and force KVL to work in changing magnetic fields by introducing changing power sources to match the drops?
It is perfectly valid to use KVL when the circuit and measurements are fully understood.
.99
NOTE: Apologies to Harvey for losing some of his post (http://www.overunityresearch.com/index.php?topic=739.msg11597#msg11597). I hit Modify rather than Quote, and I did not reply to some of the post. Sorry, it was not intended. :-[ Let me know if you have a backup of your posts, and I can put the original text back in.
Sorry for the whirlwind of posts....I'm catching up after having given things more thought. ;)
Have at it if you wish.
.99
Quote
I would disagree, first you are not using a real world test only a simulator, second you have incorrectly used a coil acting as a point source--why not justify your point by just simulating a battery as the source and be done with it?. Changing all the parameters to get the results you like is not what I would call good science and when you do a real experiment with the correct parameters then we can talk about facts.
@allcanadian
I agree with your points regarding the simulator.
@poynt99
Why are you so convinced that a simulation does not lie? You can't make a valid scientific argument from a computer simulation if the simulation is contradicted by an experiment! Protein-folding calculation (a simulation) is now in very advanced stages of research, but it still makes mistakes due to unknown phenomenon (water charge interactions, chaperon proteins, trace ions ,etc)
In the same sense, I've mentioned that simulations do not take into account EM scalar waves, but the existence of these waves can be experimentally proven. Yes I agree simulation is useful, but in reality...
The end truth comes from the bench. The experiment is where science begins.You must always reconsider your theories when contradicted by experiment -- even 'truths' like special relativity.
The physicists Günter Nimtz and Alfons Stahlhofen, of the University of Koblenz, claim to have violated relativity experimentally by transmitting photons faster than the speed of light.[38] They say they have conducted an experiment in which microwave photons—relatively low energy packets of light—travelled "instantaneously" between a pair of prisms that had been moved up to 3 ft (1 m) apart. Their experiment involved an optical phenomenon known as "evanescent modes", and they claim that since evanescent modes have an imaginary wave number, they represent a "mathematical analogy" to quantum tunnelling.[38] Nimtz has also claimed that "evanescent modes are not fully describable by the Maxwell equations and quantum mechanics have to be taken into consideration."[39] Other scientists such as Herbert Winful and Robert Helling have argued that in fact there is nothing quantum-mechanical about Nimtz's experiments, and that the results can be fully predicted by the equations of classical electromagnetism (Maxwell's equations).[40][41]
Nimtz told New Scientist magazine: "For the time being, this is the only violation of special relativity that I know of."
http://en.wikipedia.org/wiki/Faster-than-light (http://en.wikipedia.org/wiki/Faster-than-light)
I will clarify my questioning of the Lewin demonstration:
I had doubts he performed it correctly to demonstrate the reality of the applications of modified KVL. There still isn't enough evidence available to me, that he did perform it correctly. I have known top electrical engineers and a professor of physics who could not perform the experiment correctly.
For those classifying me as sided on this subject.... Of those posting on the misapplication of KVL, I do agree. Some of the other things stated, I do not agree.
I won't be performing the experiment. There is no point. I know the result will confirm the same things it did when I was required to perform it to earn another stripe. Since then, I have learned there is no amount or quality of data possible to convince most people of anything beyond their experiences short of them actually performing the physical experiment themselves.
When you see me shut-up when references are made to the inductor/spring analogy or... EMF, V/m, C or C/m
2 are being transposed, just know I'm avoiding what might become like this thread.
I may be wrong on some posts I made in this thread. I am not discontinuing participation in this thread because I am wrong about the voltage sources or the voltage drop across the inductors.
KVL, as it is most properly taught, does not apply to any circuit including induction.
Since majority rules on this subject (the majority being the lazy teachers and idiotic book writers) I will not argue with the use of modified KVL as long as it comes up with the correct answer. The same goes for a busted $2 calculator.
Some are satisfied with the correct bottom line. I am not.
Quote from: WaveWatcher on 2011.02.21, 04:54:22
.....I think you may wish to dig a bit deeper into the physics.....
.99, I tried to warn you. We are at that impasse again ;D I should have made the above quote my last post on this subject.
Quote from: WaveWatcher on 2011.03.13, 06:29:11
.99, I tried to warn you. We are at that impasse again ;D I should have made the above quote my last post on this subject.
Quote from: WaveWatcher on 2011.02.21, 04:54:22
.....I think you may wish to dig a bit deeper into the physics.....
Thanks WW, but the warning is not necessary. However, I will offer your own advice back at ya. ;)
.99
Quote from: feynman on 2011.03.13, 06:06:54
@poynt99
Why are you so convinced that a simulation does not lie?
SPICE Simulators don't lie, they produce results based on what you give them. If your source data is bad, then the results will be bad. If your source data is good, the results will be good. Source data being the models, and the circuit and sim settings you have input.
Quote
You can't make a valid scientific argument from a computer simulation if the simulation is contradicted by an experiment!
It can if the experiment has been performed with an error or hidden flaw. You are assuming there is no flaw. That is bandwagon science.
.99
.99,
Quote from: poynt99 on 2011.03.13, 13:18:25
SPICE Simulators don't lie, they produce results based on what you give them. If your source data is bad, then the results will be bad. If your source data is good, the results will be good. Source data being the models, and the circuit and sim settings you have input.
We should not forget the capabilities endowed and methods used in the application by the software designer C.C
Quote
That is bandwagon science.
As my hair continues to fall out and loose color, I'm learning much of science is bandwagon science.
All,
I'm going to paraphrase myself from several years ago (Why not? Over the past couple of weeks I've found modified or strangely similar versions of my original 1996 ideas dealing with design procedural flaws in power distribution design).
The good thing... modified KVL in SPICE (Simulation Program with Integrated Circuit Emphasis) allows less rigorous analog or digital folks quicker and easier circuit analysis. The problem is, any flavor of KVL does not explain or allow us to understand what really happens in the circuit (especially confusing for those well familiar with Faraday's works). The reason is that physics doesn't allow an equivalent circuit where the incorrectly applied KVL works. It can't work, because moving the meter physically, without changing the connection points, changes the measured voltage. Digital and poorly informed analog circuit theories do not allow this effect. Faraday does.
This can be demonstrated with any simple loop circuit where two or more loop sections act as secondaries of a transformer with resistors between. Attach a single meter to what we have been calling A & D using connections which allow rotation of that meter and meter circuit around the horizontal axis of the circuit loop. Even if SPICE could handle spatial movement it would fail. As far as I know, Faraday's laws are not included as an algorithm. Only Faraday's Laws (applied with the meter & meter loop left, right and above or below) can answer that question. The meter will read differently in all four of those positions and every position between without changing the meter's connection points.
All, without swapping connections or meter polarities! :o
KVL/KCL only pertain to static situations. They are extremely important and valuable but only fit as a subset of the Laws of Faraday when working with non conservative fields. They are only correct when looking for electric field values (NOTICE: I did not say 'EMF'. If you don't know the difference that is the first problem you MUST address. :()
In the end, the functions of the 'Real Circuit' will obey Faraday's laws because EMF falls in the realm of Faraday's work, not Kirchhoff, Faraday trumps Kirchhoff, and the Laws of Physics demand obedience. SPICE, as I know it, will fail to provide accurate results. It was not designed to handle the more wild issues possible. Hell, the first time I used it, the voltage drops across inductors was correct. They were handled as shorts. O0
KCL/KVL was only attributed to Kirchhoff during his eulogy and thereafter . You'll have to research his history to find this info.
I'm not going to put more time into this as the result is always the same. The statement will be repeated "KVL HOLDS!" when it doesn't even apply. When I see that, it registers in my mind as 'Faraday's laws are invalid for this experiment' . Let's leave the issue where Faraday's laws fall apart to a much later time.
When I see statements about 'voltage drop across inductors', I know Kirchhoff didn't consider time so the speaker is confusing the issue. Then, I await the use of the 'inductor/spring analogy' contradicting the use of 'inductor voltage drop of the electric field'.
.99,
May I ask what flavor of SPICE you are using?
If it is open-source I would be interested in seeing the actual algorithm used to simulate and inductor.
This was an easy one: http://www.compliance-club.com/pdf/Issue82.pdf
I would think the folks at Agilent Technologies know a thing or two about network analysis.
Attached is a clipping from that document.
Hi Darren,
It is unfortunate that my post was altered - I won't hold it against you ;)
They say a picture is worth a thousand words . . . so, attached is a KVL diagram for Lewin's experiment that should illustrate the problem. I think after you review it, you will see the 1/2 V differentials between A and D.
However, the reality is that we measure a 1V closed loop integral from D to D (or from A to A) NOT a 0V integral as this diagram would imply. How do you suppose we use KVL to properly show the reality of the 1V closed loop integral from T0 to T1? Remember, for KVL to hold, we must have a closed loop integral of zero for all T values between T0 and T1 and then we must extrapolate the difference between any two intervals.
It can be seen that each wire represents 1/2 V of induced EMF. These wires are in series as the current in the loop is always in the same direction, so the total induced EMF is 1 V and the loop current is 1 mA. Using ohms law, the wires then exhibit a negative resistance as the EMF is internal for the current flowing within - typical source configuration.
So, the reality is that we have 1/2 V from D to A and another 1/2 V from A to D for a total loop of 1V. Using KVL, we get a negative 1/2 V so that the two EMF values cancel to zero. Can you see any problem with this?
With regards to your Helmholtz assertions, you may want to review Faraday's law (http://en.wikipedia.org/wiki/Faraday%27s_law_of_induction) which requires a change in flux. ;)
IMHO, I think the only way you are going to understand what Lewin is trying to teach you is if you remove the wires from the circuit entirely as I have suggested wrt the resistive ring. This way A and D would truly be at the resistive nodes. All that is needed then is for Faraday's law to tell you how much flux is required to produce the 1 mA of current in the ring. Once that eddy current is established, the rest of the reality should become very evident and then I think you will grasp how KVL fails. As long as the wires are there, they keep serving as a crutch as to how the circuit gets the energy. Once they are removed, then the energy must be supplied by some other means, and when you evaluate that I think you will be forced to see Lewn's perspective on this matter.
I wish you the best in your struggle to understand this, and I wish I could help you in some way to understand it, but at this time I am at a loss of how else to illustrate it in a way you will comprehend.
Cheers,
Harvey
Quote from: WaveWatcher on 2011.03.13, 21:07:03
.99,
We should not forget the capabilities endowed and methods used in the application by the software designer C.C
I guess I will have to keep repeating this until my dying day; There is absolutely nothing wrong with SPICE. It does what it is designed to do, and as I said in a recent post, it is garbage in, garbage out. Obviously SPICE has it's limitations, but applied correctly with certain details in mind, it serves quite well here in highlighting the limitations of the measuring equipment and methods used by those doing this experiment.
Quote
This can be demonstrated with any simple loop circuit where two or more loop sections act as secondaries of a transformer with resistors between. Attach a single meter to what we have been calling A & D using connections which allow rotation of that meter and meter circuit around the horizontal axis of the circuit loop. Even if SPICE could handle spatial movement it would fail. As far as I know, Faraday's laws are not included as an algorithm. Only Faraday's Laws (applied with the meter & meter loop left, right and above or below) can answer that question. The meter will read differently in all four of those positions and every position between without changing the meter's connection points.
All, without swapping connections or meter polarities! :o
This is getting somewhere. What are the ramifications of this?
Quote
KVL/KCL only pertain to static situations. They are extremely important and valuable but only fit as a subset of the Laws of Faraday when working with non conservative fields. They are only correct when looking for electric field values (NOTICE: I did not say 'EMF'. If you don't know the difference that is the first problem you MUST address. :()
In the end, the functions of the 'Real Circuit' will obey Faraday's laws because EMF falls in the realm of Faraday's work, not Kirchhoff, Faraday trumps Kirchhoff, and the Laws of Physics demand obedience. SPICE, as I know it, will fail to provide accurate results. It was not designed to handle the more wild issues possible. Hell, the first time I used it, the voltage drops across inductors was correct. They were handled as shorts. O0
More bandwagon science.
Quote
I'm not going to put more time into this as the result is always the same. The statement will be repeated "KVL HOLDS!" when it doesn't even apply. When I see that, it registers in my mind as 'Faraday's laws are invalid for this experiment' . Let's leave the issue where Faraday's laws fall apart to a much later time.
When I see statements about 'voltage drop across inductors', I know Kirchhoff didn't consider time so the speaker is confusing the issue. Then, I await the use of the 'inductor/spring analogy' contradicting the use of 'inductor voltage drop of the electric field'.
Your cynicism is duly noted.
Quote
.99,
May I ask what flavor of SPICE you are using?
If it is open-source I would be interested in seeing the actual algorithm used to simulate and inductor.
OrCad PSpice. I think Harvey posted a paper on how SPICE works. There is nothing wrong with the way inductors are modeled in SPICE. Good grief, SPICE has been around since the 80's or so, I would hope that someone would have pointed out the flaw by now if there is one. Besides, I've not seen it disagree with reality in all the time I've been using it. I wouldn't waste your time WW.
Quote
This was an easy one: http://www.compliance-club.com/pdf/Issue82.pdf
I would think the folks at Agilent Technologies know a thing or two about network analysis.
Attached is a clipping from that document.
Appears to be another example of bandwagon science.
WW, you say you're done discussing this, and that you are not going to perform the experiment.
You may be interested to know that as of yesterday, I completed the build of the apparatus based on Lewin's design, and I have also completed several experiments, including a number of them today.
I am happy to report that certain experimental results completely validate the results of my simulation, and that the total loop voltage does sum to zero volts as prescribed by KVL. This renders the electric field as being conservative, contrary to popular belief.
I am also happy to report that certain experimental results correspond precisely with what Lewin has asserted regarding the loop voltage.
It is a matter of perspective, and to a certain degree, philosophy, as to which results
you see.
I will be posting a paper, tests, photos, and videos soon, and as time permits. All will be explained, but for now, perhaps a sneak peak of the completed build:
.99
Quote from: Harvey on 2011.03.13, 22:10:33
Hi Darren,
I wish you the best in your struggle to understand this, and I wish I could help you in some way to understand it, but at this time I am at a loss of how else to illustrate it in a way you will comprehend.
Cheers,
Harvey
Thanks Harvey, but despite the overwhelming non-confidence vote here, I'm doing quite well on my own. O0
Stick around and see what unfolds; it's quite possible that you may even learn a thing or two. :) I can assure you that it will be both interesting and controversial.
Darren
Quote from: poynt99 on 2011.03.14, 03:30:37
Thanks Harvey, but despite the overwhelming non-confidence vote here, I'm doing quite well on my own. O0
Stick around and see what unfolds; it's quite possible that you may even learn a thing or two. :) I can assure you that it will be both interesting and controversial.
Darren
I must disagree, you have fallen into the same trap as many before you.
If you can understand the attached, then perhaps you can understand KVL as it was intended to be used. He never intended for it to be used in anything other than a static
E field.
:P
Quote from: Harvey on 2011.03.14, 03:54:17
I must disagree, you have fallen into the same trap as many before you.
If you can understand the attached, then perhaps you can understand KVL as it was intended to be used. He never intended for it to be used in anything other than a static E field.
:P
Thanks Harvey. I've already read that paper a couple of times.
You should know that although KVL comes into play as a consequence of the results from one perspective, KVL is not the real issue at hand here. Sorry if I have not made that clear enough.
In other words, and to reiterate, the crux of the problem has nothing to do with KVL.
.99
When I read the Lewin's paper, I remembered a bit more recent paper from Maxim (ics) that I saw some years ago:
http://www.maxim-ic.com/app-notes/index.mvp/id/2238
I had been very impressed by how easy it is to lead wrong measurements!
The Lewin's paper is not original. His point was already emphasized in 1982 by Robert H. Romer in: "What do 'Voltmeters' Measure?" http://ajp.aapt.org/resource/1/ajpias/v50/i12/p1089_s1. Unfortunately pdf not found for free on the web. I had read it after reading the Maxim's paper but unfortunately I have not kept the file. It should be of interest here.
KVL isn't really the issue. The misapplication of physics laws is one of the issues. Another is perspective.
The position sensor example is only one application of this very basic rule. I have individuals working for me who don't really understand something as base as a motor encoder founded upon this example. They all have far more formal civilian education than me. The greatest problem they face is they can't understand that all laws of physics either have examples of where they don't apply or are at risk of being misapplied.
The ramifications of the different meter readings depending upon relative location? I won't hazard a guess. The implications are more than one. The first being that Lewin was correct in everything contained in that lecture. Also the experiment, like all of his experiments covering the lecture, contains every thing emphasized in the lecture.
Beyond that, I'll refrain to avoid effecting any conclusions from .99.
It is incredibly simple.
Quote from: exnihiloest on 2011.03.14, 11:22:03
When I read the Lewin's paper, I remembered a bit more recent paper from Maxim (ics) that I saw some years ago:
http://www.maxim-ic.com/app-notes/index.mvp/id/2238
I had been very impressed by how easy it is to lead wrong measurements!
The Lewin's paper is not original. His point was already emphasized in 1982 by Robert H. Romer in: "What do 'Voltmeters' Measure?" http://ajp.aapt.org/resource/1/ajpias/v50/i12/p1089_s1. Unfortunately pdf not found for free on the web. I had read it after reading the Maxim's paper but unfortunately I have not kept the file. It should be of interest here.
Yes exnihiloets, indeed I've read those papers and more. Most are incorrect in their conclusions, and the authors seem oblivious to the ramifications. There are papers that go back even farther than Romer's, but he seems to be the one that "popularized" the error (and the erroneous, misleading conclusions stemming from it), and ever since then most have simply jumped on his bandwagon.
.99
Quote from: WaveWatcher on 2011.03.14, 12:55:23
KVL isn't really the issue. The misapplication of physics laws is one of the issues. Another is perspective.
The position sensor example is only one application of this very basic rule. I have individuals working for me who don't really understand something as base as a motor encoder founded upon this example. They all have far more formal civilian education than me. The greatest problem they face is they can't understand that all laws of physics either have examples of where they don't apply or are at risk of being misapplied.
The ramifications of the different meter readings depending upon relative location? I won't hazard a guess. The implications are more than one. The first being that Lewin was correct in everything contained in that lecture. Also the experiment, like all of his experiments covering the lecture, contains every thing emphasized in the lecture.
Beyond that, I'll refrain to avoid effecting any conclusions from .99.
It is incredibly simple.
From your first sentence, I see we are in agreement (essentially a repeat of what I said), at least on that. Now we are getting somewhere. ;) It seems now that I have built the apparatus and have actual measurements, now you are starting to change your tune? ???
If it is "incredibly simple", why has no one here (except myself) yet offered any answers? C.C One of the answers is that Lewin's measurements are correct (as he is measuring across the resistors, and not points A and D), but like many other authors, his conclusions are incorrect, and the experiment misleading in general.
.99
OMG -- this discussion is still going! ...the very first SPICE model I made of this problem (results posted) gave results that agree with experimental evidence.
The trick with any simulator is to know how to accurately simulate reality closely enough to be useful... if your simulations are way off then your missing some fundamental understanding. In this case it was easy to get the simulator to yield results that model reality.
Quote from: exnihiloest on 2011.03.14, 11:22:03
When I read the Lewin's paper, I remembered a bit more recent paper from Maxim (ics) that I saw some years ago:
http://www.maxim-ic.com/app-notes/index.mvp/id/2238
I had been very impressed by how easy it is to lead wrong measurements!
The Lewin's paper is not original. His point was already emphasized in 1982 by Robert H. Romer in: "What do 'Voltmeters' Measure?" http://ajp.aapt.org/resource/1/ajpias/v50/i12/p1089_s1. Unfortunately pdf not found for free on the web. I had read it after reading the Maxim's paper but unfortunately I have not kept the file. It should be of interest here.
Here is a link where you can download the Romer paper freely:
EDIT: Sorry Gyula, in this case, due to the numerous possible public viewing of this hot topic, we wish not to have links to copyright material on the forum. If you want this paper, pay for it as I did. Thanks. .99
Here is an extended version of the Maxim Application Note 2238: http://powerelectronics.com/mag/411pet24.pdf
rgds, Gyula
.99,
I'm not sure what you mean by changing my tune.
As far as I'm concerned, Spice calculates inductance in a length of snapshots instead of di/dt way then fudges the math to make the overall result correct. I just looked at the source code for one used at work. That is indeed how it works.
I have no problem with that. The problem I have is this usage leads to mis or non understanding of what actually happens.
I am looking forward to corrected reasons for this issue. It wouldn't surprise me if you found one.
Quote from: marksnoswell on 2011.03.14, 14:15:26
OMG -- this discussion is still going! ...the very first SPICE model I made of this problem (results posted) gave results that agree with experimental evidence.
The trick with any simulator is to know how to accurately simulate reality closely enough to be useful... if your simulations are way off then your missing some fundamental understanding. In this case it was easy to get the simulator to yield results that model reality.
This discussion is only beginning to reach fruition. 8)
There is absolutely nothing wrong with SPICE in general, nor with my simulation of the Lewin experiment. Please hold off all commentary about SPICE and my simulation of the experiment until you have had a chance to read my full explanation. Otherwise this back and forth nonsense will go on forever. ;) Thanks.
.99
I'm going to be preemptive here and close the thread until I am able to present my test data and explanations etc.. In terms of the arguments, there is a deadlock here anyway, so I see this doing more good than harm. If you need to express your thoughts to me anyway, I'd be happy to receive your PM. Thanks for your patience. :)
Guys, this experiment and the debate stemming from it has nothing to do with KVL, nor SPICE.
Hold on to your chairs! :D
.99
Quote from: WaveWatcher on 2012.02.16, 17:09:03
Usually, the types that Prof. Lewin mentioned as unable to really understand the basics of Faraday over Kirchoff.
No one is infallible, including our beloved Prof. Lewin. Unfortunately, the good professor Lewin has had an oversight, and come to an erroneous conclusion as a result.
The actual fact is that BOTH Kirchoff AND Faraday are satisfied by Lewin's experiment. (presuming that is what you were referring to)
Quote from: poynt99 on 2012.02.16, 19:43:30
No one is infallible, including our beloved Prof. Lewin. Unfortunately, the good professor Lewin has had an oversight, and come to an erroneous conclusion as a result.
The actual fact is that BOTH Kirchoff AND Faraday are satisfied by Lewin's experiment. (presuming that is what you were referring to)
Unfortunately, I must still disagree. Both
are satisfied and Lewin's description is 'almost completely' agreeable for me. It is all about understanding what a meter actually measures and how angle does indeed affect induction and how the common formulas for induction really do not include relative angle. When you include the relative angle it is clear why the meter reads zero directly above and below the coil and why voltage drop from one view is negative and the other view positive.
My point was.... The PMH is only extraordinary because strikes strong interest only in those not understanding basic magnetics - the basics before they were generalized.
There is no current flow unless you magnetize the device with the keeper applied and you don't see more current flow until you remove the keeper. The simple nature of magnetic fields keep the closed magnetic circuit magnetized. It is not much different from preserving the strength of a horse-shoe magnet by applying a keeper bar.
Don't get me wrong... There is a lot to Leedskalnin's work that still interests me. The PMH isn't one, anymore.
Quote from: WaveWatcher on 2012.02.16, 21:19:28
Unfortunately, I must still disagree. Both are satisfied and Lewin's description is 'almost completely' agreeable for me.
How can you disagree with me when you AGREE that both are satisfied? :D
I must disagree about Lewin's analysis being "almost complete". The oversight results in a completely false conclusion, and this is being touted as the absolute TRUTH (and to make matters worse, it's being taught as such at university level physics).
.99
Quote from: poynt99 on 2012.02.16, 22:17:13
How can you disagree with me when you AGREE that both are satisfied? :D
I must disagree about Lewin's analysis being "almost complete". The oversight results in a completely false conclusion, and this is being touted as the absolute TRUTH (and to make matters worse, it's being taught as such at university level physics).
.99
Your statement following mine shows the disagreement ;D
Have you had a chance to actually perform the experiment?
At this date that subject hasn't changed anything. I'm quite sure you and I don't agree on that experiment. I also think this disagreement will not change. So, there is no point in my bringing the subject up, I apologize.
Quote from: WaveWatcher on 2012.02.16, 23:03:20
Your statement following mine shows the disagreement ;D
Have you had a chance to actually perform the experiment?
At this date that subject hasn't changed anything. I'm quite sure you and I don't agree on that experiment. I also think this disagreement will not change. So, there is no point in my bringing the subject up, I apologize.
WW,
I doubt that few if any agree with me.
I did perform the experiment, yes. And my findings do not support the commonly accepted conclusions, including your own. The test is scientific enough that there is no requirement to agree or disagree, nor offer opinions. The results are conclusive, and that's what actually counts.
@Harvey
QuoteAs a side note regarding Poynt99's comment regarding professor Lewin - I disagree wholeheartedly. Professor Lewin did not make a mistake, and he clearly showed that Kirchhoff's Law is a subset and 'special case' of Faraday's law. Any engineer that understands path dependency will fully agree with these statements. Kirchhoff's law simply will not hold where path dependency is present and real world examples of this were previously provided in a prior thread with a link to a well written magazine article on that specific subject.
I would agree with your post 100%, the lengths to which some people are willing to go to conform reminds me of the guy trying to drive a square peg into a round hole which seems kind of fanatical in my opinion.
Quote: "A fanatic is one who can't change his mind and won't change the subject" -- Winston Churchill
Hmm, that quote reminds me of a few of our resident experts, lol.
Regards
AC
I don't know .99 as a fanatic of any kind except the rule of truth.
If he says his results are conclusive then they must be.
I just can't imagine how any results could overturn those Lewin supplied and have been used since inductive position and direction sensors were invented.
@Wavewatcher
QuoteI don't know .99 as a fanatic of any kind except the rule of truth.
I would not say .99 is a fanatic either however if he has no proof to substantiate his claim then really he is no better or worse than any other person in this field of technology making "claims", we cannot say one person making an unproven claim is a nut and the other not because that is absurd.
QuoteIf he says his results are conclusive then they must be.
They are simply unproven claims until proven otherwise ---- as I said no better no worse.
QuoteI just can't imagine how any results could overturn those Lewin supplied and have been used since inductive position and direction sensors were invented.
You are using the wrong perspective, let's start with a question -- it is well know that many physicists, Nobel Prize winning physicists, disagree on many issues. Now if these experts are supossedly the best of the best and light years ahead of anyone here and always correct or moreso believe they are correct then why do they disagree?. You see someone must be right and someone must be wrong, it does not matter how smart they think we are, their credibility or experience they all have the capacity to be right or wrong. Which kind of put's a new perspective on this I'm an expert so I must be right bullshit doesn't it?, if a Nobel Prize winning physicist can be wrong then all bets are off which is why we require proof, explicit proof which applies in every case and in the proper context.
As well the fact that so many supposedly mature professional people have gotten so upset over contradictions to their opinions should tell us something not about science but people. There is no such thing as objective science when people start dragging their emotions and ego's into the fray, if they are mature and responsible adults then maybe they should start acting like it, as they say ---- prove it. I have always found this comical when some self-proclaimed expert starts ranting and then the name calling starts then personal attacks then attacks on credibility. It is so not professional it boggles the mind, it is more like something we might see from children in a playground yet time and time again we see adults doing this very thing and there is no sense to it.
It should be no suprise that this hubub about Kirchhoff's rule is the same hubub as Betz law which has limitations, not unlike the Earnshaw's theorem hubub or the homopolar generator hubub where almost none of the experts can seem to agree. You see it is not so much about expertise or science but people and of course the quote from my last post ---"A fanatic is one who can't change his mind and won't change the subject" -- Winston Churchill.
Regards
AC
Quote from: Harvey on 2012.02.17, 01:34:29
As a side note regarding Poynt99's comment regarding professor Lewin - I disagree wholeheartedly. Professor Lewin did not make a mistake, and he clearly showed that Kirchhoff's Law is a subset and 'special case' of Faraday's law. Any engineer that understands path dependency will fully agree with these statements. Kirchhoff's law simply will not hold where path dependency is present and real world examples of this were previously provided in a prior thread with a link to a well written magazine article on that specific subject.
8)
Just copied this part over from another thread. Applies before Reply #112 above.
Quote from: WaveWatcher on 2012.02.17, 03:09:39
Harvey,
Right on! on your post about the Lewin experiment. Kirchhoff is satisfied because his laws do not apply. His work was done when Faraday's laws over ruled him. You use the term 'sub-set' and that is completely correct. KVL only applies to path independent potential distribution. Path dependency is the telltale sign.
It is a very good thing this works as Lewin described or there would be dozens of different type devices that would never work.
Sorry .99. I would love to see your conclusive results some time. They must be very convincing.
I came up with almost exactly the same conclusion offered by Lewin a very long time ago. Since then, I've performed variations a couple of times. The conclusions never change.
Just copying this over from another thread. Applies before Reply #113 above.
Quote from: WaveWatcher on 2012.02.16, 21:19:28
Both are satisfied and Lewin's description is 'almost completely' agreeable for me. [snip] When you include the relative angle it is clear why the meter reads zero directly above and below the coil and why voltage drop from one view is negative and the other view positive.
I would appreciate some clarification/expansion on your above statements WW.
Thanks,
.99
I will be happy to do so. It is a bit late for me right now.
Basically, induction can be calculated to determine the resulting EMF or the resulting current. Both formula are virtually the same.
The difference is one uses the 'apparent' Gaussian surfaces facing each other (two loops parallel to one another = full surfaces facing each other).
The other uses the previous plus the COS and/or Theta or 'the comparative angle' between them.
It has been stated that this angle is already part of common formula for induction. It is but it assumes changing the relative angle only varies the relative surface area. I should say, it assumes the change in relative surface area is the only thing causing a variation of coupling.
When the meter and the loop of wire connecting it to the two points mid to the resistors is directly above or below the coil zero voltage is measured because there induction is cancelled by the opposing two meter branches. If the resistor loop had its own battery then the meter would measure the same all the way around - above, left, below & right.
Since we are only dealing with induction (& path dependent) the meter reading depends upon the path the current takes.
Phase is shifted.
The difference between the two readings (-.1 & +.9) is the expected 1V. A measured potential is the difference between two points not the total readings taken across segments in series - when it is path dependent.
The meter measures what is on the wires connected to the meter - not what is at the other end of the wires. In EM induction the measured values will depend upon the location of the wires/meter of the measuring device because we are dealing with non-conservative fields(path dependent fields).
With such a circuit the most important thing is to look at the difference between measured values. The answer is the same as if KVL is applied to a path independent(conservative) circuit.
KVL is satisfied because it can not be applied to path independent fields. KVL was only meant to be applied to static potentials or snippets of time showing static potentials of conservative fields.
Induced EMF across points in a series loop (secondary of a transformer) may not be added together to match the applied potential. The only way to get the right answer is to find the difference.
Like I said, it is late and I rambled. This isn't even the tough one. I still have problems where we find Faraday's induction is a subset of Lorentz and not always correct (at least, when it is applied where it shouldn't be applied, like KVL - so I'm told :o )
This is an odd position for me. Here I am.... in defense of accepted theory. Very strange feeling ;D
Quote from: WaveWatcher on 2012.02.21, 03:54:55
When the meter and the loop of wire connecting it to the two points mid to the resistors is directly above or below the coil zero voltage is measured because there induction is cancelled by the opposing two meter branches.
WW,
If I understand you correctly, you are saying that in the depiction attached, there would be zero voltage (or very miniscule) reading on the oscilloscope while "B" is increasing?
Do I also understand correctly that your reason given for this zero voltage reading is because the induced current in each scope wire cancel each other?
.99
I can't say from experience with 6 feet of loop height.
It shouldn't matter. The loop create by the scope leads is the wrong orientation to B to create much induction, in the first place. If there is any measured it would be due to a slightly imperfect vertical orientation and slightly different leg lengths.
If everything is exactly balanced (wire lengths, etc.) you will measure the actual potential difference between the two points(this may be raising a question - I think the proper term for the potential difference is the word 'undefined') - also because induction on that scope loop is cancelled. Making all so well balanced was difficult so I won't be surprised, if anyone does actually perform the experiment, and then cry foul.
I've found the primary coil I used for that and other experiments some years ago. I'm tempted but already know most will claim it is faked.
DC,
There really isn't anything special about what is discussed here. Actually, it is in most higher level books on the subject. I know the contents of these books to be correct. Others say they aren't correct. If this is true and can be proven by experiment, then I want to know.
It is the same old argument. The only thing odd is the participants switched sides :)
The only time induction is special is when you don't quite understand it.
Quote from: WaveWatcher on 2012.02.22, 00:49:25
I can't say from experience with 6 feet of loop height.
It shouldn't matter. The loop create by the scope leads is the wrong orientation to B to create much induction, in the first place. If there is any measured it would be due to a slightly imperfect vertical orientation and slightly different leg lengths.
This is
relatively clear; I'll take this as a YES then, agreed?
Quote
If everything is exactly balanced (wire lengths, etc.) you will measure the actual potential difference between the two points(this may be raising a question - I think the proper term for the potential difference is the word 'undefined') - also because induction on that scope loop is cancelled. Making all so well balanced was difficult so I won't be surprised, if anyone does actually perform the experiment, and then cry foul.
This is unclear. What are you trying to say here? It's NOT zero volts? ???
.99
The increasing B field induces a voltage in the loop such that when integrated equals 1V from A to A or from D to D or anywhere in the complete loop from start to finish the total voltage will be 1V. This is satisfied by Faraday's Law of Induction:
(http://upload.wikimedia.org/wikipedia/en/math/e/b/b/ebb99f3be0e674d27898c3fe8cbf9475.png)
(Image taken from http://en.wikipedia.org/wiki/Faradays_law#Flux_through_a_surface_and_EMF_around_a_loop (http://en.wikipedia.org/wiki/Faradays_law#Flux_through_a_surface_and_EMF_around_a_loop) )
Understanding that the induced voltage of 1V is measured from A to A or from D to D is the first step in comprehending why KVL fails in this instance.
8)
Quote from: poynt99 on 2012.02.22, 02:04:13
This is relatively clear; I'll take this as a YES then, agreed?
Agreed.
Quote
This is unclear. What are you trying to say here? It's NOT zero volts? ???
.99
It will not be 'a' voltage measured. IF you are able to measure a voltage the result will vary from -.1 to +.9 on a scope. An analog meter will show zero volts. The effective measurement will be zero or, more correctly identified as 'undefined'. Some measuring devices may show the average between -.1 and +.9 .
Harvey is correct. The only way to measure voltage is both probes on one point. Even then, the placement of meter conductors is crucial and some of the measuring process must be done in one's head to have the final result. In other words, you must place both probes at one point and visualize what the voltage would be if you opened the circuit between the probes.
You don't normally concern yourself with the voltage drop across a component after cutting the component into different length sections. That voltage drop across each section will change with time.
Quote from: WaveWatcher on 2012.02.22, 10:53:27
Agreed.
That being; you assert that the voltage measured between points A and D in the diagram will be "zero" at any point in time, regardless if "B" is changing or not.
Quote
It will not be 'a' voltage measured. IF you are able to measure a voltage the result will vary from -.1 to +.9 on a scope. An analog meter will show zero volts. The effective measurement will be zero or, more correctly identified as 'undefined'. Some measuring devices may show the average between -.1 and +.9 .
Is this not a contradiction of the above? Please clarify your assertion; will the voltage measured with an
oscilloscope be "zero", OR some value between "-.1 and +.9"? (I am assuming you are not saying that it can be BOTH).
.99
I am afraid you have the same understanding I had before performing the physical experiment. It made no sense that KVL may not apply in all instances. At that point, I would also find my previous statements to be a contradiction.
"At any point in time" is a valid statement for the application of KVL and can't be applied to non-conservative fields in motion. In such fields you must consider the point in time and other factors as the measurement results may be different for each point in time, relative position and path current has taken.
The point in time and current path may determine if the measuring device is shorted, in parallel, shorting or functionally unconnected to the circuit or portions of the circuit.
The same discussion arises when I tell someone the digital world is only a very small subset of the analog world and they believe the digital world can define all in the analog world. At best, digital can only provide a good approximation of the analog.
KVL vs. Faraday is the same thing. Mangled KVL can provide good approximations of such circuits but simply does not apply on its own. When we apply KVL, in what we think is the pure form, we get the wrong answers to questions like these.
If we apply it in the pure form, i.e. only to path independent situations with summations of measured charge, there is no problem.
WW,
In the interest of brevity and focus:
With reference to the attached diagram, and assuming that the scope leads are largely normal to the loop, will the voltage measured with the oscilloscope be "zero", OR some value between "-.1 and +.9"?
.99
You will measure zero EMF.
Since EMF and PD (Potential Difference) are not the same but both are measured as Volts I can't guess at what you will measure as PD.
If I remember correctly you saw no difference between EMF and PD. If that is true then I have no way to make my answer more clear since the difference between PD and EMF is one thing that defines this experiment.
The only thing else I can say is there is no point to continue until you perform the actual experiment.
Sorry, I should have let the sleepy dog lie.
I'll keep asking essentially the same question until you answer it as it is worded. I have a fair bit of patience. ;)
With reference to the diagram, and assuming that the scope leads are largely normal to the loop, will the oscilloscope measure and display:
a) zero (i.e. nothing, no change), OR
b) some appreciable non-zero value?
a) or b)?
.99
Unfortunately, it looks as though WW has decided to withdraw from this discussion. ???
Is anyone else interested in continuing where we left off with this important debate?
.99
Not withdrawn - swamped.
Above and below a significant non-zero VOLTAGE will be measured but it will not be from induction into the scope probe wires.
I haven't figured the details, yet.
The 'above' and the 'below' measurements will be opposite sine and offset from mid-point (.5V)
http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/lecture-notes/lecsup41.pdf (http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/lecture-notes/lecsup41.pdf)
8)
(Also http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/lecture-notes/lecsup315.pdf (http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/lecture-notes/lecsup315.pdf) )
Harvey,
What is your opinion of the possible answer to .99's question?
I've never seen a paper covering this detail so I doubt you'll find another link. The link you last provided does cover this question but folks tend to think on one plane at a time.
Quote from: WaveWatcher on 2012.02.21, 03:54:55
When the meter and the loop of wire connecting it to the two points mid to the resistors is directly above or below the coil zero voltage is measured because there induction is cancelled by the opposing two meter branches.
Quote from: WaveWatcher on 2012.02.24, 03:37:30
Above and below a significant non-zero VOLTAGE will be measured but it will not be from induction into the scope probe wires.
OK.
Since the two above quotes contradict each other, are you now retracting your prior statement shown in the first quote?
No.
The problem is the question doesn't fit the experiment - in the most strict sense.
My last answer is what I expect you will see should you perform the experiment. I would see the same results but translate them differently than I am sure you would.
Not trying to be an ass. Your perspective is still clearly KVL when I know KVL doesn't apply unless your meter is at a very precise position above or below. 'Generally normal' means nothing in this case or the world I work in.
It is a simple question, but it is being unnecessarily complicated by your responses. (you're not being an ass; more like "difficult" or "unreasonable").
Regarding the question at hand, there can only be one of those two answers which are correct, a) or b).
As such, you must retract one or the other of those two quotes. If you do not, then you are obviously undecided on the question.
The question IS INDEED PERTINENT to the experiment, Lewin's lecture, and all the papers and books written about this.
I can assure you that this is not a fanciful indulgence on my part; I ask every question with a specific purpose, and although it may be interpreted by some as a "side track", it is actually a means to an end. I ask for some patience and cooperation, that's all.
Quote from: WaveWatcher on 2012.02.24, 04:09:41
Harvey,
What is your opinion of the possible answer to .99's question?
I've never seen a paper covering this detail so I doubt you'll find another link. The link you last provided does cover this question but folks tend to think on one plane at a time.
His question implies a time-changing magnetic field, but the necessary information regarding that aspect is missing - therefore we are left to assume he is referring to time-changing magnetic field associated with the title of this thread. In that case, the closed loop integral for the circuit is 1V. Since Ohms law holds, the current is the same throughout the circuit with 1V across the series resistance of 900 + 100 Ohms we get 1 mA. Therefore, depending on which path you take you will read 0.9V from A to D across the 900 Ohm resistor (E=IR) or 0.1V from D to A across the 100 Ohm resistor.
Of course Lewin's actual experiment was reduced by a factor of 10, but the video clearly shows both voltages being measured by two separate probes simultaneously.
The actual physics involved shows that the E-field is non-uniform throughout the loop and that charges are naturally created at the resistive junctions in order to balance the net result. This is counter intuitive because standard textbook theory does not provide a mechanism for introducing these electric charges where they need to be.
In order to get the simulators to respond properly, KVL is modified for inductance and sources must be inserted in the loop to provide the necessary drops to bring the net to 0V. But reality shows the net to be 1V.
When we put inductors onto the substrates in chips, you can be sure that Faraday's law is used. I'm sure you can see how critical that would be in very low voltage applications.
8)
Undecided?
Yes. I haven't performed the experiment using 100 and 900 ohm resistors.
I don't consider your question a side track.
In my experiment the orientations shown on your diagram resulted in zero volts measured above and below. I used galvanometers and continuous pulses with difference between the resistors but I don't recall the values. Since a scope can show far more precise values I can only think it will show some fraction of 1V at that position, probably positive voltage and probably slightly more than .5V. The measurement from below will be the opposite with the sum of the two readings being zero.
To provide you with a concise and correct answer I will need more than the usual five minutes I can apply to the problem or I'll need to perform the experiment.
I don't have personal time allowed on MatLab & Femm.
So I can't recant either. I measured zero but you and your scope will probably measure + above and - below with a sum of zero.
Harvey,
No need to repeat what you and I see as the foundation of the thread.
WW,
Logic dictates that both your answers can not be true.
Either there will be no change on the oscilloscope display, OR the oscilloscope will measure and display some appreciable non-zero value.
Until you are willing to firm up your assertion and commit to ONE of these answers, you ARE undecided.
Respectfully,
.99
Quote from: Harvey on 2012.02.24, 07:57:34
His question implies a time-changing magnetic field, but the necessary information regarding that aspect is missing [snip]
The diagram clearly indicates a magnetic field and its direction. When I first posted the diagram (http://www.overunityresearch.com/index.php?topic=739.msg21093#msg21093) I stated that the field was increasing.
I believe WW would agree that we are assuming an induced loop emf of 1V (peak), but the value does not matter when it comes to my question, as long as some appreciable emf is induced.
Then let it be.
I am undecided until I can perform the experiment with a scope.
Considering the much higher impedance of the scope sensing circuit, compared to a galvanometer, my best guess is the graphic position of the scope shown will yield a measurement of positive .4 volts.
It is not my nature to make such guesses. If I don't know for sure I push the questions aside until I find a definite answer.
If I'm wrong, fine. I'll run the test when I have a chance.
Thank you for the answer WW. :)
Harvey, would you care to provide an answer to the same question?
Quote from: poynt99 on 2012.02.24, 13:03:11
The diagram clearly indicates a magnetic field and its direction. When I first posted the diagram (http://www.overunityresearch.com/index.php?topic=739.msg21093#msg21093) I stated that the field was increasing.
I believe WW would agree that we are assuming an induced loop emf of 1V (peak), but the value does not matter when it comes to my question, as long as some appreciable emf is induced.
Fair enough. While we are assuming, let us also assume that the drawn ellipse is intended as an isometric view of a true circle. Let us also assume that the 100 Ohm resistor is farther back from the viewer than the 900 ohm resistor such that the diagram resembles Lewin's setup. And lastly, and mostly pedantic, let us assume the scope connection is a standard BNC wherein the ring is the reference and the tip is the signal input.
With these assumptions in place, we can properly ascertain the polarity and magnitude of the voltage and current induced in the loop. O0
Harvey,
YES to all the assumptions. I might add one other; the wire connecting the two resistors has a uniform resistance along its length totaling no more than 0.1 Ohms each leg.
Are you interested in offering an answer to the question I posed?
Here is a slightly more accurate isometric view (I am not much of an artist) of the diagram (smaller far resistor).
.99
Quote from: poynt99 on 2012.02.24, 19:20:35
Harvey,
YES to all the assumptions. I might add one other; the wire connecting the two resistors has a uniform resistance along its length totaling no more than 0.1 Ohms each leg.
Are you interested in offering an answer to the question I posed?
With the assumptions in place we know from Lenz's law the direction current will flow in the loop do the increasing flux denoted by Vector B which in most equations is called Phi ( Φ ). Therefore, the conventional current will flow clockwise through the loop such that if we start at A it will first reach the 100 Ohm resistor, then D then the 900 Ohm resistor and back to A.
Consequently, we can expect a voltage (EMF) to be present directly at the leads of the 100 ohm resistor proportional to that current and polarized such that the A side is positive. That voltage would be 0.1V. Similarly, we can expect a voltage to be present directly at the leads of the 900 Ohm resistor such that the D side is positive. That voltage will be 0.9V.
If both resistors are the same in size, and the leads between them are identical for the A side and the D side, then we are forced to realize that at that exact instant in time whereby the induced loop voltage is 1V that a differential exists across those leads 'totaling no more than 0.1 Ohms each leg' such that if we if we move our probe and its reference from the 900Ω resistor where it reads zero point nine volts (0.9V), over to the 100Ω resistor where it reads negative zero point one volt (-0.1V) we know that for the pair of leads (both the reference and the signal) we have 1V differential. And this is exactly that 1V that is induced.
Therefore, when we position the reference at A and the signal probe at D we expect to see the midpoint voltage of zero point four volts (0.4V).
We can illustrate this by dividing the length of the leg into 10 sections using 11 points where point 1 is the -0.1V position and point 11 is the 0.9V position. If the EMF is equally distributed, we can expect 0.1V differential at each point. Therefore, point 6 is exactly midpoint thus representing 0.4V.
Why do we have such a high differential on a wire that measures only 0.1 Ohm? Because of inductive reactance. In order for the experiment to work as stated, the time-changing field must be done with such a time frame so as to allow the wire to act as a resistor to the 1 mA induced for the loop. Remember, this wire is functioning as a generator and thus appears as a supply to the circuit replacing the 1V battery previously referenced in Professor Lewin's video. Thus the reactance is part of the internal source impedance.
8)
Edit to correct reference polarity :P
Yea, but what would the measured voltage be if the scope and leads were directly below the loop? :D
I'll give you one guess ;)
(without disconnecting the scope probe from its current points)
Gibbs just did some superposition of the equivalent circuit. Looks like it is in line with some of the thinking in here. The voltage/battery used is 1 V ignored impedance. The two mid points are measured with scope. 1 and 2 Ohms are used to simplify the situation.
The one on the left would reads 0 V.
The one in the middle would reads 1 V.
The one on the right would reads .33V.
One can play around with the voltage induced and resistance to calculate any similar situation.
Sorry Gibbs. That concept isn't in line with my thinking.
Without changing the circuit we can make the scope register various voltages and either polarity simply by changing the position of the scope relative to the loop.
Please review the following 3 diagrams.
Are we in agreement with the scope reading in each?
No.
I disagree with the second and third pictures.
All three would register .4V
Quote from: WaveWatcher on 2012.02.25, 01:25:13
No.
I disagree with the second and third pictures.
All three would register .4V
Harvey, are you in agreement with all 3 diagrams?
Quote from: WaveWatcher on 2012.02.24, 22:48:15
Yea, but what would the measured voltage be if the scope and leads were directly below the loop? :D
I'll give you one guess ;)
(without disconnecting the scope probe from its current points)
If the move is completely symmetrical, then the results are the same.
Because of the geometry of the leads, any induced value in them can more than likely be dismissed as being within the margin of error. Three sides of the lead loop are parallel to the changing magnetic field and thus produce no emf. The circuit under consideration makes up the fourth side. Even if there were some angular offset, we must consider that the distance indicated for the horizontal portion of those leads is 9 times that of the circuit loop. That places the flux value at about 0.0123 times (1/x²) that which the circuit loop experiences which is nearly negligible by comparison even if the entire probe loop were perfectly perpendicular.
Apply Lenz's Law to those probe leads to see that the magnetic field is not affecting the signal within the leads.
;)
WW,
Are you in agreement with Harvey on his response regarding the measurement taken from below vs. above?
Yes. It seems I'm asking the easy questions :)
I'll stop, now. It doesn't become interesting until the metering circuit is no longer normal to the plane of the loop under measurement.
Quote from: WaveWatcher on 2012.02.24, 03:37:30
The 'above' and the 'below' measurements will be opposite sine and offset from mid-point (.5V)
You may wish to review this in light of your last post?
Quote from: poynt99 on 2012.02.25, 01:32:04
Harvey, are you in agreement with all 3 diagrams?
Yes, that does seem correct from my point of view - the labels at the resistors is a bit odd, but I suppose you needed some moniker for your leads attachment.
It is important to state that those pictures represent a single snapshot in time where the potential is at its maximum.
In essence we have presented here what Professor Lewin demonstrated at the end of his video with the dual probe readings.
8)
Here is another way to see the problem:
(http://tinyurl.com/7sy93b7)
In the same manner that the displacement currents are added to Kirchhoff's current law so that it still applies, we see that the hidden induced emf generators must also be added to Kirchhoff's voltage law in order for it to work. In the present case, it is easy: the "observer" is the voltmeter and we want only know what it measures. But if we had several embedded loops with different induced emf due to varying B fields through each one, I presume that the problem would become very complicated because observer dependent. The Faraday's law is more general and more physical so imho it is the best one to deal with such problems.
Quote from: Harvey on 2012.02.25, 07:39:43
Yes, that does seem correct from my point of view - the labels at the resistors is a bit odd, but I suppose you needed some moniker for your leads attachment.
1 and 1' for the 100 Ohm, and 9 and 9' for the 900 Ohm. I expected the association to be somewhat obvious, and precludes the need to see the diagram to know which resistor is being measured while referring to these designators.
Quote
It is important to state that those pictures represent a single snapshot in time where the potential is at its maximum.
Of course. Has this ever NOT been the case while discussing this experiment?
Quote
In essence we have presented here what Professor Lewin demonstrated at the end of his video with the dual probe readings.
I disagree. Lewin never makes a measurement with the probe leads normal to the loop. All the diagrams I've posted recently show the probe leads normal to the loop being measured.
In light of the above, you may wish to review those 3 diagrams again and reconsider your agreement with them.
Quote from: WaveWatcher on 2012.02.25, 01:25:13
No.
I disagree with the second and third pictures.
All three would register .4V
Quote from: Harvey on 2012.02.25, 07:39:43
Yes, that does seem correct from my point of view [snip]
You two are in disagreement regarding the 3 diagrams and their associated indicated voltages. Once you two resolve this disagreement, we can move on.
Quote from: poynt99 on 2012.02.25, 03:33:07
You may wish to review this in light of your last post?
I'll expand upon it.
The meter will display the same above or below, when normal to the measured loop. The sum is zero because the measurements below must be subtracted since the viewpoint of the meter is opposite.
This is a sticking point for many as it boils down to translating the observation. I can't say I agree with the practice but that is the method taught to me and it works.
Quote from: poynt99 on 2012.02.25, 13:46:21
You two are in disagreement regarding the 3 diagrams and their associated indicated voltages. Once you two resolve this disagreement, we can move on.
No change here.
The perspective of the observing meter doesn't change (Farday). The wire resistance is a minor factor so there is no Ohmic change (Ohm).
Moving the meter lead connections doesn't change the area crossed by B since the meter leads are normal to the loop in all three instances.
I disagree this experiment represents the one by Lewin except both are prime examples of the important differences between EMF and PD (Potential Difference).
Quote from: WaveWatcher on 2012.02.25, 14:35:39
I'll expand upon it.
The meter will display the same above or below, when normal to the measured loop. The sum is zero because the measurements below must be subtracted since the viewpoint of the meter is opposite.
This is a sticking point for many as it boils down to translating the observation. I can't say I agree with the practice but that is the method taught to me and it works.
Why are you "summing" them? The measurement is "as it is". It has one value when the measurement device is decoupled from the experiment, as it is in these two cases, i.e. "above" and "below".
Quote from: WaveWatcher on 2012.02.25, 14:41:59
No change here.
The perspective of the observing meter doesn't change (Farday). The wire resistance is a minor factor so there is no Ohmic change (Ohm).
Moving the meter lead connections doesn't change the area crossed by B since the meter leads are normal to the loop in all three instances.
In effect what you are saying then if you disagree with the indicated readings, is the following:
1mA x 900 Ohms = 0.4V and
1mA x 100 Ohms = 0.4V
Are you sure you want to stick with your theory?
OK guys,
I think we're making this harder than it really is. I think that professor Lewin actually put the probes near the resistors on each side as diagram 2 and 3. Regarding the angle whether is is normal or not makes no different in the reading although it contains some induction point of view.
Even if we tell him all this, he probably still laugh at us. IMO
It is true that the actual experiment was reduced by a factor of 10, so that instead of 1V we saw 0.1V overall induced voltage.
Also, it is true that Lewin did not need to have his reference and signal leads separated, nor did he have a need to place them six feet away vertically - however the image attached does show the scope probes being nearly normal to loop and positioned at a point in the magnetic field that little influence would be on them.
8)
Quote from: Harvey on 2012.02.25, 15:48:34
It is true that the actual experiment was reduced by a factor of 10, so that instead of 1V we saw 0.1V overall induced voltage.
For the sake of keeping in line with the values Lewin uses on the board, do you mind if we stick to his numbers? The proportions are identical.
Quote
Also, it is true that Lewin did not need to have his reference and signal leads separated, nor did he have a need to place them six feet away vertically - however the image attached does show the scope probes being nearly normal to loop and positioned at a point in the magnetic field that little influence would be on them.
Lewin probably has the probes directly across the resistors; I stated this at the beginning of the discussions on this a year or so ago.
The probes in Lewin's demonstration are highly coupled to the experiment. This is supported by the fact that he illustrates the probes and scope being to the side of the loop, laterally out from the resistor. So there is no question that the probes are largely parallel to the plane of the loop. This is in contrast to the measurements I am discussion right now.
Had you actually performed this experiment, you should be aware that the length of scope wire that has to be parallel to the plane of the loop in order to be coupled to the experiment, only has to be long enough to span the resistor. The probe used by Lewin would have satisfied this case. And also there are 2 or 3 inches of acrylic sheeting around the solenoid supporting the probes in a horizontal position.
Now having said that, are you in agreement with the 3 diagrams and the associated voltages indicated on the scope, are you in agreement with WW, or perhaps you have your own set of values?
The probes are nearly vertical, normal to the time-changing magnetic field.
You may want to review the image attached to my previous post.
It is my contention that do to the manner in which the probes are setup in Lewin's experiment (vertical - normal), no appreciable signal is induced in the probe loops by the time-changing magnetic field. Therefore, the probes are only measuring the voltage drop across the resistors.
Whether or not his probes were AC coupled or DC coupled is not stated (or I missed it) but either way they are physically connected for a certainty. The probes are not magnetically coupled in this experiment and they are purposely configured so as not to be.
As far as you drawings, I think I already stated that they looked correct to me.
As far as any perceived disagreement between my statements or WaveWatcher's statements, each stand as stated and I see no need to argue our individual understandings.
My statements are based on the empirical proof provided by Professor Lewin and the associated laws that support that proof as formerly stated.
I think its time that you state what you think you have learned from your tests and evaluation and how they measure up to your 3 drawings. O0
Quote from: Harvey on 2012.02.26, 06:59:18
The probes are nearly vertical, normal to the time-changing magnetic field.
You may want to review the image attached to my previous post.
You are mistaken. Please review the image I've attached. You meant to say "parallel" vs. "normal" I think.
Quote
It is my contention that do to the manner in which the probes are setup in Lewin's experiment (vertical - normal), no appreciable signal is induced in the probe loops by the time-changing magnetic field.
This is not correct.
Quote
Whether or not his probes were AC coupled or DC coupled is not stated (or I missed it) but either way they are physically connected for a certainty. The probes are not magnetically coupled in this experiment and they are purposely configured so as not to be.
No one is arguing if the probes are connected or not, of course they are.
You can't know for certain that Lewin was purposely trying to decouple the probes from the experiment. I can tell you from experience however, that he was not.
Quote
As far as you drawings, I think I already stated that they looked correct to me.
As far as any perceived disagreement between my statements or WaveWatcher's statements, each stand as stated and I see no need to argue our individual understandings.
Obviously, at least one of you is incorrect on the matter.
Quote
My statements are based on the empirical proof provided by Professor Lewin and the associated laws that support that proof as formerly stated.
Lewin's setup and results are what they are, but they are being misinterpreted in more ways than one.
Quote
I think its time that you state what you think you have learned from your tests and evaluation and how they measure up to your 3 drawings. O0
We are getting there, but we're not yet on the same page.
Quote from: Harvey on 2012.02.26, 06:59:18
The probes are nearly vertical, normal to the time-changing magnetic field.
Just for clarity, my recent discussion here is based upon the recent drawings by .99, not Lewin's demonstration.
Quote
I think its time that you state what you think you have learned from your tests and evaluation and how they measure up to your 3 drawings. O0
If debate is thought to be required then that debate should happen between the bench and the experimenter. I may be wrong since Ohm's law is said to always prevail and I've never performed the experiment exactly as pictured by .99. Neither has Lewin, AFAIK.
@Poynt99
QuoteLewin's setup and results are what they are, but they are being misinterpreted in more ways than one.
I would agree, maybe a simple analogy is in order.
In the picture posted there are two pumps, one a point source in which the pump and the restrictions determine the flow and pressure measured at any given point in the system. We should understand that in pump 1 any measure is solely dependant on the pressure and flow through pump 1 itself.
Next we have pump 2, now we can imagine that if we rapidly rotate the whole of the ring filled with water CCW then every single drop of the water in the closed pipe will experience an equal force which will produce a pressure differential across any restriction. If we could imagine that each restriction dissipates energy in proportion to the flow through it not unlike a resistance then some things should become apparent.
First there is no point source and that every single drop of water in the pipe is in itself a source. That R1 is the greatest resistance and due to it's resistance will become a partial source in and of itself by restricting flow which will act on the rest of the system. This is not a normal series loop like pump 1 because the resistance of R1 will create a pressure gradient, a gradient of force, acting in both directions towards R2 thus the standard series calculations have no application. In pump 2, R1 high acts towards R2 low and R1 low acts towards R2 high. The primary focus of pressure is at R1 acting outwards and the primary focus of flow at R2 acting inwards.
It is obvious the two systems are not the same and do not act the same because in pump 2 we have "relative motion" to consider which tends to lead to all kinds of confusion. The concepts are easy however one must have a firm grasp on the mechanics of dynamic systems with relative properties.
Regards
AC
Quote from: WaveWatcher on 2012.02.26, 13:35:13
Just for clarity, my recent discussion here is based upon the recent drawings by .99, not Lewin's demonstration.
If debate is thought to be required then that debate should happen between the bench and the experimenter. I may be wrong since Ohm's law is said to always prevail and I've never performed the experiment exactly as pictured by .99. Neither has Lewin, AFAIK.
With reference to the 3 diagrams and their associated indicated voltage readings on the oscilloscope,
they are correct.
There is an induced emf of 1V, so
we know (and will measure) that the voltages across the two resistors has to be +.9 and -.1.
Quote from: poynt99 on 2012.02.26, 14:34:21
With reference to the 3 diagrams and their associated indicated voltage readings on the oscilloscope, they are correct.
There is an induced emf of 1V, so we know (and will measure) that the voltages across the two resistors has to be +.9 and -.1.
Have you proven that by experiment?
The meter only measures what is at the meter terminals not at the probe.
If pictures two and three are meant to convey the scope is directly above the resistor then you will measure the voltage drop across the distant resistor, not the resistor the probes are directly connected to.
I took the pictures as the scope not changing position from middle with the probes being moved to each resistor. In that case. the scope will continue to measure .4V.
>>Edit...
To be precise....
The position of the scope relative to the circuit is what matters. Considering the probes cables are 6' long and the scope was moved above each resistor then the angular difference from 'normal to the loop' (and directly above the mid-point) would change very little. I'll correct my statement to say there would be some minor deviation from the .4V reading when the scope was directly over the mid-point of the loop.
Quote from: WaveWatcher on 2012.02.26, 16:07:40
Have you proven that by experiment?
The meter only measures what is at the meter terminals not at the probe.
For the third time,
YES I HAVE DONE THE EXPERIMENT,
AND VERIFIED EVERY STATEMENT I HAVE MADE THUS FAR.
Agreed regarding what the meter indicates comment.
Quote from: WaveWatcher on 2012.02.26, 16:15:36
If pictures two and three are meant to convey the scope is directly above the resistor then you will measure the voltage drop across the distant resistor, not the resistor the probes are directly connected to.
I took the pictures as the scope not changing position from middle with the probes being moved to each resistor. In that case. the scope will continue to measure .4V.
>>Edit...
To be precise....
The position of the scope relative to the circuit is what matters. Considering the probes cables are 6' long and the scope was moved above each resistor then the angular difference from 'normal to the loop' (and directly above the mid-point) would change very little. I'll correct my statement to say there would be some minor deviation from the .4V reading when the scope was directly over the mid-point of the loop.
The voltages measured are as per the diagram (within 10% error or less), REGARDLESS if there is a slight angle on the scope leads or not. In fact, you will measure +0.9 and -0.1V at any and every angle you can imagine, and this makes perfect sense when you understand the dynamics of what is going on. :D
Sorry WW, 0.4V is simply not correct for the resistor measurements, and you will not measure that. You're just bound and determined to re-write Ohm's law aren't you? :P
Good. Ohm's law stands correct.
I'm anxious to hear the dynamics involved. I'm also anxious to hear how KVL still stands since the total voltage drop is .8V.
Yes, I already know the EMF is 1.0V and the PD is 1.0V.
Did Lewin ever say which resistor was at each side? I'm wondering since in his explanations I understood each scope shows the voltage drop across the distant resistor, not the resistor on the same side of the loop as the scope.
Quote from: WaveWatcher on 2012.02.27, 01:21:32
I'm also anxious to hear how KVL still stands since the total voltage drop is .8V.
What do you mean exactly by "the total voltage drop is 0.8V"?
Quote
Did Lewin ever say which resistor was at each side?
I think the resistors are positioned as he has them on the board. But since the apparatus is turned around facing him, not the audience, the resistors are reversed for the audience. I surmised this because while Lewin is motioning towards the apparatus, he refers to the resistor on our left as "nine hundred Volts", and the resistor on our right as "one hundred Volts".
Quote
I'm wondering since in his explanations I understood each scope shows the voltage drop across the distant resistor, not the resistor on the same side of the loop as the scope.
How did you get that impression?
The total voltage drop is -.1 + .9 = .8
-.1V PD across the 100 ohm resistor
.9V PD across the 900 ohm resistor
1.0V EMF applied to the circuit
----------
.2V EMF
AS far as my impression of which side each resistor was on... I'll watch it again to see where I had that impression. It is possible I assumed positions because his experiment is a repeat of one presented to me as a student some decades before. KVL didn't hold true in that one, either.
Quote from: WaveWatcher on 2012.02.27, 10:52:07
The total voltage drop is -.1 + .9 = .8
-.1V PD across the 100 ohm resistor
.9V PD across the 900 ohm resistor
1.0V EMF applied to the circuit
----------
.2V EMF
Good grief WW, you're forgetting your basic electronics. ???
When you go to calculate the total voltage drops in a series circuit like this, you travel around the circuit/loop IN ONE DIRECTION (usually the direction of the current), and add all the drops and/or gains until you get back to your starting point.
So, finding the
resistor voltage drops with this circuit starting at point "9" and traveling clockwise on the drawing, we have: 0.9V + 0.1V = 1.0V. This is of course the result we expect, because
we know the induced emf is 1V.
I'm not.
The only mistake I made was repeating the idea put forward by another much earlier in this thread. I now see that post has either been edited or removed.
So, I'm not able to use it to make my point.
For now, we must keep in mind that there aren't that many similarities between the current discussion and the earlier on in this thread.
i.e. test points in the middle when they were never there in the offending Lewin experiment. Recently they were moved to each resistor (clearly seen by me now - don't know why I didn't perceive it this way before).
As for angles not making a difference when probes are connected directly to the resistors.... no argument there. I'm seeing my mistakes.
As a matter of fact, I'm seeing enough things that don't jive I pulled the old coil out. I just need to rig a pulser so I can refresh my memories - this time with scopes instead of galvanometers.
My only argument with Lewin is where he describes an added induction loop formed by the meter circuit.
I notice that the 3 drawings by .99 have been viewed over 150 times so I feel impelled to focus on why KVL cannot be applied to "Professor Lewin's Non-conservative Fields Experiment" - especially in light of the various digressions and permutations that have arisen in this thread.
I think everyone will agree that when we take an instantaneous snapshot in time, that at that moment if the fields are not changing between the beginning and ending of the snapshot, then we will have reduced the period to a state whereby we CAN apply KVL comfortably and with confidence. Much of the discussion has been based on this instantaneous nature, focusing on the "Peak" or maximum value obtained - and this peak is quite readily visible in the scope shots presented.
However, and this is of the utmost importance to this thread, Dr. Lewin's experiment was not an instantaneous experiment. No, his experiment covered a period of time and is representative to explicitly show a CHANGE in magnetic flux - extremely important factor to this very particular experiment. When we take the entire time period stated to be "About 10 Milliseconds" (which is quite long in today's nanosecond economy), we CANNOT apply KVL to the problem. In fact, there is NO time snapshot during the experiment where the beginning and ending periods do not see some small change in the field and resulting induced voltage.
Why do I say that KVL cannot be applied? Because KVL requires two things: 1. A static field and 2. A closed loop of zero volts (0V)
However, by applying Faraday's law, we can derive the various induced voltages at various stages in time and quite closely approximate the "source" needed to balance the KVL equations. And this is precisely what some of the new simulators do. Alternatively, we could arbitrarily assign the source convincing ourselves that it must match the drops so as to net to zero. But even then, we are kluging together a series of stair stepped values to approximate the reality, whereas Faraday's formula gives us a means to smoothly transition from each point to the next and map out the real curves with as high a resolution we choose to adopt.
So like many of the lessons provided by the well experienced Professor, there are deep nuances embedded in his lectures that are designed to be thought provoking and illicit a deep contemplation leading to solid understanding. That is what makes him one the best teachers on this planet.
What are some of the finer points of this experiment?
1. Results are different for Static and Dynamic fields
2. Because of inductive reactance a wire is not always just a wire
3. KVL is a subset or special case of Faraday's Law
4. Measurement equipment can become part of the circuit in very non-intuitive ways
5. The majority of textbooks and schools overlook the importance of this subject.
6. Real world situations will demand that the proper method be used
There may be more points that you have learned.
One of the most profound things I took away from this experiment is a magnetic field that is changing is no longer conservative. When a magnetic field is not conservative, energy can be extracted from it. Ponder that.
After realizing that truth, it was easy to see that the same thing holds true for other "conservative" fields, like gravity. It is because we are able to break the field(s) into non-conservative parts (think tides and energy extraction (http://www.darvill.clara.net/altenerg/tidal.htm) ) that we are able to extract energy from them.
8)
Much of that can be applied to Ohm's law.
Just think about how many of these rules are based upon zero time length. Sure, you can string a bunch of snap-shots together or let the software do it for you ;)
Quote from: WaveWatcher on 2012.02.27, 18:06:53
I'm not.
Sorry WW, your computation of 0.8V is not the correct total for the loop resistor voltage drops. That was my point.
Do you understand why the correct total is 1V?
QuoteAs a matter of fact, I'm seeing enough things that don't jive I pulled the old coil out. I just need to rig a pulser so I can refresh my memories - this time with scopes instead of galvanometers.
By all means check it yourself if you have doubts about my results.
Now let's try and move the discussion forward and look at the experiment from a bird's eye view (overhead perspective) for a moment.
With the oscilloscope leads in the same plane as the loop, and as shown, what voltage will the oscilloscope indicate on CH1 and CH2?
Quote from: Harvey on 2012.02.27, 19:35:12
...
I think everyone will agree that when we take an instantaneous snapshot in time, that at that moment if the fields are not changing between the beginning and ending of the snapshot, then we will have reduced the period to a state whereby we CAN apply KVL comfortably and with confidence. Much of the discussion has been based on this instantaneous nature, focusing on the "Peak" or maximum value obtained - and this peak is quite readily visible in the scope shots presented.
However, and this is of the utmost importance to this thread, Dr. Lewin's experiment was not an instantaneous experiment. No, his experiment covered a period of time and is representative to explicitly show a CHANGE in magnetic flux - extremely important factor to this very particular experiment. When we take the entire time period stated to be "About 10 Milliseconds" (which is quite long in today's nanosecond economy), we CANNOT apply KVL to the problem. In fact, there is NO time snapshot during the experiment where the beginning and ending periods do not see some small change in the field and resulting induced voltage.
...
I will comment on this later. For now, let's give Poynt our undivided attention.
Quote from: poynt99 on 2012.02.27, 23:49:16
Now let's try and move the discussion forward and look at the experiment from a bird's eye view (overhead perspective) for a moment.
With the oscilloscope leads in the same plane as the loop, and as shown, what voltage will the oscilloscope indicate on CH1 and CH2?
I'll give you a hint:
1. The inductive reactance of all three loops is critical to deriving the answer.
2. The scope impedance is absolutely necessary to determining the induced probe currents.
End result, probes don't play any significant part in the measured results - it's still 0.4V and 0.9V +/- 0.1V margin of error.
Quote from: Harvey on 2012.02.28, 01:46:08
I'll give you a hint:
1. The inductive reactance of all three loops is critical to deriving the answer.
2. The scope impedance is absolutely necessary to determining the induced probe currents.
End result, probes don't play any significant part in the measured results - it's still 0.4V and 0.9V +/- 0.1V margin of error.
That's interesting, but not correct actually. Think about it and try again.
WW, what do you think?
Quote from: poynt99 on 2012.02.28, 01:50:37
That's interesting, but not correct actually. Think about it and try again.
WW, what do you think?
No, your comment is not correct - actually. You need to think about it, review Lecture 16 (http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-16-electromagnetic-induction/) from about 24 minutes onward and then consider an apology. O0
Quote from: Harvey on 2012.02.28, 01:56:56
No, your comment is not correct - actually. You need to think about it, review Lecture 16 (http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-16-electromagnetic-induction/) from about 24 minutes onward and then consider an apology. O0
Well Harvey, your answer is not correct. It's that simple.
I only asked about one side/resistor, but since you've provided an answer for both sides/resistors, you have two values correct, and two values incorrect.
EDIT: Indeed I mistook the +/- 0.1 as part of the answer. But 0.4V and 0.9V is still incorrect, even with 0.1 margin of error..
Quote from: poynt99 on 2012.02.28, 01:59:30
Well Harvey, your answer is not correct. It's that simple.
I only asked about one side/resistor, but since you've provided an answer for both sides/resistors, you have two values correct, and two values incorrect.
Harvey is busy, but he said you need to look at his post a little closer.....you are mistaken...
:)
Quote from: catlady on 2012.02.28, 02:35:41
Harvey is busy, but he said you need to look at his post a little closer.....you are mistaken...
:)
For the question I posed reference the diagram, I understood his answer to be the following:
CH1 = 0.9V
CH2 = 0.4V
Agreed?
If so, then his answer is not correct.
Quote from: poynt99 on 2012.02.27, 23:49:16
Now let's try and move the discussion forward and look at the experiment from a bird's eye view (overhead perspective) for a moment.
With the oscilloscope leads in the same plane as the loop, and as shown, what voltage will the oscilloscope indicate on CH1 and CH2?
No straightforward answer, it depends on the connection of probe 2.
When we see B pointing toward us at the center of the circle, we must not forget that outside of the circle, B is pointing away from us because the magnetic flux is looped and the field at the center is obtained from a not infinite solenoid (if it was infinite, the field outside could be neglected which is not the case in a real experiment).
Therefore, if there is no magnetic flux through the surface delimitated by the two wires of the probe 2 and the half circle D-9-9'-A because this surface would be minimized by putting the wires connecting the probe 2 as near as possible along the circle, then the probe 2 will measure the same value as probe 1.
Otherwise, the magnetic flux outside the circle will induce some reversed emf in probe 2 by crossing the surface of the loop E-D-9-9'-A-E, E being the end of the coaxial cable of probe 2. This will reduce the voltage measured by probe 2 when comparing to probe 1. The probe 2 measures the emf generated by the B flux in the loop E-D-1-1'-A-E minus the emf generated in the loop E-D-9-9'-A-E by a part of the reversed flux crossing it (I didn't care of the drop R*I in the resistances, which is the same viewed from CH1 as from CH2).
.8V wasn't my computation. That was my point.
I couldn't bring the point to bear. My mistake was not making sure the posts by another were still there.
My thoughts are that KVL may only be applied concerning PD (potential difference) NOT EMF (electromotive force). There is and always has been a difference between a force and a potential.
The PD between -.1V and .9V is 1V but that only makes sense if you disregard the fact that one meter is reversed since it is on the opposite side of the loop. In either case (the PD between -.1V to .9V) or (the total of .1V plus .9V) is still 1V.
For me, it is just cleaner to figure that KVL only applies to potential and Faraday applies to force even though both are measured in Volts.
The greatest problem here is we keep boiling this down to a static situation, which it is not.
In any case, these principles are used in real-world applications. They aren't unique. The experiment tells more about the experimenter than the physics. (Not pointed only at .99)
Quote from: poynt99 on 2012.02.28, 01:50:37
That's interesting, but not correct actually. Think about it and try again.
WW, what do you think?
I 'think' channel 1 will measure .9V but channel 2 will measure -.1V. (Again... if galvanometers are used - I haven't performed this part with a scope)
I'm doubting my memory, not your results. I can't declare your results false unless my memory is true and confirmed.
Quote from: WaveWatcher on 2012.02.27, 01:21:32
I'm anxious to hear the dynamics involved. I'm also anxious to hear how KVL still stands since the total voltage drop is .8V.
Quote from: WaveWatcher on 2012.02.27, 10:52:07
The total voltage drop is -.1 + .9 = .8
-.1V PD across the 100 ohm resistor
.9V PD across the 900 ohm resistor
1.0V EMF applied to the circuit
----------
.2V EMF
WW, I don't think the above could be any clearer as to what you were saying when you posted it. If that isn't really what you were trying to say, then please don't post it, regardless of what might exist in some other post elsewhere. Agreed?
Could I ask in the future that you post what you mean? It would be appreciated I'm certain. This doesn't have to be difficult, but this sort of flip-flopping makes progressing any argument quite arduous...unless that is your purpose?
Quote from: WaveWatcher on 2012.02.28, 10:12:01
The experiment tells more about the experimenter than the physics. (Not pointed only at .99)
Please do tell.
Quote from: WaveWatcher on 2012.02.28, 10:28:23
I 'think' channel 1 will measure .9V but channel 2 will measure -.1V. (Again... if galvanometers are used - I haven't performed this part with a scope)
I'm doubting my memory, not your results. I can't declare your results false unless my memory is true and confirmed.
Your answer is also incorrect.
One doesn't need to perform this experiment to know the answers to these relatively simple questions, if one truly understands the dynamics involved. What has become clear is that neither you nor Harvey have a good understanding of the dynamics involved in this experiment. It's therefore ironic that you both believe I am wrong about the statements I've made thus far. I encourage both of you to perform this experiment and make the measurements I have outlined here. Prove it to yourself.
exnihiloest has the correct values, but the experiment is not nearly as sensitive to error as he has alluded to.
The correct answer to the most recent question regarding the RIGHT side of the experiment, is:
CH1 = 0.9V
CH2 = 0.9V
Of course when we move the probes to the LEFT side, the answer will be:
CH1 = -0.1V
CH2 = -0.1V
Quote from: poynt99 on 2012.02.28, 13:11:51
Your answer is also incorrect.
One doesn't need to perform this experiment to know the answers to these relatively simple questions, if one truly understands the dynamics involved. What has become clear is that neither you nor Harvey have a good understanding of the dynamics involved in this experiment. It's therefore ironic that you both believe I am wrong about the statements I've made thus far. I encourage both of you to perform this experiment and make the measurements I have outlined here. Prove it to yourself.
exnihiloest has the correct values, but the experiment is not nearly as sensitive to error as he has alluded to.
The correct answer to the most recent question regarding the RIGHT side of the experiment, is:
CH1 = 0.9V
CH2 = 0.9V
Of course when we move the probes to the LEFT side, the answer will be:
CH1 = -0.1V
CH2 = -0.1V
Not while using galvanometers.
Yes, I now wish more than ever to perform the experiment.
Agreed, sarcasm is problematic on text based forums. I'll quote and comment rather than repeat and hope folks get it.
Ex is wrong simply because there is no reversal of magnetic polarity on a setup identical to the Lewin experiment anywhere outside the coil except above and below the coil.
Harvey is correct about impedance being an important consideration.
Apparently, the dynamics aren't as simple as stated.
With that said, Yes, I will start assembling the experiment and report back here as soon as possible.
One additional question......
Are you using differential probes on that single scope? Two scopes with isolated chassis? One probe at a time?
The reason I ask is I believe the Lewin experiment is using differential amplifiers between the probes and the scopes.
Regarding the most recent question I posed, Ex is correct actually.
Both scope channels will indicate the SAME voltage.
btw, I have been asking questions about indicated voltages, NOT theories. For now, I would ask that all please answer the questions accordingly.
Quote from: WaveWatcher on 2012.02.28, 13:50:39
One additional question......
Are you using differential probes on that single scope? Two scopes with isolated chassis? One probe at a time?
The reason I ask is I believe the Lewin experiment is using differential amplifiers between the probes and the scopes.
Not shown, but yes. It should be assumed that there is no interaction between the probes, i.e. complete isolation.
I agree on the diff amp/probe, but I was hesitant to say so in fear of another arduous diversion. Obviously if one uses one probe at a time, there is no such isolation problem.
In my tests, I used one conventional passive 10x probe only at all times. For the decoupled measurements, I used twin-lead wire strung straight up to the ceiling, then over and down to the scope a few feet away. The twin-lead then was connected to the same 10x probe.
Quote from: poynt99 on 2012.02.26, 17:47:59
The voltages measured are as per the diagram (within 10% error or less), REGARDLESS if there is a slight angle on the scope leads or not. In fact, you will measure +0.9 and -0.1V at any and every angle you can imagine, and this makes perfect sense when you understand the dynamics of what is going on. :D
...
Please clarify what do you mean by " any and every angle". The way I understand it is that when the position is normal, let's say it's 90 degrees. When you have CH2 on the right, it's 0 degrees and when you have it on the left it's 180 degrees. Is this correct?
Quote from: GibbsHelmholtz on 2012.02.28, 14:03:29
Please clarify what do you mean by " any and every angle". The way I understand it is that when the position is normal, let's say it's 90 degrees. When you have CH2 on the right, it's 0 degrees and when you have it on the left it's 180 degrees. Is this correct?
I should have been more specific.
Imagine a vertical plane normal to the loop and intersecting the 900 Ohm resistor. By "any angle" this means any angle including and to the right of this plane, with reference to the probes connected to points 9 and 9'.
Quote from: poynt99 on 2012.02.28, 16:04:19
I should have been more specific.
Imagine a vertical plane normal to the loop and intersecting the 900 Ohm resistor. By "any angle" this means any angle including and to the right of this plane, with reference to the probes connected to points 9 and 9'.
So this means any angle to the left of this plane would give a different value?
Yes.
Quote from: poynt99 on 2012.02.28, 16:18:29
Yes.
I see. So you're saying if the plane is to the right from 0 - 90 degrees, the value would be .9V . When it pass to the left 90-180 degrees, it suddenly change its value. And what do you suggest this value is?
Quote from: GibbsHelmholtz on 2012.02.28, 16:39:40
I see. So you're saying if the plane is to the right from 0 - 90 degrees, the value would be .9V . When it pass to the left 90-180 degrees, it suddenly change its value. And what do you suggest this value is?
You tell me. I've already given you enough information to formulate an answer.
Quote from: poynt99 on 2012.02.28, 17:08:35
You tell me. I've already given you enough information to formulate an answer.
I don't know. That's why I ask. I think it's the same value because if not, we have to rewrite Ohm's law. But why speculate when you already done the experiment.
You already have all the necessary information to formulate an answer.
However, it won't hurt to make another diagram, which will help you and provide more insight.
I'll post that shortly.
What will be the indicated voltage on CH2?
Quote from: poynt99 on 2012.02.28, 13:52:26
Regarding the most recent question I posed, Ex is correct actually.
Both scope channels will indicate the SAME voltage.
btw, I have been asking questions about indicated voltages, NOT theories. For now, I would ask that all please answer the questions accordingly.
I see now why my answers are no better than theories and why Ex is correct. Ex's comment pertains to your setup and my 'theories' do not.
All along, I've been trying to imagine how you were rotating your scope around the DUT with a 6' probe wire. That would have been silly anyway. It would be easier to rotate the DUT and leave the scope on the bench.
I do appreciate the mental gymnastics but I should have stuck to my guns when I said I couldn't say what you would measure.
So, no more theories from me :)
The time for the theory behind the dynamics involved is approaching, but it's necessary to look at the empirical results and feel out everyone's perspectives first, which is what I have been trying to do up to this point.
Quote from: poynt99 on 2012.02.28, 17:39:09
What will be the indicated voltage on CH2?
I suppose .1V .
Let's wait for Harvey and WW to post their answers before I respond.
Quote from: WaveWatcher on 2012.02.28, 17:43:22
I see now why my answers are no better than theories and why Ex is correct. Ex's comment pertains to your setup and my 'theories' do not.
I have performed a number of additional measurements compared to Lewin, but aside from that "my setup" is not all that different. And it seems you disagree?
Quote
All along, I've been trying to imagine how you were rotating your scope around the DUT with a 6' probe wire. That would have been silly anyway. It would be easier to rotate the DUT and leave the scope on the bench.
I do appreciate the mental gymnastics but I should have stuck to my guns when I said I couldn't say what you would measure.
The tendency does seem to be that folks want to over-complicate things, and this is and has been the case in this discussion as well.
Quote from: poynt99 on 2012.02.28, 18:00:16
Let's wait for Harvey and WW to post their answers before I respond.
I have no desire to contribute any further to this nonsense - .99 has boldly claimed that my comments are incorrect, so obviously my experience and expertise in this area are meaningless to this forum - at least from his perspective. It is very clear that he has not paid careful attention to my comments and has digressed the discussion to such an extent that the topic has very little to do with the thread title.
What is even worse, he is misleading the readers based on his own measurement errors.
My comments are supported not only by University (MIT) empirical evidence, but also by Farday's Law (not theory or conjecture) and is also supported by documents within high tech engineering circles.
The scope impedance is so high in these tests that the induced current in those loops produces no appreciable EMF regardless of their orientation, size of the loop or distance from the solenoid. However it is needed to provide an accurate evaluation. But I am confident that it is well below my +/- 0.1V error margin offered in my answer.
So my answer stands - final answer. C.C
Quote from: Harvey on 2012.02.29, 01:50:40
I have no desire to contribute any further to this nonsense - .99 has boldly claimed that my comments are incorrect, so obviously my experience and expertise in this area are meaningless to this forum - at least from his perspective. It is very clear that he has not paid careful attention to my comments and has digressed the discussion to such an extent that the topic has very little to do with the thread title.
What is even worse, he is misleading the readers based on his own measurement errors.
My comments are supported not only by University (MIT) empirical evidence, but also by Farday's Law (not theory or conjecture) and is also supported by documents within high tech engineering circles.
The scope impedance is so high in these tests that the induced current in those loops produces no appreciable EMF regardless of their orientation, size of the loop or distance from the solenoid. However it is needed to provide an accurate evaluation. But I am confident that it is well below my +/- 0.1V error margin offered in my answer.
So my answer stands - final answer. C.C
I know how you feel, but I don't think Poynt is misleading the readers. If he saids .9/.9V on both channel, then that is the measurement. It is also surprising to me but I'm glad to know new things. Let's give him a chance to express his view.
Quote from: Harvey on 2012.02.29, 01:50:40
I have no desire to contribute any further to this nonsense [snip]
That's disappointing Harvey, but I'm not surprised.
Quote
What is even worse, he is misleading the readers based on his own measurement errors.
You have provided no proof of either allegation. I've done the measurements (and more), so I have empirical proof. Where is your own empirical evidence?
Quote
My comments are supported not only by University (MIT) empirical evidence, but also by Farday's Law (not theory or conjecture) and is also supported by documents within high tech engineering circles.
Evidently, your comments are supported only by dogma. Try giving some actual measurements a go,
then you'll be in a position to formulate an analysis based on experience, just as I have.
So we will continue moving forward Gibbs, with or without Harvey, and WW too it seems.
Quote from: Harvey on 2012.02.29, 01:50:40
So my answer stands - final answer. C.C
Gibbs, would you agree that Harvey is saying the answer to the latest question and associated diagram, is -0.1V?
If so, then that is two, yourself and Harvey that say the indicated voltage would be -0.1V.
Quote from: poynt99 on 2012.02.29, 04:15:52
So we will continue moving forward Gibbs, with or without Harvey, and WW too it seems.
I hope they will continue following. I learn a lot from their point of view. Ok, the last diagram I said .1V . I suppose this is wrong. I'm all ears. Let's do this. lol
I said .1V because the positive probe is at the bottom, but doesn't matter unless it's correct. lol
Quote from: GibbsHelmholtz on 2012.02.29, 04:34:41
I hope they will continue following. I learn a lot from their point of view. Ok, the last diagram I said .1V . I suppose this is wrong. I'm all ears. Let's do this. lol
I said .1V because the positive probe is at the bottom, but doesn't matter unless it's correct. lol
The positive probe is on top, so the answer, if correct, would be -0.1V. ;)
Quote from: poynt99 on 2012.02.29, 04:37:59
The positive probe is on top, so the answer, if correct, would be -0.1V. ;)
Fair enough, I'll buy that.
Here's a hint:
Changing the diagram slightly, what would the indicated voltage be if the scope probes were placed across points 9 and D?
Quote from: poynt99 on 2012.02.29, 04:47:34
Here's a hint:
Changing the diagram slightly, what would the indicated voltage be if the scope probes were placed across points 9 and D?
Hm..... :( ..... 0 V ?
Correct, 0V.
And what would be indicated if the probes were placed on points 9 and 1?
Quote from: poynt99 on 2012.02.29, 05:05:22
Correct, 0V.
And what would be indicated if the probes were placed on points 9 and 1?
Depends,
If the scope loop eat all the resistors (contain within its loop), 1V.
If the scope loop leave the resistors outside its loop, 0V.
We've moved one probe from point D to point 1. That's all.
So 0V is again the correct answer.
Now, what is the voltage indicated when the probes are on points 1 and 1' when taken from the right side as per the diagram?
Quote from: poynt99 on 2012.02.29, 05:26:34
We've moved one probe from point D to point 1. That's all.
So 0V is again the correct answer.
Now, what is the voltage indicated when the probes are on points 1 and 1' when taken from the right side as per the diagram?
I see... .9V
Quote from: WaveWatcher on 2012.02.28, 13:45:37
..
Ex is wrong simply because there is no reversal of magnetic polarity on a setup identical to the Lewin experiment anywhere outside the coil except above and below the coil.
...
I gave the general case and Lewin's experiment is a particular case that of course is included in the first one. There is
always somewhere reversal of magnetic polarity because a magnetic flux is looped. Naturally outside of the coil the B field is reduced but the flux is the same, just a question of crossed surfaces much larger outside than inside. As the emf depends on the flux crossing by unit of surface, the emf around the same loop surface will be much lower outside that inside of the coil. So outside of the coil, a more sensitive measurement must be done to show emf.
The Lewin's experiment doesn't show that the flux is non conservative because every observed thing is explainable quantitatively and qualitatively according to the current laws of electromagnetism.
Quote from: Harvey on 2012.02.29, 01:50:40
...
The scope impedance is so high in these tests that the induced current in those loops produces no appreciable EMF regardless of their orientation, size of the loop or distance from the solenoid. However it is needed to provide an accurate evaluation. But I am confident that it is well below my +/- 0.1V error margin offered in my answer.
...
Emf is a voltage around a looped pass, it doesn't imply any current. So the question of the probe impedance is irrelevant in this discussion (except if it was too low and made the voltage drop).
Quote from: exnihiloest on 2012.02.29, 07:43:21
The Lewin's experiment doesn't show that the flux is non conservative because every observed thing is explainable quantitatively and qualitatively according to the current laws of electromagnetism.
Very good point. Had he used only one meter with a fixed middle connection and simply moved the meter he would prove the non-conservative aspect. However, non-conservative fields ARE part of the current laws of electromagnetism - at least... this is what I was taught. Has that changed?
If so, who has the task of rewriting Maxwell's interpretation of Faraday's laws?
I am following the thread. Theories aren't a good thing to offer when the results are empiric. So, I'll comment on experimental results with experimental results as soon as possible :)
Until then, assigning potential (a static aspect) to fields in motion (induced EMF) only leads to a contradiction. So, 'potential' means nothing with induced electric fields. Therefore, KVL can not be applied as it only deals with potential, not changing fields.
The difference between the non-motional EMF and motional EMF is the first produces current from an imbalance of charges and the latter produces current according to Lorentz. (I don't like the term 'non-motional EMF'. It makes no sense but is common).Above: Incorrect in almost any sense. A quick check in the books indicates the term 'emf' was started by Volta and varied in real meaning since then. In the most agreed upon correct and current usage it has nothing to do with current and is not the voltage measured in a circuit. Circuit voltage is always less than emf (the potential source creating the circuit voltage).
Just thought I would correct comments that I suspected may be incorrect.
Quote from: WaveWatcher on 2012.02.29, 10:37:39
Until then, assigning potential (a static aspect) to fields in motion (induced EMF) only leads to a contradiction. So, 'potential' means nothing with induced electric fields. Therefore, KVL can not be applied as it only deals with potential, not changing fields.
The fact that we are dealing with changing fields has little to no relevance with certain aspects of the analysis, and discussing it is an unnecessary complication that only muddies the waters. It is important to address the dynamic aspect of the experiment, but not for the reasons you and Harvey have espoused. And we're getting to the theory part of this discussion very soon.
QuoteIf so, who has the task of rewriting Maxwell's interpretation of Faraday's laws?
There is no need. Farady and Maxwell are fine.
Quote from: GibbsHelmholtz on 2012.02.29, 05:45:27
I see... .9V
Correct. Now that's non-intuitive!
So going back to your original question posed here (http://www.overunityresearch.com/index.php?topic=739.msg21364#msg21364), what is your answer now?
Quote from: poynt99 on 2012.02.29, 13:01:18
Correct. Now that's non-intuitive!
So going back to your original question posed here (http://www.overunityresearch.com/index.php?topic=739.msg21364#msg21364), what is your answer now?
The original question was what is the value as the loop pass into the changing magnetic field. It would depends on how much flux the loop can contain. It can have value range from .9V to -.1V .
Quote from: GibbsHelmholtz on 2012.02.29, 13:30:28
The original question was what is the value as the loop pass into the changing magnetic field. It would depends on how much flux the loop can contain. It can have value range from .9V to -.1V .
O0
So what are your thoughts regarding the dynamics involved with that particular situation?
Why does Ohm's law not hold in this particular case when the angle is beyond 90 degrees as we discussed?
Why DOES Ohm's law appear to hold when the angle is within that first 90 degrees?
Quote from: poynt99 on 2012.02.29, 14:38:24
O0
So what are your thoughts regarding the dynamics involved with that particular situation?
Why does Ohm's law not hold in this particular case when the angle is beyond 90 degrees as we discussed?
Why DOES Ohm's law appear to hold when the angle is within that first 90 degrees?
I think Ohm's law holds.
One can either use Faraday or KVL and the answer is the same. When the angle is within 90 degrees, one solve it in two ways and either one would give correct answer:
1/ The loop that contain no magnetic field (probe through 900 Ohms back to probe)
2/ The loop that contain all magnetic field (probe through 100 Ohms back to probe)
For no magnetic field, voltage of the scope has to equal the 900 Ohms. No magnetic flux in that loop so wherever you move the loop, as long as it does not contain flux, the answer is always .9V .
For the magnetic flux case, the voltage has to be 100 Ohms and compensated by the induced voltage by the flux, depends on how much flux. If we contain all the flux, then it's 1V minus 100Ohms potential which is still .9V.
The reason it departs from .9V is the flux in the loop that we set up is partial. This happens when we move the loop pass 90 degrees.
Either case, if you do KVL/Faraday on any closed loop, you will get the right answer.
You're getting the idea Gibbs, but you're not quite correct yet.
For the following discussion, unless otherwise stated, assume the probes are in the same plane as the loop. Now, when we place the probes in such a way that they encircle part or all of the loop, what are we actually measuring?
Quote from: poynt99 on 2012.02.29, 17:58:32
You're getting the idea Gibbs, but you're not quite correct yet.
For the following discussion, unless otherwise stated, assume the probes are in the same plane as the loop. Now, when we place the probes in such a way that they encircle part or all of the loop, what are we actually measuring?
I'm not sure what is your thinking but I think the probe plane always has a component in the same plane as the loop. It's the cosine part of that plane. When 90 degrees, it means cos 90= 0 .
When it encircle part or all of the loop, we measure the total EMF minus resistance produced by the field flux, depends on how you define your loop. Other than that, I'm not sure what you mean.
Reference my question of what exactly we are measuring;
The answer is, we are measuring the electric field E (or part thereof) produced by the changing magnetic field B.
That's why Ohm's law does not hold in some cases.
I see. I think Ohm's law is just that V=IR .
The boundary condition for this system is encircle just the resistor and not anything outside of that.
So yes, we're measuring the E field due to changing B, but it's outside of the boundary of how we defined Ohm's law.
Quote from: GibbsHelmholtz on 2012.02.29, 22:49:33
I see. I think Ohm's law is just that V=IR .
Yes that's Ohm's law. But Ohm's law
appears to be broken when we measure the voltage as shown in the diagram, agreed?
Quote
The boundary condition for this system is encircle just the resistor and not anything outside of that.
The boundary condition of this experiment is the entire loop when considering that plane.
Quote from: poynt99 on 2012.02.29, 23:56:22
The boundary condition of this experiment is the entire loop when considering that plane.
Yes, the boundary condition for this experiment is the entire loop, but the boundary condition for Ohm's law is just the resistor. That's why if you look into the entire loop you have a value of the scope, but when you reduce the boundary to just the resistor, you have the potential across the resistor. It is indeed different from scope and obey Ohm's law.
OK, I'll buy that.
So do you now have a clear understanding of the dynamics involved when the measurements are taken in-plane vs. decoupled? Can you with confidence predict the voltage indicated on the scope across any two points and in both scenarios? Do you understand why the indicated scope voltage is the same for points 9-9' and D-A when measured in-plane?
Quote from: poynt99 on 2012.03.01, 01:30:25
So do you now have a clear understanding of the dynamics involved when the measurements are taken in-plane vs. decoupled? Can you with confidence predict the voltage indicated on the scope across any two points and in both scenarios? Do you understand why the indicated scope voltage is the same for points 9-9' and D-A when measured in-plane?
Not sure what decoupled means, but yes, if you put the probes on any two points and show your probe wiring geometry, I can give you the exact value. The indicated value is the same for 9-9' and D-A are all in accordance with KVL/Faraday. So if one understand what KVL/Faraday is, then he should understand why.
Quote from: GibbsHelmholtz on 2012.03.01, 01:43:03
Not sure what decoupled means,
This is in reference to how the first 3 diagrams are depicted. The measurement device's leads are normal to the loop, therefore
decoupled from the experiment, vs. the in-plane configuration, where the leads ARE coupled to the experiment.
Quote
but yes, if you put the probes on any two points and show your probe wiring geometry, I can give you the exact value. The indicated value is the same for 9-9' and D-A are all in accordance with KVL/Faraday. So if one understand what KVL/Faraday is, then he should understand why.
Indeed? Alright. 8)
What will be the indicated voltage on the scope for this decoupled configuration?
Quote from: GibbsHelmholtz on 2012.03.01, 01:43:03
The indicated value is the same for 9-9' and D-A are all in accordance with KVL/Faraday. So if one understand what KVL/Faraday is, then he should understand why.
Please explain it. :)
If you knew KVL/Faraday and that is all you require to know the dynamics of this experiment, why did you get most of the answers wrong? ???
Quote from: poynt99 on 2012.03.01, 01:55:53
What will be the indicated voltage on the scope for this decoupled configuration?
I see. Decouple means not letting the magnetic field effect it. Thanks.
;D Testing me out? It's fair... lol
.5V (forget polarities, I hate it lol )
I need to formulate my thoughts again for KVL/Faraday. Maybe sometimes later.
BTW Poynt.... Thanks. :)
I got it wrong because I didn't trust your experiment. But it's absurd to think you would lie. lol
0.5V is correct with the way the probes are. O0 Good job.
Now, of course there will also be 0.5V across points 9' and 1' for a total of 1V, agreed? Is this 1V in-phase or out-of-phase with the combined resistor voltages of 1V?
Quote from: poynt99 on 2012.03.01, 02:12:41
Now, of course there will also be 0.5V across points 9' and 1' for a total of 1V, agreed? Is this 1V in-phase or out-of-phase with the combined resistor voltages of 1V?
I just see it as a 1V battery connected to a 1k Ohm resistor. It's in phase in my opinion.
Quote from: GibbsHelmholtz on 2012.03.01, 02:27:25
I just see it as a 1V battery connected to a 1k Ohm resistor. It's in phase in my opinion.
Are you certain? What happened to Faraday and Lenz?
Quote from: poynt99 on 2012.03.01, 02:29:33
Are you certain? What happened to Faraday and Lenz?
You just got ahead of me. lol There is also parasitic inductance in the wire that will not effect by the B field. Those will have some effect of Lenz. Looks like I'll talk about it now anyway.
Faraday original equation is contain within an absolute value. He doesn't care much about polarities. When we take a close loop integral path dependence, we get two answers, one is positive and one is negative.
dB/dt = 1V
dB/dt = -1V
If you apply KVL with this form, it won't work. One will add up to zero and one is doubled. What makes it works is we have to assign a sign value to dB/dt . This is where Lenz come in. The purpose of Lenz is to assign a polarity to dB/dt to make KVL works.
Now Lenz is not like everyone think it is. Lenz superimposed a current so we can know the direction of EMF. The assumption of current does not reflect the reality situation. It is just an assumption to obtain EMF polarities. Current can flow against or with the EMF.
FOCUS Gibbs: 8)
Will the voltage indicated on the oscilloscope for the wires be in or out of phase with the voltage indicated for the resistors?
Quote from: poynt99 on 2012.03.01, 04:03:37
FOCUS Gibbs: 8)
Will the voltage indicated on the oscilloscope for the wires be in or out of phase with the voltage indicated for the resistors?
;D At least I don't have to write the last post tomorrow. lol
Voltage against voltage? Hm... I never thought of that. Well, if a battery is connected to a resistor, the positive of the battery and resistor join at one junction. Is this what you mean by out of phase? lol Then yes.
No, not what I mean.
Phase or relative polarity.
Using decoupled measurements and going around the loop and summing the 4 voltage drops (or PD's as WW calls them), will the total sum be 0V or 2V? Hint; it can't be both.
Quote from: poynt99 on 2012.03.01, 04:52:29
No, not what I mean.
Phase or relative polarity.
Using decoupled measurements and going around the loop and summing the 4 voltage drops (or PD's as WW calls them), will the total sum be 0V or 2V? Hint; it can't be both.
0 V , but both probes has to move in 1 direction when measuring PD on 4 voltage drops.
Quote from: poynt99 on 2012.03.01, 02:29:33
Are you certain? What happened to Faraday and Lenz?
It's sure. The voltages are in phase, provided that you see each of them by turning clockwise or counterclockwise. There is nothing that could change the relative phase of the voltages because they are synchronous with the same dB/dt, which is the sum of the main field and of the opposing field due to Lenz's law.
What is measured is always the voltage induced by the magnetic flux crossing the surface delimited by the 2 points connecting the probe, and the rest of the circuit. We don't measure an independent circuit, we measure the emf of a circuit constituted by both the circuit to be tested and the circuit of the measurement apparatus.
When the probe is connected to two diametrically opposed points, there are 2 half disk surfaces each side, crossed by half the magnetic flux, and contributing equally to half the emf, i.e. 0.5v, provided that the total emf is 1v. It doesn't even depend on which side of the resistance we connect the probe.
Gibbs, you are now correct. 0V.
And of course the probes are kept the same polarity orientation as we travel around the loop, that IS how it is always done in ALL cases, agreed?
@Ex: You will NOT measure an in-phase PD across the wires, relative to that across the resistors. You need to pay closer attention to the question and diagram being discussed. Your diagram is in reference to the in-plane measurements, while the last question and diagram reference the decoupled mode of measurement, which btw, is the only method of the two which provides the TRUE PD measurement of all the components in a dynamic circuit such as this.
The E field of the EMF/emf in all circuits, whether placed there physically (such as with a battery) or induced through induction, are always, and without exception, IN OPPOSITION to the E fields of the loads, such as the resistors in our case.
Therefore, the TRUE PD sum of the loop is 0V at any instant of time, or total length of time.
While measuring the in-plane E field however, it is non-conservative, and is always 1V.
The notion that the true PD sum can be 2V (vs. 0V) is ludicrous. You would have a FE generator at your fingertips, if it was. :P
Quote from: exnihiloest on 2012.03.01, 09:05:51
,,,
When the probe is connected to two diametrically opposed points, there are 2 half disk surfaces each side, crossed by half the magnetic flux, and contributing equally to half the emf, i.e. 0.5v, provided that the total emf is 1v. It doesn't even depend on which side of the resistance we connect the probe.
Even if it is decoupled, the reading would be .5V, .4V
Quote from: poynt99 on 2012.03.01, 12:58:10
The notion that the true PD sum can be 2V (vs. 0V) is ludicrous. You would have a FE generator at your fingertips, if it was. :P
I think we can have FE generator even with 0V sum , but still trying to formulate an argument for it.
Wait.... you guys know how to calculate reading for any two points decoupled or in-plane, right? Decoupled or in-plane are just the same and used the same method.
Why don't you show us your method Gibbs?
I posted a picture below. Both of these are in-plane, but the one on the right can also be recognized as decoupled. The first step is to convert it into in-plane. If we look at it from the top (bird's view), they are all in-plane. With this view we know exactly how much magnetic flux passing through the loop we choose.
For the picture on the right, we go around the loop contain the scope to determine the reading. This loop contain only half of the flux so we automatically know the total voltage is .5V. Then we compensate that with the resistor also in the loop. For right loop, compensate with .9V, for left loop, compensate with .1V. Either will give .4V . This is just taking a close loop integral.
However you set up your probes geometry, by looking it from bird's view, you can attain the answer by doing a loop integral.
@exnihiloest
QuoteIt's sure. The voltages are in phase, provided that you see each of them by turning clockwise or counterclockwise. There is nothing that could change the relative phase of the voltages because they are synchronous with the same dB/dt, which is the sum of the main field and of the opposing field due to Lenz's law.
What is measured is always the voltage induced by the magnetic flux crossing the surface delimited by the 2 points connecting the probe, and the rest of the circuit. We don't measure an independent circuit, we measure the emf of a circuit constituted by both the circuit to be tested and the circuit of the measurement apparatus.
When the probe is connected to two diametrically opposed points, there are 2 half disk surfaces each side, crossed by half the magnetic flux, and contributing equally to half the emf, i.e. 0.5v, provided that the total emf is 1v. It doesn't even depend on which side of the resistance we connect the probe.
Well put, fundamentally we are dealing with external forces and charge density at any given point which dictates the phenomena, equations do not dictate reality it is vice versa.
Regards
AC
@Poynt99
QuoteThe notion that the true PD sum can be 2V (vs. 0V) is ludicrous. You would have a FE generator at your fingertips, if it was
I guess that depends on ones perspective, does your getting a measurement you do not expect mean the conservation of energy has been violated or that you simply do not understand what is happening in reality?. Personally I have measured many things I did not expect nor have many others and a violation of the conservation of energy never crossed my mind --- not once. As such I think using the COE as a mental crutch to justify our personal opinions is and has always been a losing proposition. Every phenomena must stand on it's own two feet without the COE otherwise the concept is completely biased which is anything but scientific, it must prove the COE not vice versa.
Regards
AC
If anyone can give the reading on this one, then I think you understand the idea. Total induced voltage still 1V.
Quote from: GibbsHelmholtz on 2012.03.01, 17:29:18
If anyone can give the reading on this one, then I think you understand the idea. Total induced voltage still 1V.
From a bird's eye view, and with the leads in an in-plane configuration, the voltage will be 0.9V. Keep in mind that your right lead must travel outside the loop, because the solenoid is roughly the same diameter. You are showing that lead cutting across, which is not possible with an in-plane configuration.
Quote from: poynt99 on 2012.03.01, 19:30:05
From a bird's eye view, and with the leads in an in-plane configuration, the voltage will be 0.9V. Keep in mind that your right lead must travel outside the loop, because the solenoid is roughly the same diameter. You are showing that lead cutting across, which is not possible with an in-plane configuration.
But my lead is not outside the loop. I cut the loop like that on purpose. lol You're right, if it's outside the loop, it will be .9V . I'm sure if you configured like the picture in a real experiment, you'll get a value, what is that value? Let's round it to the thousandth placed, so you can't cheat with scope. lol You can use the scope to guide.
BTW, the two measured point are top and bottom, the right intersection does not contact each other.
You can cut through the loop by wire it up and over the top of the solenoid.
I'll give a hint. This one reads .4V .
@GibbsHelmholtz
QuoteBut my lead is not outside the loop. I cut the loop like that on purpose. lol You're right, if it's outside the loop, it will be .9V . I'm sure if you configured like the picture in a real experiment, you'll get a value, what is that value? Let's round it to the thousandth placed, so you can't cheat with scope. lol You can use the scope to guide.
I find this absolutely remarkable, page after page of in plane/out plane probe lead nonsense and nobody has the common sense to simply stick their probes inside a couple pieces of iron pipe to shield from the influence of an external magnetic field and induced electric fields because the probe leads are effectively inside a faraday cage. Not to mention a little fact called the inverse square law as well as a failure to understand exactly what occurs in a simple loop of conductor with a couple resistors. 11 pages and counting and nobody can seem to explain this phenomena in anything resembling simple terms, Really?, Im really not feeling that expert vibe in fact it seems more like amature night.
Regards
AC
.675V - roughly.
Second thought... .425V
AC,
I've been watching and hoping you would offer your explanation :)
Quote from: WaveWatcher on 2012.03.01, 21:08:18
.675V - roughly.
Yep, but that's too rough. lol A smoother one is .8... V lol
AC, I suspect that is a complement but I have no idea and will try to understand the content. Thanks. lol
In your examples Gibbs, are your leads in-plane or decoupled, or a mixture of both? I would appreciate the designation of each lead.
Quote from: allcanadian on 2012.03.01, 21:07:48
11 pages and counting and nobody can seem to explain this phenomena in anything resembling simple terms, Really?, Im really not feeling that expert vibe in fact it seems more like amature night.
Regards
AC
What "phenomena" specifically are you referring to?
Quote from: poynt99 on 2012.03.01, 22:04:43
In your examples Gibbs, are your leads in-plane or decoupled, or a mixture of both? I would appreciate the designation of each lead.
I'm afraid this is where we're not clear. In the experiment where decoupled reading .4V. The probes wiring are actually cutting half of the loop if from top looking down. I think you when you say in plane, you actually referring to covering the entire field.
Gibbs,
I think it is prudent that we adhere to the actual experimental configuration. While it is interesting to look at other hypothetical lead configurations without the presence of a big solenoid coil, this would be departing from the essence of the experiment. Moreover, it serves to obscure the salient points I've had to wrestle you guys on in making my points in this discussion.
May I ask that for now we stick to the diagrams and their associated lead configurations I have uploaded for discussion?
Quote from: poynt99 on 2012.03.01, 22:55:24
Gibbs,
I think it is prudent that we adhere to the actual experimental configuration. While it is interesting to look at other hypothetical lead configurations without the presence of a big solenoid coil, this would be departing from the essence of the experiment. Moreover, it serves to obscure the salient points I've had to wrestle you guys on in making my points in this discussion.
Yeah, if this is too confusing , we just need to stop.
Sure, let's just stick to your config.
@Poynt99
QuoteWhat "phenomena" specifically are you referring to?
The phenomena of many experts not being able to agree on what happens when a simple loop of wire with two resistors is induced by an external magnetic field, I have no issue with the circuit however the disagreement between the experts is somewhat mindboggling. Let me put this in perspective, we suppose to understand and debate any number of complex subjects here and in the scientific community and yet we still find disagreement on even the simplest things such as this circuit, can you not see the irony in this?. If all of us supposedly understand so much then why do we disagree?, If we understand everything so well then why can't we give a simple explanation anyone could understand? as I said all I see is 11 pages on "non-answers", that is it kind of sounds like an answer but it just ain't because it only leads to more questions.
I believe you have claimed or at least implied you understand why Professor Lewin's experiment gives an answer we would not normally expect, now either you can justify this claim in simple terms anyone can understand or you cannot and it is an unsubstantiated claim --- not unlike the kind you remind so many others here not to make. I simply want to know which one it is as it is impossible to debate an answer which was never actually given.
On a side note keep in mind I am the designated Critics Critic and generally hated by all because I tend to touch on subjects which many consider taboo. Like Milehigh's "Bingo" fiasco, one too many pages of -- no, no thats not the right answer, no that's not the answer Im looking for. Then I find his extremely profound Bingo moment is something I knew some 30 years ago and could find in any grade school textbook. And I thought -- thats your Bingo moment--Really? It made me want to have a chug a 66 of rye and drive my car into a brick wall then vomit kind of moment to ease the pain of having to endure these kinds of posts. I think we can do better, I think we can say what we mean in plain english and debate it like rational adults, I think we can justify what we say using simple terms everyone can understand but that remains to be seen.
Regards
AC
Poynt,
You have a theory too how to attain these values. Let's hear it, maybe we can work it out from there.
Quote from: allcanadian on 2012.03.01, 23:58:14
I believe you have claimed or at least implied you understand why Professor Lewin's experiment gives an answer we would not normally expect, now either you can justify this claim in simple terms anyone can understand or you cannot and it is an unsubstantiated claim --- not unlike the kind you remind so many others here not to make. I simply want to know which one it is as it is impossible to debate an answer which was never actually given.
I indeed may not be finished with my treatise on this experiment, but you should have been able to draw some conclusions of your own by now based on what has been presented that raise questions about the experiment and the conclusions Lewin's audience departed with. And what exactly did you have in mind in terms of providing an explanation that
everyone can understand? It seems to me that you are saying you don't understand what has been discussed thus far, correct? Or is it that you don't agree with what has been said? Or both? Please clarify your stance.
And regarding what my claim is about, I don't think you've got that right. What you stated as your belief of what my claim is, I don't understand. Could you please clarify that as well?
Quote from: GibbsHelmholtz on 2012.03.02, 00:16:34
Poynt,
You have a theory too how to attain these values. Let's hear it, maybe we can work it out from there.
There is only
one value that is not so straight forward and easy to come up with just by simple observation, and that is the
decoupled measurement when the probes are anywhere other than directly across each resistor. The others I've presented are quite simple to come up with without having to do the actual measurement.
That being said, this "not-so-straight-forward" decoupled value is not that difficult to intuit really, when one knows and understands the following:a) The induced emf is 1V.
b) The induced emf of 1V is primarily induced in the conductors, not in the resistive material.
c) The induced emf will exhibit a polarity opposite to that of the PD's across each resistor.
d) The radial angle along the conductor for each probe (there will be a slight error because the resistors are of finite length).
e) The length of each conductor is equal.
The in-plane values are easy to come up with when one knows and understands the following:a) Measurements taken "in-plane", measure the circular electric field (or part thereof) produced by the changing magnetic field.
b) The total electric field produced is 1V, i.e. the sum around the loop is 1V.
c) The circular electric field is present regardless if a conducting loop exists in its path or not.
d) A uniform conductor in the E field's path retains a uniform-intensity circular E field within it because of an even distribution of charge carriers along its length.
e) Rresistors >> Rconductors
f) The introduction of series resistors into the conductor loop causes a proportionate re-distribution of the E field intensity due to an accumulated "lack" and "excess" of charge carriers across the resistors.
@Poynt99
As far as I know this was your statement or claim from page 1--
QuoteWe've discussed this experiment and concluded that Professor Lewin is incorrect in his assessment of his observations made regarding this experiment. The fields are indeed fully conserved, and it is more of a measurement error combined with an assumption which allowed him to come to this conclusion that the voltages measured are different depending on the direction taken.
My first concern was a statement made by professor Lewin in regards to a "non-conservative" field and I did not believe he made this in reference to the conservation of energy as I believe he knows all too well what this implies. Think about that, you don't honestly believe he was referring to the conservation of energy do you?. I believe he was making reference to "conserved" as in the dictionary definition "Conserved:To keep (a quantity) constant" and in his case we could say the fields are not conserved as he believes their qualities have changed and are not constant, that is not consistent with what we would normally expect.
I know there is an impulse to make assumptions especially when someone even mentions non-conservative fields because people get very emotional over these kinds of things because it challenges their beliefs however much of science is about being impartial and not reading something into things which may never have been there. I know all about this because I still do it all the time and thankfully with experience I have learned to catch it through logic and a little intuition.
As well there is the issue of measurement error which would depend solely on the actual equipment Lewin used to measure the variables, it's accuracy and how it was configured. I would assume you must have all of this information do you not? and I would also assume you have exercised caution and contacted someone directly related to this experiment in regards to how the equipment was configured in the actual experiment. If not then what do we actually know of substance?, that which can be proven or disproven based on the facts we have on hand. If their is one thing I do know as a fact it is you had damn well better have your facts in order if you are going to challenge someone as being incorrect if not misguided as it tends to blow up in ones face rather quickly. Let's put it this way, if all this is based solely on a simulation of an experiment which can be proven wrong by the actual experiment then this is definately not going to turn out well in the end.
Im not saying anyone is right or wrong here, Im just saying making assumptions can have a way of turning around and biting us in the ass when we least expect it.
Regards
AC
How you misconstrued my statement of "conserved" to be COE, I have no idea.
Now, I did not specify, and apologize for that. I was referring to the case when the measurement device is decoupled from the experiment.
Quote
As well there is the issue of measurement error which would depend solely on the actual equipment Lewin used to measure the variables, it's accuracy and how it was configured. I would assume you must have all of this information do you not? and I would also assume you have exercised caution and contacted someone directly related to this experiment in regards to how the equipment was configured in the actual experiment. If not then what do we actually know of substance?, that which can be proven or disproven based on the facts we have on hand. If their is one thing I do know as a fact it is you had damn well better have your facts in order if you are going to challenge someone as being incorrect if not misguided as it tends to blow up in ones face rather quickly. Let's put it this way, if all this is based solely on a simulation of an experiment which can be proven wrong by the actual experiment then this is definately not going to turn out well in the end.
Lewin did a poor job of
illustrating the experiment in terms of where exactly the probes were connected. However, for the measurement he was making, it makes no appreciable difference to the observed result anyway. And if you already didn't know why that is, you should know at this point.
Lewin's "measurement error" was largely the measurements that he did
not perform for the class. One other faux pas was that he did not explain to the class all the implications of what they had witnessed. Most if not all of Lewin's errors have been exposed in this discussion. I would encourage you to review it.
Now, perhaps you'd be so kind as to respond to the first part of my post (http://www.overunityresearch.com/index.php?topic=739.msg21508#msg21508)?
And I'll add this:
AC, if you have anything of value of your own "technically-speaking", to add to this discussion, what are you waiting for? Please do.
Thus far I've not seen anything in that regard.
If you do have a cogent argument which you believe counters my own, then present it. Address each item one by one. We're here to get to the truth and foster understanding, right?
This whole discussion has been most instructive.
It demonstrates quite emphatically how easily
deception can be accomplished.
The power of the EGO as a force of distortion.
How when we make certain assumptions it is
possible to believe nearly anything no matter
how improbable.
Professor Lewin has successfully fooled a great
many people.
Is it really any wonder that Truth is often a victim
within the world of academia?
It would be quite easy to duplicate the "experiment"
with an alternating magnetic field to make steady
state measurements and thereby isolate the flaw.
What was the Professor's trick?
Quote from: Dumped on 2012.03.02, 05:45:07
It would be quite easy to duplicate the "experiment"
with an alternating magnetic field to make steady
state measurements and thereby isolate the flaw.
It would be relatively easy, yes. However I chose to stick to a single capacitive discharge to:
a) stay true to the experiment Lewin performed.
b) get higher current and hence higher induced emf.
c) keep things simple.
It is indeed very simple!
The good Professor (a very clever man)
leads us a bit astray while still at the
chalkboard early on - the moment he
erases the electric cell (DC source) and
replaces it with a length of wire.
What is the totality of his circuit once
he includes the loop of wire, the two resistors
and the internal coil which generates the changing
magnetic field?
It's something we make use of each and every
day.
We must be very careful about the assumptions
we make regarding his "circuit."
It bears repeating: the good Professor is indeed
a very clever man.
Dumped,
You may find the work of the Prof easy to understand. That is good.
Making the experiment use repeated pulses is not so easy. Yes, the circuitry is easy. The amount of energy used makes it a bit more difficult.
I know. I'm in the process of doing just that :)
Ah yes! Now you're beginning to understand
why the good Professor chose "pulsing" to
demonstrate the phenomenon - after first
setting the audience up to mis-interpret what is
actually taking place.
It's a classic case of controlled deception.
Once you see how you've been fandangled
you'll slap your head in amazement...
(Error correction)
I must add; the points Poynt.99 has been trying to
drive home regarding what has been overlooked
or "not measured" is key.
Most recently, yes.
(Error correction)
Quote from: Poynt.99Lewin's "measurement error" was largely the measurements that he did not perform for the class. One other faux pas was that he did not explain to the class all the implications of what they had witnessed. Most if not all of Lewin's errors have been exposed in this discussion. I would encourage you to review it.
That is MY quote. C.C Clearly you doctored that....why?
http://www.overunityresearch.com/index.php?topic=739.msg21515#msg21515
You are absolutely correct!
Scrolling in too much haste has
caused the loss of sourcemanship.
My error. I'll make appropriate corrections
to earlier erroneous postings...
You are spot on!
Quote from: Dumped on 2012.03.02, 22:23:17
Ah yes! Now you're beginning to understand
why the good Professor chose "pulsing" to
demonstrate the phenomenon - after first
setting the audience up to mis-interpret what is
actually taking place.
It's a classic case of controlled deception.
....
Ummm.... Ok?
The professor hasn't deceived me. The moment he first said both meters were connected to points 'A' & 'D' after the setup clearly shows the probes connected directly to the resistors, I caught that minor mistake. I'm sure that is all it was or more probably - he knew it didn't really matter with his setup.
The only other part is where he insisted, loudly, that Ohm's law is always true. For all practical purposes it is always true. In general, as long as it is applied to the device (component, wire section, etc.) and not that device and the world around it, it should always hold true. It can be argued so I was surprised to hear him announce that Ohm's law always holds so boldly.
.99 has clearly shown another part of my objections to Lewin's experiment. There was soooo much more he could have shown with that setup but he didn't. Why? Time to dismiss the class? I doubt it.
In my experience, by the time you are shown that part you have already killed brain cells figuring the rest that could have been shown. I doubt his class needed to see the rest.
Then, why didn't he just say the battery was replaced with an indirect energy source? Well, I suppose he did, didn't he? That, alone, breaks KVL because you can't sum with voltage not in the loop.
Would it all be easier if we used the same definition of Kirchoff? My favorite is the one Maxwell massaged into working conservative or not. I don't have the quote from Maxwell but it goes something like this - the sum of potential differences around a loop is equal to the sum of EMFs around the loop - or was it 'equal to the inverse'? I don't remember.
Regardless, Maxwell and Kirchoff were around the same time frame. I'll trust Maxwell to interpret Kirchoff over dozens of re-writes, adaptations and generalizations since his time.
When you adjust KVL/KCL to work in non conservative fields you are just repeating what Maxwell already did.
Lewin said, "Kirchoff is for da birds!" He is correct. Kirchoff has been for da birds since Maxwell fixed his law that he never wrote and was only attributed to him during his funeral (I read that somewhere while researching this mess about 15 years ago).
Yes, I am working on the flavor of this experiment I am accustomed to. The whole point of that one was to show all Lewin didn't show.
The nature of the energy input to the
circuit is critical to its asymmetrical
voltage distribution.
The circuit isn't normally configured
in circular form but rather in parallel
conductor form.
Transients are tricky...
The good Professor's "mistakes" may be
much more akin to the superfluous moves
a magician makes to draw attention away
from the "magick."
This is not to say that the Professor's motives
were less than honorable. In the learning
environment deception is sometimes used as
a vehicle to stimulate thought in problem solving.
Quote from: poynt99 on 2012.03.01, 12:58:10
...
@Ex: You will NOT measure an in-phase PD across the wires, relative to that across the resistors. You need to pay closer attention to the question and diagram being discussed. Your diagram is in reference to the in-plane measurements, while the last question and diagram reference the decoupled mode of measurement, which btw, is the only method of the two which provides the TRUE PD measurement of all the components in a dynamic circuit such as this.
The E field of the EMF/emf in all circuits, whether placed there physically (such as with a battery) or induced through induction, are always, and without exception, IN OPPOSITION to the E fields of the loads, such as the resistors in our case.
Therefore, the TRUE PD sum of the loop is 0V at any instant of time, or total length of time.
While measuring the in-plane E field however, it is non-conservative, and is always 1V.
The notion that the true PD sum can be 2V (vs. 0V) is ludicrous. You would have a FE generator at your fingertips, if it was. :P
You didn't understand my previous reply. I was refering to:
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=7377;image)
As explained, what you measure is really the emf in the half circle 1-D-9 or 1-1'-A-9'-9, i.e. half the total emf.
The fact to be in plane or out of plane is irrelevant: when you see the circle from above by a projection, you see the wires from the oscilloscope connecting the points 1 and 9 as being along the diameter.
In any case each half-flux leaves each half-circle respectively right and left, therefore they cross the surface of the measurement circuit whatever the height of the oscilloscope. It doesn't a matter you put a micro-oscilloscope just between point 1 and 9 or above this diameter line.
Ex. Yes, that diagram measures half the emf. Orientation of the leads DOES matter.
Some relevant points on that diagram and on the in-plane vs. decoupled measurements, are the following:
1) the indicated scope voltage across points 9-1 are different for the in-plane and decoupled lead orientations
2) for V91: decoupled = -0.5V, in-plane = 0.9V
3) the loop emf(s) is opposite in polarity to the PD(s)
4) the sum around the loop IS always 0V, i.e. KVL always holds
5) only the decoupled lead orientation can provide the emf(s) AND PD(s) measurements simultaneously
6) measuring in-plane as Lewin does, provides only half the complete picture, i.e. the E field only
7) the measured E fields and PD's have to be, and ARE equivalent when indicated on the scope
@poynt99
How is generated the primary field? My reply supposed that the dB/dt field is obtained from a concentric coil of same diameter as the circle A-R100-D-R900.
A different case could explain the discrepancies.
Yes,
In my case (and I'm quite certain for Lewin's setup as well), the solenoid is only a slightly smaller diameter than the loop.
The loop is located mid-way along the solenoid, and the solenoid length is >> greater than the loop length.
Quote from: poynt99 on 2012.03.03, 16:44:54
...
The loop is located mid-way along the solenoid, and the solenoid length is >> greater than the loop length.
It becomes very difficult, because we are now with circuits in 3D.
My previous reasoning was based on a magnetic field as if it was generated by the circular circuit itself (i.e. generated in fact by an identical coil tighly positionned with the other).
I suggest to simplify the experimental setup by using a flat narrow coil of same diameter as the circuit with the resistances, the second being positionned just above the first one, otherwise I'm afraid that we are no more able to take into account all the parameters.
On the contrary Ex, you're on the wrong track.
Go back to the basic principles of the experiment.
The requirement is for a UNIFORM (i.e. as close to pure vector as possible) changing magnetic field passing through the loop. As such, the absolute ideal configuration is to use a solenoid of infinite length, with the short loop (one wire width long) located precisely mid-way (assuming you can divide infinity in half, which of course you can't) along the length. In practice, we simply make the solenoid length >> the loop length, and the loop is located mid-way.
There are no "discrepancies" in the results I posted. They corroborate perfectly with the theory. The setup is correct and the results accurate.
Specifically, what is your objection to these results?
Your solenoid is not infinite. You can't make the approximation of an infinite solenoid while connecting an oscilloscope above this supposed vertical infinite solenoid. In this case you have to take into account the end of the solenoid and the way that the wires escape from the proximity of the solenoid. Near a end of the solenoid, the field in not constant neither in intensity nor in direction, the field lines are curved in order to loop towards the other end. The oscilloscope doesn't measure anything else than the emf generating by the varying flux through the surface delimited by the measurement circuit (not only and not simply the circular circuit with the resistances), and this topology is not trivial.
As I said, in practice, we make the solenoid length >> the loop length.
This is sufficient to achieve a uniform field through the loop.
The decoupled lead configuration results in very minimal (<1%) induced voltage in the leads, despite the curl of the B field.
For the in-plane measurements (leads are in-plane), the scope leads measure the electric field only.
Just an update....
Yes, I am assembling the experiment. I wish it to be set up the same as Lewin's but it won't be an exact duplication.
All along, I've been sensing some major holes in my memory on this experiment (the one I performed long before the Lewin version on YouTube).
I realized today that what he demonstrated is far from ground breaking, unique or even current. Memories are returning.
The sole purpose of the same experiment (by others) was to demonstrate the difference between conservative and non-conservative fields. By partial definition, a non-conservative field is path dependent. In other words, the value measured will depend upon the path the meter lead current takes, or probe orientation. This is well proven by Lewin and .99.
So, I'm left wondering what the hubbub is about. Is there an argument that the fields are not non-conservative? Is it still believed that KVL can be applied to such a field? If so, I don't and won't get it. Was it believed if the field is conservative then there is no non-conservative aspect - or vice-verse? Perhaps, there is an argument that non-conservative fields don't exist?
There are several things in physics that I prefer to check myself and a couple of things I completely disagree on. This isn't one of them. I suppose you could say that KVL still holds IF you are holding the probes the right way (& standing on one foot while whistling 'Dixie'? ;D).
I'll lean on the same post used by some others here..... 'It has been proven repeatedly for eons and hasn't been wrong, yet. So, why do I need to argue it?'
Also remembered was the fact that my first version of that experiment was in 1977. Lewin didn't invent this experiment. I am convinced the same experiment was on a list of lessons at MIT in 77 except digital bench meters were used and the power supply was a variac. The class numbering was different- H20 or H30? Maybe that was just the military shortened version? In later demonstrations I used galvanometers so I could show the real fun aspects by rotating the meters around the transformer (solenoid primary / loop secondary). They didn't work at all like a scope sitting on a table would work. There is that path dependency, again :)
Some interesting points I intend to confirm....
The ratio of voltage drop across the resistors is induction dependent.... The higher the voltage induced the wider the ratio. After all, that resistor loop is just a transformer secondary. The resistors resistances don't determine the induced fields. That is left up to Faraday with no concern for Kirchoff. They can only limit the current. Since they are part of the winding it was never a matter of simple voltage drop across the resistor because the resistor was also a section of the secondary.
Now, wouldn't that be interesting? If you crank the primary up with enough juice one resistor will show all the voltage dropped across it and apparently none across the other - all depending upon the path the current takes, of course :o (Not my discovery)
The scope probes connected directly to the resistors had no effect upon the outcome of the Lewin version. This would be why he connected directly to the resistors - he knew it didn't matter when the scopes and probes weren't going to change relative orientation to the measured loop.
The bosses keep jerking me a from one project to another. Don't be surprised if I must ask later - what are we trying to disprove?
Strange that you're not sure what all the "hub bub" is/was about. I thought you were following along, but I guess you completely missed it. :D
I think Gibbs and Dumped got it, and Ex, well he is still undecided apparently.
Tomorrow, I'll try and spell it out for those who still don't know why we're here, or where we went.
On second thought WW, I'd prefer to chat with you one on one first.
Lessons Learned: You and I always seem to go round and round in circles while discussing something.
So in an effort to mitigate the need for numerous posts on both our parts, we should chat on Skype, or Messenger or phone a couple times for an hour or so. Are you game?
Let me know via PM please, what medium you'd prefer, and when. I am available after 16:30 MST, and all day weekends.
Quote from: WaveWatcher
...
Is it still believed that KVL can be applied to such a field?
If so, I don't and won't get it.
...
Aye, KVL does apply. But we must recognize that the
circuit is not a DC circuit and therefore somewhat
different "rules" apply.
Has the pulse width of the observed waveforms been
established? And the constituent frequency domain?
Are you familiar with reflectometry?
Quote from: Dumped on 2012.03.13, 07:48:12
Aye, KVL does apply.
Only as far as KVL is a sub-set of Faraday. If you consider KVL to be the most frequent description, it doesn't apply at all.
Quote
But we must recognize that the
circuit is not a DC circuit and therefore somewhat
different "rules" apply.
Will you invent some new ones?
Quote
Has the pulse width of the observed waveforms been
established? And the constituent frequency domain?
If one can read a scope, yes.
Quote
Are you familiar with reflectometry?
More than 'familiar'.
That has little to do with this experiment.
Quote from: poynt99 on 2012.03.13, 04:23:24
On second thought WW, I'd prefer to chat with you one on one first.
Lessons Learned: You and I always seem to go round and round in circles while discussing something.
So in an effort to mitigate the need for numerous posts on both our parts, we should chat on Skype, or Messenger or phone a couple times for an hour or so. Are you game?
Let me know via PM please, what medium you'd prefer, and when. I am available after 16:30 MST, and all day weekends.
I posted since is saw you say something about this subject on the shout box, earlier yesterday.
I don't spend a lot of time on the web lately. So, maybe I did miss the point.
Ex will be back on the subject. I think it just takes a long time for him to filter everything through Google Translate.
It is obvious this subject will be another one of those not needing further discussion. What it needs is full demonstration of all aspects. When that is done there should be absolutely no doubt that the 'static/original' form of KVL applied is like measuring AC with a DC meter.
I intend to finish this demonstration on YouTube. With that built, I can have a reference point for a detailed conversation.
Sorry, Skype sucks on my connection.
I think the point was indeed missed.
QuoteIt is obvious this subject will be another one of those not needing further discussion. What it needs is full demonstration of all aspects. When that is done there should be absolutely no doubt that the 'static/original' form of KVL applied is like measuring AC with a DC meter.
Is this a round-about way of saying you don't believe my treatise thus far, and want to
see proof? That's not a problem, will do.
Your other statement proves further that you are either not aware of, or don't understand the "problem" and the reason for this undertaking.
Do you have a phone that works? ???
Quote from: WaveWatcher on 2012.03.13, 02:36:26
Some interesting points I intend to confirm....
The ratio of voltage drop across the resistors is induction dependent.... The higher the voltage induced the wider the ratio.
??? That's truly a wild speculation. Absurd to be blunt.
Quote from: poynt99 on 2012.03.13, 12:26:06
??? That's truly a wild speculation. Absurd to be blunt.
Very well. That will be part of the demonstration.
Quote
Is this a round-about way of saying you don't believe my treatise thus far, and want to see proof? That's not a problem, will do.
Absolutely not. I have never had doubts about your honesty.
Quote
Your other statement proves further that you are either not aware of, or don't understand the "problem" and the reason for this undertaking.
My comparison to using a DC meter to measure AC is not one of equipment misuse. It is the fact that a DC meter can be used to measure AC with a bit of circuitry. I'm saying KVL can be applied but only when you consider the battery source is replaced with the source of induction and probe orientation is a certain way. Yes, the sum can be zero.
I know my comments appear to be posted to p you off but that isn't the case.
If you don't mind... PM your time zone. I'm on 12H days right now but we'll work a call in somehow. I'm willing to try Skype but I'm on satellite. The lag is more than most can bear.
BTW:
I made the comment about the changing voltage drop ratio because I found part of my notes from my first attempt. The changes are there and they were replicated - and expected by the prof. (I only remember him as our pet name - 'Bernie'.
Tomorrow, I will try to post a summary of things that might help in our conversation.
We can always chat here in the chat engine if voice is not practical.
But this comment is getting there:
QuoteI'm saying KVL can be applied but only when you consider the battery source is replaced with the source of induction and probe orientation is a certain way. Yes, the sum can be zero.
And to comment briefly: Indeed, the battery source IS replaced by an induced emf in the experiment. But again, Lewin is mixing apples and oranges, and as a result of this oversight, his audience departs with an understanding that KVL never applies with induced circuits. Clearly it does. Faraday AND KVL are satisfied at the same time. Lewin does NOT espouse this fact at all with his treatise. In fact he diminishes KVL, when the truth is they are equally correct.
Hint: One approach to getting better clarity on this, is to forget about the measurement device and its leads, and focus on the physics and EM theory.
Quote from: WaveWatcher
Quote from: Original CommentAye, KVL does apply.
Only as far as KVL is a sub-set of Faraday. If you consider KVL to be the most frequent description, it doesn't apply at all.
Quote from: WaveWatcher
Quote from: Original CommentBut we must recognize that the
circuit is not a DC circuit and therefore somewhat
different "rules" apply.
Will you invent some new ones?
No, the rules are now well established.
Quote from: WaveWatcher
Quote from: Original CommentHas the pulse width of the observed waveforms been
established? And the constituent frequency domain?
If one can read a scope, yes.
Those questions were meant to stimulate
thought. Could the pulse width or the
representative frequencies in any way
account for the unusual nature of the
circuit performance?
Quote from: WaveWatcher
Quote from: Original CommentAre you familiar with reflectometry?
More than 'familiar'.
That has little to do with this experiment.
That question was meant to stimulate some
unorthodox or unconventional thinking.
Does the circuit strangeness in any way
resemble a length of transmission line
unbalanced at each end with a transient
standing wave pattern in progress?
Do you recollect Tesla's experiments with
discharge excited parallel lines and the
resultant voltage patterns?
Quote from: poynt99 on 2012.03.14, 02:29:48
...
Faraday AND KVL are satisfied at the same time.
...
Maybe we can save KVL but at the price of a complicated methodology. In KVL, the voltages are obtained from potential differences between two nodes. But a voltage from induced emf doesn't derive from a potential. It must be calculated by an intregral over a closed path. The mixing of the two forms emerges neither straightforwardly nor clearly from KVL.
All true Ex, but that's not a problem, nor THE problem. ;)
Lewin is brilliant with his delivery of this lecture. He has all the grace and sly of a good magician or illusionist. Most were captivated. But not all were fooled. ;)
At the very beginning, Lewin plants a seed that what you are about to see is "a very non-intuitive result". In reality, this of course is not the case, but you have already been set up to believe this is so. Call it "preconditioning".
Lewin begins by illustrating and discussing a simple DC circuit that contains an EMF source, and two potential drops across resistors. Take careful note that BOTH the EMF and the potential drops are illustrated. Also note that the potential differences calculated across the resistors is based on the circuit current. Ostensibly, his focus is on the potential difference between points D and A (VD-VA). See "DC_cct01.png" below.
Next, Lewin erases the battery and replaces the EMF source with an increasing magnetic field inside the loop. Here, Lewin notes that the resulting induced loop emf (peak) is 1V.
Lewin goes on to explain that the circuit current is still 1mA because the induced emf is 1V. Then he asks again; "what is VD - VA?" At this point, most viewers do not notice the slight of hand so brilliantly performed by the good professor. Did you?
Lewin is again calculating VD-VA using 1V emf and 1mA current, yet he has NOT illustrated the 1V emf source on his diagram. Why did he not illustrate the induced circuit emf?
This is in fact the faux pas that invalidates the rest of his lecture. Why? Because professor Lewin boisterously exclaims that Kirchhoff's rule (or law) does not hold in this case. He is absolutely incorrect with his assertion, and had he illustrated the induced circuit emf, it would be obvious why.
You can NOT have any potential difference across a circuit load, unless you ALSO have a circuit emf (or EMF in cases of DC) present. How can you have an emf induced on a piece of wire? I would ask, how can you not? This is Faraday's law.
The "induct_cct01.png" diagram illustrates what Lewin SHOULD have drawn on the board before he went off leading his audience down the garden path. Take careful note that the polarity of each emf source (red lines and polarity markers) is in opposition to the potential differences (PD) across the loads (green polarity markers). This is very important to understand. This is ALWAYS the case in every circuit, regardless if it is a DC or an induction circuit such as this. The total induced emf is equal to the total potential drops across the circuit non-reactive loads (i.e. the two resistors); the sum of the two being equal to zero. In other words, the closed loop integral equals 0V. And at ANY instant of time.
What this means, is that Kirchhoff's Law DOES hold in this case, contrary to what Lewin erroneously espouses in his lecture.
The simple matter of HOW to prove this and properly measure the potential drops AND emf's in the loop have been illustrated in several posts prior to this. It's important to realize that Lewin never made one emf nor PD measurement in his experiment. He only measured the electric field, and that's all he could measure considering the way his probes were situated. When you see this, you realize professor Lewin was confused about what he was measuring. It is clear he was mixing apples and oranges.
This should answer the question as to what the purpose of all this testing and discussion has been about.
btw, what is the closed loop integral of the induced electric field in this experiment? It is 1V of course! O0
.99
Quote from: poynt99
...
You can NOT have any potential difference across a circuit load, unless you ALSO have a circuit emf (or EMF in cases of DC) present. How can you have an emf induced on a piece of wire? I would ask, how can you not? This is Faraday's law.
...
And there you have it!
Well done.
Quote from: poynt99 on 2012.03.29, 23:14:40
You can NOT have any potential difference across a circuit load, unless you ALSO have a circuit emf (or EMF in cases of DC) present. How can you have an emf induced on a piece of wire? I would ask, how can you not? This is Faraday's law.
......
tw, what is the closed loop integral of the induced electric field in this experiment? It is 1V of course! O0
Great work and fine detail, as I have come to expect from you. O0
However ( :D ), with the problems in nomenclature, in general, and the lack of agreement for the definition of 'emf' throughout all facets of the sciences and industries, I can only express my complete agreement with Ex:
"Maybe we can save KVL but at the price of a complicated methodology. In KVL, the voltages are obtained from potential differences between two nodes. But a
voltage from induced emf doesn't derive from a potential. It must be calculated by an intregral over a closed path.
The mixing of the two forms emerges neither straightforwardly nor clearly from KVL."
My problem with all of this ( and I do see it as my problem ) is that I understand KVL as only being applicable to electric fields. There are no electric fields from one end of an inductor to another except where winding resistance is considered.
To me, the closed loop integral of the electric field produced by induction is 1V. Therefore, KVL is not applicable.
Sorry :)
Quote from: WaveWatcher on 2012.03.30, 10:25:49
My problem with all of this ( and I do see it as my problem ) is that I understand KVL as only being applicable to electric fields. There are no electric fields from one end of an inductor to another except where winding resistance is considered.
To me, the closed loop integral of the electric field produced by induction is 1V. Therefore, KVL is not applicable.
Sorry :)
"emf" is any source of voltage or current, manifest either by physical placement (eg. battery) or induced. It is unambiguous. In this case it results from the
induced electric field. Yes, the CLI of the induced electric field is 1V, I've stated that.
You believe KVL only applies to electric fields? I would disagree. The CLI of the electric field in this case is 1V, and is therefore non-conservative.
KVL applies only to emf's and PD's, NOT to E-fields. Lewin didn't measure emf's nor PD's in his experiment, so he is incorrect when he states that KVL does not hold. btw, Lewin does not mention electric fields at all; he relates everything to the circuit current and potential drops. In that, it would seem you are in disagreement with the professor.
What is your definition of an electric field?
You may disabuse yourself of your beliefs or understanding by making the same measurements I have made and shown here in this thread. Bench measurements trump all beliefs and theories.
.99:
Quote
Bench measurements trump all beliefs and theories.
Well said, poynt.
"Theory may guide, but experiment decides".
My simple proposal is that we need to find out why measurement difference between decoupled and in plane. What method can we use to solve both cases?
Gibbs,
The reason for the difference in the measurements has already been explained. What is it that you don't understand?
In terms of making the measurement, there is no one way to get the same results for both all around, but the decoupled measurement is the only TRUE measurement method IF you are interested in the emf's and PD's.
Clearly professor Lewin was interested in illustrating the existence of a non-conservative field, and he would have completely succeeded had he not decided to mix in PD calculations.
To re-emphasize; professor Lewin was measuring nothing more than the induced electric field in his experiment.
Surely an MITPHD would know this.
Maybe he is making his students think?, Make them question? Find a flaw in what he is telling them like what has been done in this thread?
Quote from: giantkiller on 2012.03.30, 19:21:00
Surely an MITPHD would know this.
Every MITPHD SHOULD know this. Evidently, not all do.
Quote
Maybe he is making his students think?, Make them question? Find a flaw in what he is telling them like what has been done in this thread?
That is EXTREMELY unlikely.
It's clear Lewin intended everything he said to be taken as the indisputable truth.
I would then consider any time spent at MIT would be counter intuitive to thinking freely. Is this the type of professoring they cloisture?
Quote from: poynt99 on 2012.03.30, 19:48:56
Every MITPHD SHOULD know this. Evidently, not all do.
That is EXTREMELY unlikely.
It's clear Lewin intended everything he said to be taken as the indisputable truth.
Quote from: poynt99 on 2012.03.30, 16:53:33
Gibbs,
The reason for the difference in the measurements has already been explained. What is it that you don't understand?
In terms of making the measurement, there is no one way to get the same results for both all around, but the decoupled measurement is the only TRUE measurement method IF you are interested in the emf's and PD's.
Clearly professor Lewin was interested in illustrating the existence of a non-conservative field, and he would have completely succeeded had he not decided to mix in PD calculations.
To re-emphasize; professor Lewin was measuring nothing more than the induced electric field in his experiment.
I'm not sure if all got the explanation. If we understand how it works, we should be able to calculate every situation that can throw at us. It could be a mix of decoupled and in plane. I told you that decoupled is just the same as in plane, so there is no true measurement... or more correctly, every measurement is correct if it cover by 1 method.
The difference between measurement depends on how much flux passing through the loop. Decoupled is no exception. Half of the flux go up, but since the wire is 6' up, the field lines split half left and half right to return to the south pole.
Quote from: GibbsHelmholtz on 2012.03.30, 21:55:00
I'm not sure if all got the explanation.
Sometimes it's good to think about the things that were presented and figure out the details yourself.
Quote
If we understand how it works, we should be able to calculate every situation that can throw at us.
Yes, and we do.
Quote
It could be a mix of decoupled and in plane.
Anything in between fully decoupled or fully in-plane is irrelevant as far as I am concerned. What relevance does it have to you or this experiment?
Quote
I told you that decoupled is just the same as in plane, so there is no true measurement... or more correctly, every measurement is correct if it cover by 1 method.
The decoupled and in-plane measurements are worlds apart. They are measuring completely different things.
Quote
The difference between measurement depends on how much flux passing through the loop. Decoupled is no exception. Half of the flux go up, but since the wire is 6' up, the field lines split half left and half right to return to the south pole.
The amount of flux passing through the measurement wire while it is decoupled (i.e. vertical), is inconsequential. The result is very little, if any induced emf.
Quote from: poynt99 on 2012.03.30, 22:15:01
...
The amount of flux passing through the measurement wire while it is decoupled (i.e. vertical), is inconsequential. The result is very little, if any induced emf.
...
It's not little. The decoupled measurement of .4V contained half of the flux through its
loop. It doesn't matter how big the loop is, the amount of emf is how much flux through that loop. We can't say flux pass through a wire, it only makes sense if it pass through a loop.
Here is an example. This is decoupled. Half of the flux pass through the red loop.
Gibbs,
The measurement wires are NOT part of the induction loop when they are vertical, i.e. normal to the loop.
Quote from: poynt99 on 2012.03.30, 23:05:00
Gibbs,
The measurement wires are NOT part of the induction loop when they are vertical, i.e. normal to the loop.
My opinion is yes it is. This is how I based all my calculations on anyway.
Quote from: GibbsHelmholtz on 2012.03.30, 23:39:48
My opinion is yes it is. This is how I based all my calculations on anyway.
Opinions can't change the physics of reality.
From Wiki:
QuoteMichael Faraday formulated that electromotive force (EMF) produced around a closed path is proportional to the rate of change of the magnetic flux through any surface bounded by that path.
Flux can not pass through a surface area that does not exist. When a loop is normal to the flux path, NO emf is induced in that loop, because there is NO surface area. You may wish to review your basic EM theory. It is well known that two coils situated 90º from each other have no coupling. This is no different
Here's an experiment for you to try or think about:
Take an air-core solenoid, roughly 4 inches long and 1/2' in diameter. Now loop a wire through the centre of the solenoid about 10 loops, say 15 inches in diameter. You now have two coils, one looped through the other, and they are normal to each other. Drive one coil with a sinusoid while monitoring the other "output" coil. What will you see on the "output" coil?
.99
Quote from: poynt99 on 2012.03.31, 12:18:08
Opinions can't change the physics of reality.
From Wiki:
Flux can not pass through a surface area that does not exist. When a loop is normal to the flux path, NO emf is induced in that loop, because there is NO surface area. You may wish to review your basic EM theory. It is well known that two coils situated 90º from each other have no coupling. This is no different
Here's an experiment for you to try or think about:
Take an air-core solenoid, roughly 4 inches long and 1/2' in diameter. Now loop a wire through the centre of the solenoid about 10 loops, say 15 inches in diameter. You now have two coils, one looped through the other, and they are normal to each other. Drive one coil with a sinusoid while monitoring the other "output" coil. What will you see on the "output" coil?
.99
The reality is I see a surface area and flux pass through it.
The experiment you propose would yield no emf induced I know, but the in the Lewin's experiment the connection of measuring wire is physically connected to the loop with two resistors. It itself formed its own loop. The geometry is weird, but it's a loop and has surface area.
I'll try again with some picture. The first one has no emf since it is normal to the loop. The second one has some emf since it tilted. The third one has half emf.
Quote from: GibbsHelmholtz on 2012.03.31, 12:46:22
The reality is I see a surface area and flux pass through it.
The third one has half emf.
Incorrect on both counts.
The vertical portions of the wire (the measurement wires in the vertical plane)
contribute NO emf to the measurement. Therefore, you are measuring ONLY what is induced in the horizontal loop at those two points.
Quote from: poynt99 on 2012.03.31, 13:19:05
Incorrect on both counts.
The vertical portions of the wire (the measurement wires in the vertical plane) contribute NO emf to the measurement. Therefore, you are measuring ONLY what is induced in the horizontal loop at those two points.
Then how come in plane contribute emf?
Gibbs,
I say, you are on the right track.
I will also say, the majority of my posted opinions ARE based upon actually performing the experiments. I have no theories to promote regarding KVL. Faraday did that and Maxwell corrected Kirchhoff and Faraday while misusing the Volta's 'emf'.
No more time, at the moment. I'll be back.
No contribution to emf, yes.
It does contribute to the change of the surface area AND redefines that original loop area into TWO loop areas.
Quote from: GibbsHelmholtz on 2012.03.31, 13:27:19
Then how come in plane contribute emf?
Gibbs.
You seem lost in regards to how induction works?
Also, you seem to be ignoring the fact that the in-plane measurement becomes part of the experiment, and that the probes/leads in that position can ONLY measure the induced electric field, or if you want to think of it another way, the measurement leads sense what the electric field is doing between those two points when approached from a particular side.
Quote from: WaveWatcher on 2012.03.31, 13:38:16
I will also say, the majority of my posted opinions ARE based upon actually performing the experiments.
Wavewatcher, based on your past performance throughout this thread, I have to say that your opinions are just that. Either it's been too long since you did the experiments, or you didn't perform the experiments I did, or you didn't realize the ramifications of the results you saw, or all of the above.
I'm guessing "all of the above".
Quote from: poynt99 on 2012.03.31, 18:46:41
Gibbs.
You seem lost in regards to how induction works?
Also, you seem to be ignoring the fact that the in-plane measurement becomes part of the experiment, and that the probes/leads in that position can ONLY measure the induced electric field, or if you want to think of it another way, the measurement leads sense what the electric field is doing between those two points when approached from a particular side.
Poynt,
The reason I say it like that because the measurement is base on induction and not in plane or decoupled. I agree with WW. The loops now contain 2 planes. One is vertical and the other is horizontal. There is no flux through vertical plane but horizontal, yes. I can't seems to find a better explanation. Maybe someone else can describes it better.
Quote from: poynt99 on 2012.03.31, 18:54:00
Wavewatcher, based on your past performance throughout this thread, I have to say that your opinions are just that. Either it's been too long since you did the experiments, or you didn't perform the experiments I did, or you didn't realize the ramifications of the results you saw, or all of the above.
I'm guessing "all of the above".
For your first guess, I'll just say I managed to find some notes that refreshed my memory.
Second, yes I didn't perform the experiments you did. The first time it was with two digital bench meters and a variac driving the primary. The last time, the measurements were with a galvanometer. In the first case the test circuit could be rotated between the measuring devices while they were connected to the secondary loop with slip-rings on the pivoting axis. The second case, the galvanometer rotated around the test circuit.
The third, well, I suppose more than one professor could be wrong. The ramifications were beaten into our heads by a very displeased associate professor (supposedly). He was always cranky after driving from Boston to Ft.Devens. It wasn't Lewin. We only knew him as 'Bernie'.
Gibbs,
With your current understanding (more than one plane, etc.), what do you think the measurement would be if the vertical section (normal to the transformer secondary loop) was still vertical but below the measure loop? (This would be with the measurement points midway between resistors - not across the resistors like Lewin shows, just for clarity.)
Should it still be +.4V, -.4V or another number entirely different?
No hurry. As soon as my double shifts are over I'll provide the answer with something better than opinions.
Because of those fine memories (now), I seriously doubt Lewin was doing more than reciting a lecture script from the 70's.
Gibbs,
As an attempt to assist you with your thoughts on the many ways to change the value of the measurements....
If you are familiar with Faraday's sliding bar experiment, look at your latest posted drawing showing the two planes and imagine what it would look like from above.
To create induction or change the measurement you need not only change B. You may also choose to change A by sliding the bar (the Faraday sliding bar experiment) or change the relative angle between the loop or the test probe circuit and the flux.
Quote from: WaveWatcher on 2012.04.01, 04:30:41
Gibbs,
With your current understanding (more than one plane, etc.), what do you think the measurement would be if the vertical section (normal to the transformer secondary loop) was still vertical but below the measure loop? (This would be with the measurement points midway between resistors - not across the resistors like Lewin shows, just for clarity.)
Should it still be +.4V, -.4V or another number entirely different?
No hurry. As soon as my double shifts are over I'll provide the answer with something better than opinions.
Because of those fine memories (now), I seriously doubt Lewin was doing more than reciting a lecture script from the 70's.
I think the value is still +/- .4V . I'm just thinking would it reverse polarity, but my conclusion is it should not.
Quote from: WaveWatcher on 2012.04.01, 05:11:42
Gibbs,
As an attempt to assist you with your thoughts on the many ways to change the value of the measurements....
If you are familiar with Faraday's sliding bar experiment, look at your latest posted drawing showing the two planes and imagine what it would look like from above.
To create induction or change the measurement you need not only change B. You may also choose to change A by sliding the bar (the Faraday sliding bar experiment) or change the relative angle between the loop or the test probe circuit and the flux.
Thanks WW,
This is the main point of my argument. In-plane and decouple are just the extreme points between 0-90 degrees. They still obey induction law no matter which position they're in. We just need to know how much flux passing through our define loop and use KVL/Faraday on the defined loop.
Quote from: GibbsHelmholtz on 2012.03.31, 23:53:08
Poynt,
The reason I say it like that because the measurement is base on induction and not in plane or decoupled. I agree with WW. The loops now contain 2 planes. One is vertical and the other is horizontal. There is no flux through vertical plane but horizontal, yes. I can't seems to find a better explanation. Maybe someone else can describes it better.
I really don't understand you Gibbs. You appear to be a answering your own question every time you say that there is no flux through the vertical plane. :-\
I don't understand what your confusion is. ???
Also, I don't understand what you mean by this:
Quotethe measurement is base on induction and not in plane or decoupled
That makes no sense to me whatsoever. Throw me a bone here man, I'm trying to understand whats preventing you from understanding this very simple concept. :)
.99
Gibbs,
I'll also emphasize this once more:
I have no interest in any measurement angle other than 0 degrees, and 90 degrees. Anything in between is irrelevant to this experiment.
Quote from: WaveWatcher on 2012.04.01, 04:30:41
...what do you think the measurement would be if the vertical section (normal to the transformer secondary loop) was still vertical but below the measure loop?
Should it still be +.4V, -.4V or another number entirely different?
The answer is +0.4V. The same as it was before. Nothing has changed.
And that is the whole point I am trying to get across to Gibbs; a vertical (decoupled) measurement lead is not influenced by the experiment, therefore it does not alter the true (emf and PD) readings taken from the conductive loop.
If there is no coupling (that's why I call it "decoupled"), how can the leads be getting influenced by the experiment? They can't. If they can't, then placing them above or below the solenoid at 90 degrees makes absolutely no difference in the measurement.
.99
.99,
Your use of the term 'decoupled' is a fair usage for this case. Translation of it for a none native English speaker may be an issue.
When normal to the loop, the metering leads are not inductively coupled to the solenoid coil.
I'll leave the other angles and relative positions alone until I can provide more than an opinion. That information really goes beyond the Lewin demonstration of a non-conservative field, anyway.
Don't worry about it Poynt. :) I'm just getting into everything and try to understand and unify things. I think we'll understand each other someday. Let's just have some fun.
Here I post a picture. The set up is as discussed in this section. The secondary loop is now squared. All measurement are decoupled. Can you guys guess what is the measurement at point 1,2, and 3? If we have disagreement, then we can check it back with experiment.
My opinion is:
123-1=0
123-2=.9v
123-3=0
Quote from: GibbsHelmholtz on 2012.04.01, 13:23:50
Here I post a picture. The set up is as discussed in this section. The secondary loop is now squared. All measurement are decoupled. Can you guys guess what is the measurement at point 1,2, and 3? If we have disagreement, then we can check it back with experiment.
I would, but I don't understand your question, nor how it relates to your diagram. :-\
Quote from: WaveWatcher on 2012.04.01, 15:44:32
My opinion is:
123-1=0
123-2=.9v
123-3=0
you are probably correct if the probes are like Lewin's, but this is vertical/decoupled/normal to the measured loop.
Poynt,
let's say point 123 is the one reference, move the other one around to obtains different measurements.
Refering to:
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=7679;image)
For me Gibbs is right.
One can't simply consider the two diametrically opposed points of connection as being the terminals of a battery, presenting a potential difference. There is no objective potential difference, not even a potential difference. The voltage is observer dependent.
Induced Emf must be evaluated only along a closed path because it is defined only along a closed path. There is no particular voltage between the two diametrically opposed points until we define the path for the measurement. The fact that the wires connecting the voltmeter are along the B field don't disqualify the rule that want that the emf is defined only along a closed circuit.
If we had a resistance in series with the circular loop and want to know the voltage at the resistance terminals, i.e. the voltage felt by the resistance, we would have to put the voltmeter in parallel with the resistance with no flux crossing in between, in order the resistance and the voltmeter have the same viewpoint, being the same and alone "observer" viewing the same closed path which is both the resistance circuit and the measurement circuit.
The voltmeter being the "observer", it can see only the emf along the closed path of its own measurement circuit. The fact that there is no flux crossing the vertical part of the circuit and the vertical wires are along the B field is an irrelevant data because the emf is given only by dφ/dt, φ being the crossing flux through the surface of the whole circuit.
Here we have 2 loops.
1) We see obviously that some flux crosses the surface of the measurement circuit, as represented by the arrows (presuming the voltmeter is connected to the top).
2) We see that for symmetry reasons, the same flux crosses in equal quantity left and right, so from the voltmeter viewpoint, the resultant flux cancels because each half flux crosses each identical half circuit but in reverse directions. The measure will be zero (1/2*dφ/dt-1/2*dφ/dt).
If one connected a voltmeter directly through the diameter, it will also show zero, for the same reason, and in despite its wires would be othogonal to the B field .
The important point is that an induced emf is defined only on a closed path and that we must never consider that the path could be broken down into independent sections, each one being as a partial source of emf, all being in series. Nature doesn't work this way.
A simple experiment can prove what I say: an single electron near a conductor loop carrying a varying current, doesn't feel any force. It don't move. But if you place billions of electrons around the first loop while imposing them a closed path, they will move. Principle of a transformer. I'm the first surprised.
Quote from: GibbsHelmholtz on 2012.04.01, 16:04:13
Poynt,
let's say point 123 is the one reference, move the other one around to obtains different measurements.
There isn't enough information on the diagram to answer the question.
What are the resistor values, and which one is which? What is the length of each wire segment? What is the induced emf?
Quote from: poynt99 on 2012.04.01, 16:15:48
There isn't enough information on the diagram to answer the question.
What are the resistor values, and which one is which? What is the length of each wire segment? What is the induced emf?
All parameter is the same as before. 1V total induced EMF, 100Ohm, 900Ohm, one segment is twice the other which is the diameter of the primary. I don't think it matter which resistor is which in this case.
Gibbs,
In order to ask your question properly, you must state that the length of the resistors are negligible relative to the length of the wire segments. When all the segment lengths are equal, this is not a requirement. In your example, the segment lengths are NOT equal, therefore you MUST assume that the resistors are negligible in comparison.
Let's assume the resistors are 0.01 inches long. Now your question can be answered.
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=7683)
Alright, assuming the scope NEG lead is on terminal "123" (as you stated) and the current CW, the correct answers are as follows:
V1 - V123 = -0.1666V
V2 - V123 = -0.2333V
V3 - V123 = +0.3333V
Note these are calculated values, not opinions. ;)
Quote from: poynt99 on 2012.04.01, 20:15:52
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=7683)
Alright, assuming the scope NEG lead is on terminal "123" (as you stated) and the current CW, the correct answers are as follows:
V1 - V123 = -0.1666V
V2 - V123 = -0.2333V
V3 - V123 = +0.3333V
Note these are calculated values, not opinions. ;)
O0
So now you are the one saying Ohm's law doesn't always apply? ;)
Quote from: GibbsHelmholtz on 2012.04.01, 16:04:13
you are probably correct if the probes are like Lewin's, but this is vertical/decoupled/normal to the measured loop.
Ah! Ok.
Then I will change my .9V to .4V ;)
Quote from: WaveWatcher on 2012.04.01, 20:45:29
So now you are the one saying Ohm's law doesn't always apply? ;)
Nope. You're not paying careful enough attention to the diagrams WW.
Look one post above this one of yours where you object to the indicated readings:
http://www.overunityresearch.com/index.php?topic=739.msg21215#msg21215
You will see the diagrams in question above this post of yours.
You state that all 3 voltages will read 0.4V. That is incorrect.
What you are apprently NOT paying attention to on my diagrams, and Gibbs' diagram, is where the probes are connected. Are they directly across the resistors as in my diagrams "V11'" and "V99'"?, or is there a wire segment in between the measurement probes as shown in my diagram "Vda" and Gibbs' "de.jpg" diagram?
Hopefully, once you realize this oversight, you will see that Ohm's law still holds in this case.
Gibbs has reminded you, but in case you missed it, the measurement probes in his diagram are decoupled, even though they don't appear to be.
.99
PS. Ohm's law can
appear to be broken when making certain measurements, such as the one shown here in the diagram:
http://www.overunityresearch.com/index.php?topic=739.msg21406#msg21406
But this can only happen when the measurement is made in-plane and as shown.
In light of the way things have gone so far for some of the participants in this thread, I thought it fitting to re-post these words of wisdom:
Quote from: WaveWatcher on 2012.02.28, 10:12:01
The experiment tells more about the experimenter than the physics. (Not pointed only at .99)
I couldn't agree more. O0
Quote from: poynt99 on 2012.04.01, 20:15:52
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=7683)
Alright, assuming the scope NEG lead is on terminal "123" (as you stated) and the current CW, the correct answers are as follows:
V1 - V123 = -0.1666V
V2 - V123 = -0.2333V
V3 - V123 = +0.3333V
Note these are calculated values, not opinions. ;)
I'm still trying to figure out how you obtains these values lol. I think I see it. You divided 1V into the total segment, and gives voltage proportional to the length of the measurement.
I have to agree with WW that the value at V2 is .4V . We'll ignore all polarities.
My answer is
V1 = .25V
V2 = .4V
V3 = .25V
Gibbs,
Had you drawn a perfect square, your values would be correct.
What you have there is a 2:1 rectangle.
Explain why you think that the two would produce equal results.
Quote from: poynt99 on 2012.04.01, 23:13:24
:D
LOL
My opinion is we have a disagreement. ;D
I based my calculation on the angles from primary rather than length of the secondary. As shown in the picture below, point 1 and 3 have equal amount of primary area radiate out. That is both have 1/4 flux. Each 1/4 flux represent .25V . At point 2, it's 1/2 flux, which is .5V, but if the resistor takes away .1 or .9V, it's still .4V .
If you were to measure 123- 4 or 5. You can use the amount of flux based on its angles. Point 4 is .125V and point 5 is .166V.
Change your diagram to a square.
Now what do you think the voltages will be?
Quote from: poynt99 on 2012.04.02, 00:05:43
Change your diagram to a square.
Now what do you think the voltages will be?
The exact same as double squares. There will be some loss as we get further, but it's basically the same.
So the debate appears to about the two segments of wire; 1-123 and 3-123.
You say that the voltage measured across each will be the same magnitude, while I say they will not. Correct?
Quote from: poynt99 on 2012.04.02, 00:27:02
So the debate appears to about the two segments of wire; 1-123 and 3-123.
You say that the voltage measured across each will be the same magnitude, while I say they will not. Correct?
yes
Aha!
I agree with Gibbs that they are the same magnitude but I say zero volt drop on the first and third measurements because the resistance of the wire between the two measuring points for each measurement is effectively zero Ohms.
The fact that those sections also have inductance is no matter since there is no actual voltage drop across an inductor. --- I think this is another example of the magnetic field line scenario - where it is said one way to allow for understanding but it isn't that way at all.
>>Edit
Quote from: exnihiloest on 2012.04.01, 16:06:23
The important point is that an induced emf is defined only on a closed path and that we must never consider that the path could be broken down into independent sections, each one being as a partial source of emf, all being in series. Nature doesn't work this way.
I agree with Ex on his statement and it applies to my 'choice' of choosing zero volts on the first and third measurements.
In fact... the meter circuit must exist decoupled or not. The actual voltage in existence for the first and third measurements will be zero volts. The 'measured' voltage will NOT be zero volts because the attachment of the meter circuit changes the area of the measured loop.
Yes, I understand these things but tend to jump from actual vs. measured so much I confuse myself.
When I said change my opinion to .4V I was referring to the second value, not all three.
Gibbs,
.99 has a good point ( :D). But I see how the area of the loop is a concern while I understand the area of the flux within the loop is the important factor.
All very interesting :)
--
What a refresher from today! I just spent the entire day on the roof of my house.
Hehehe .... That reminds me.... When I was a kid I worked as a roofer to put myself through school. The boss fired me because he said I wasn't smart enough to be a roofer. So, I took engineering jobs after that ;D
Well,
I got out the rig again and built/tested Gibbs' rectangular loop.
Gibbs & WaveWatcher, you are incorrect with your assertion that both measurements are equal; they are not. WaveWatcher you are incorrect as neither measurement is 0V (you really need to get past that silly notion of yours ;) ).
However, I was not correct either, and now seeing what it actually is, it does makes sense.
Measurement V3-V123 is precisely 1/2 the voltage of V1-V123.
Possible explanation: The V3-V123 measurement has the same amount of flux, but through twice the surface area compared to the V1-V123 measurement?
I'll measure and post all the actual voltages from my experiment in the next day or two.
Gibbs,
Your value of .25 for measurements 1 & 3 indicate you are using the sector area of the circular changing flux?
If so, you do know that would require portions of the metering loop to not be normal to the loop but normal to the field lines?
Even so, I would think your measurement would show .5V as opposed to .25V. Why? Because the rest of the secondary is also being measured by the same meter. The voltage of that remaining section would be .75V and opposite polarity to the section you intended to measure (.25V) so the actual displayed voltage would e .5V.
Since I would have figured all of the meter circuit being normal to the secondary loop.... I would calculate using the area of the segment, not the sector.
That would mean the calculated induced voltage of that segment area would be .090845V. That leaves .909155V for the remainder of the secondary loop also being measured with reverse polarity.
.909155 - .090845 = .81831V to be displayed on the meter. Now, move one meter lead back to where we originally decided the measured voltage would be .4V.
We now know another way to find the measured value of .4V :D
It was easier seeing it as angles in my head :o
Quote from: poynt99 on 2012.04.02, 02:20:21
Possible explanation: The V3-V123 measurement has the same amount of flux, but through twice the surface area compared to the V1-V123 measurement?
Very Plausible!
Wouldn't that be an odd thing to miss in basic induction formulae? I don't recall if it is there or not?
-----
Never mind.
It is there. Increasing the area of the loop would decrease the coupling coefficient - perhaps to half in this case?
Quote from: poynt99 on 2012.04.02, 02:20:21
Measurement V3-V123 is precisely 1/2 the voltage of V1-V123.
Possible explanation: The V3-V123 measurement has the same amount of flux, but through twice the surface area compared to the V1-V123 measurement?
The explanation only works if WW is correct that it looses coupling as the area goes up. If you make Lewin's secondary bigger, it would still induce total 1V, right? The fact that it
precisely 1/2 shows that there is an exact mathematics relationship in my opinion. I calculate mine based on 3 dimensional visualization of field lines. Maybe my 3D is not precise, I'll think about it.
Quote from: GibbsHelmholtz on 2012.04.02, 03:02:32
The explanation only works if WW is correct that it looses coupling as the area goes up. If you make Lewin's secondary bigger, it would still induce total 1V, right? The fact that it precisely 1/2 shows that there is an exact mathematics relationship in my opinion. I calculate mine based on 3 dimensional visualization of field lines. Maybe my 3D is not precise, I'll think about it.
Gibbs,
I'm sure I'm correct about the coupling coefficient. That alone should not change the peak voltage induced. I suspect it only makes the peak 1V for the complete loop more difficult to display due to the weaker coupling and subsequent stronger effect of self-loading (closed loop).
Yes, making Lewin's loop larger should still create the same induced 1V.
The same goes for primary not centered in the secondary loop. There should be no change in induced voltage.
Quote from: WaveWatcher on 2012.04.02, 09:35:23
Yes, making Lewin's loop larger should still create the same induced 1V.
The same goes for primary not centered in the secondary loop. There should be no change in induced voltage.
I ignored my gut instinct when I came up with my numbers for Gibbs' loop. I should have went with it.
WW, I am 99% certain that the induced emf will NOT be 1V with a larger loop. I will prove that tonight if I have the time.
Quote from: WaveWatcher on 2012.04.02, 09:35:23
Gibbs,
I'm sure I'm correct about the coupling coefficient. That alone should not change the peak voltage induced. I suspect it only makes the peak 1V for the complete loop more difficult to display due to the weaker coupling and subsequent stronger effect of self-loading (closed loop).
Yes, making Lewin's loop larger should still create the same induced 1V.
The same goes for primary not centered in the secondary loop. There should be no change in induced voltage.
I think you're right about coupling losses and it will change the peak voltage. This is the issue Ex. and Poynt talked about magnetic field going back in the same loop making it have less inductance.
I've revised my 3D. As the field line move outward from the primary, it also has a downward component. These downward field lines failed to pass the loop surface area reducing the voltage of the total loop and the measured segment.
Quote from: poynt99 on 2012.04.02, 12:12:02
I ignored my gut instinct when I came up with my numbers for Gibbs' loop. I should have went with it.
WW, I am 99% certain that the induced emf will NOT be 1V with a larger loop. I will prove that tonight if I have the time.
Now THAT would be interesting :o
Could it be that the 1:9 ratio changes with the amount of power induced into the secondary? ^-^
Strange things may happen on loosely coupled transformers O0
Quote from: GibbsHelmholtz on 2012.04.02, 12:53:46
As the field line move outward from the primary, it also has a downward component. These downward field lines failed to pass the loop surface area reducing the voltage of the total loop and the measured segment.
So, your thoughts are that each loop sector may not see the same polarity or intensity of the changing flux at each instant during the change?
This can be true in the real world but only on very loosely coupled arrangements like antenna elements, air core RF coupling transformers, etc.
I think many conventional transformers have the primary wound over the secondary or the magnetic loops are tightly closed to avoid this when it is a problem.
Quote from: WaveWatcher on 2012.04.02, 13:02:57
...
I think many conventional transformers have the primary wound over the secondary or the magnetic loops are tightly closed to avoid this when it is a problem.
Right, but now we stretch the secondary out so it departs from being tight. How much did we loose? I think it's proportional to the distance outward. Maybe this could explain why magnetic strength is inverse cubed. Radiate outward is inverse squared and downward component makes it inverse cubed.
Quote from: GibbsHelmholtz on 2012.04.02, 13:10:52
Radiate outward is inverse squared and downward component makes it inverse cubed.
Interesting. I've never thought of it that way.
Quote from: WaveWatcher on 2012.04.02, 12:55:44
Now THAT would be interesting :o
Okee.
I performed an experiment to test my hypothesis that the induced emf will decrease as the diameter of the circular loop increases. Here are the test results:
- loop diameter ~ solenoid diameter: emf= 400mV
- loop diameter ~ 2x solenoid diameter: emf=300mV
Summary: The induced emf decreases as a function of loop radius. In my test, doubling the loop diameter resulted in a 25% decrease in emf.
The equations out there support this result.
That explains the results as measured on Gibbs' rectangular loop. The induced emf in that case decreased even more because of the rectangular shape of the loop. I know you guys are going to ask why >:-)
O0
Bravo Zulu Poynt,
Your "take one for the team" measuring things is appreciated.
So why does the induced emf decreased even more? lol
Quote from: GibbsHelmholtz on 2012.04.03, 02:10:12
So why does the induced emf decreased even more? lol
You're not going to give it a try? How about an educated guess?
Well, the area of double diameter loop is sure greater than double square. Unless the loss function is non linear, the square should have less.
Quote from: poynt99 on 2012.04.03, 01:51:45
...
I performed an experiment to test my hypothesis that the induced emf will decrease as the diameter of the circular loop increases.
...
Not a question of principle. The induced emf is the same but the limits of a real setup disturb the measurement. For example the equivalent circuit of a small coil probe weakly coupled to the loop (due to the gap of diameters) shows an inductance in series, which represents the not coupled inductance part of the loop. With the capacitive coupling between turns, the series resistances, the capacitive impedance of the probes of the voltmeter or oscilloscope, all these effects decreases the measured voltage.
Quote from: exnihiloest on 2012.04.05, 07:29:04
Not a question of principle. The induced emf is the same but the limits of a real setup disturb the measurement.
I agree with the above statement from Ex.
I think the change in the coupling coefficient explains the lower energy transfer ability of the loop because the loop is now more loosely coupled.
The change in coupling coefficient should only change the efficiency of energy transfer between primary and secondary, not the emf.
In practice, we should find that relocating the primary anywhere within the secondary or changing the relative areas has no effect upon the induced emf until other factors become more important in the experiment.
Any variation of the original experiment, like this, simply opens a can of worms. Normally, this is a great thing in scientific experiments. You must tag and classify each worm :D
Quote from: WaveWatcher on 2012.04.05, 09:25:08
I agree with the above statement from Ex.
I'm not really certain I understand 100% what Ex is saying. However, I will say that my method of measurement has no serious affect on the theoretical induced emf. And the equations clearly indicate that the induced emf IS different at differing diameters.
Quote
I think the change in the coupling coefficient explains the lower energy transfer ability of the loop because the loop is now more loosely coupled.
I don't think the physical coupling between the primary and secondary (outside the solenoid) has any real affect on the induced emf. We know that the magnetic field outside the solenoid is practically zero.
Quote
In practice, we should find that relocating the primary anywhere within the secondary
Yes, for the most part this is the case.
Quote
or changing the relative areas has no effect upon the induced emf until other factors become more important in the experiment.
Evidently, not. And the equations disagree as well.
Quote
Any variation of the original experiment, like this, simply opens a can of worms.
Not at all. The results would simply be shifted down in relative amplitude. Everything else is the same.
Poynt,
I'm sure you know your equations but I must ask, which ones?
"I don't think the physical coupling between the primary and secondary (outside the solenoid) has any real affect on the induced emf."
Agreed. That isn't what I meant. Up to the point where other outside factors have an overriding effect, the total loop area change or coupling will not change the induced electric field. The amount of bi-directional power that may be transferred via that coupling does change.
A total loop area change implies a change in the path outside the solenoid. A larger area implies a net path farther from the solenoid, correct?
Therefore, an increase in area will decrease the induced emf.
I will post the equation later. I have to fight my way to work through 6 inches of wet snow that fell last night (and still falling). >:(
Quote from: poynt99 on 2012.04.05, 11:31:06
A total loop area change implies a change in the path outside the solenoid. A larger area implies a net path farther from the solenoid, correct?
Therefore, an increase in area will decrease the induced emf.
I will post the equation later. I have to fight my way to work through 6 inches of wet snow that fell last night (and still falling). >:(
Are you referring to the circumference of the loop? If so I think I know what the cause of the result.
Hope you're on time to work.
Yes,
Area and radius of the loop.
What's your theory?
This is what I see.
As the area approaches infinity, the voltage induced is 0.
What are we looking at there Gibbs?
Quote from: poynt99 on 2012.04.05, 20:54:30
What are we looking at there Gibbs?
Well, if the field lines go up the primary and come back though the loop, the total flux would be less, hence the induced voltage. I could have overestimate this effect.
What are the red rings in your diagram representing?
Quote from: poynt99 on 2012.04.05, 22:37:48
What are the red rings in your diagram representing?
Uh... the secondary loops.
OK, I see where you're going, and that I'll need to explain things...tomorrow.
Going to the last hockey game of the year for the Oilers tonight with me pops. O0
Quote from: poynt99 on 2012.04.05, 23:04:27
OK, I see where you're going, and that I'll need to explain things...tomorrow.
Going to the last hockey game of the year for the Oilers tonight with me pops. O0
Nothing better than that!
Quote from: GibbsHelmholtz on 2012.04.05, 22:47:36
Uh... the secondary loops.
Separate but concentric secondary loops :)
Faraday via Maxwell: "The line integral of the electric field around a closed loop is equal to the negative of the rate of change of
the magnetic flux through the area enclosed by the loop."
With concentric secondaries the rate of change is equal on all sectors. Gibbs's red circles will each have the same induced emf because all sectors of these loops are normal to the changing magnetic flux.
On the rectangular secondary there is a decrease in induced emf, not because the area of the loop has changed but because not all of the secondary conductor is normal to the changing magnetic flux. If you distort that rectangular secondary into a nice round loop, the same amount of emf will be induced as each of the above two red circles.
Here is where we get into trouble by considering 'potential differences' between any two points around the secondary'.
AND - Where we run into the contradiction :D
That is my take on it!
Quote from: poynt99 on 2012.04.05, 11:10:37
I'm not really certain I understand 100% what Ex is saying. However, I will say that my method of measurement has no serious affect on the theoretical induced emf. And the equations clearly indicate that the induced emf IS different at differing diameters.
...
What equations?
The induced emf depends only on dphi/dt through the circuit surface, it doesn't depend on the surface, therefore it doesn't depend on the diameter.
Show us an experimental setup giving a result depending on the diameter, and I could say where is the bias. For example if you use a big loop around a solenoid, of course the emf will depend on the diameter if you don't increase simultaneously the length of the solenoid.
Quote from: exnihiloest on 2012.04.06, 14:52:40
What equations?
The induced emf depends only on dphi/dt through the circuit surface, it doesn't depend on the surface, therefore it doesn't depend on the diameter.
Show us an experimental setup giving a result depending on the diameter, and I could say where is the bias. For example if you use a big loop around a solenoid, of course the emf will depend on the diameter if you don't increase simultaneously the length of the solenoid.
Even with an infinitely-long solenoid, the induced emf decreases as a function of increasing loop radius.
I will elaborate shortly.
Calm down guys,
It's a good time to review your thinking. What other possibilities did you or Poynt missed. I have one other possibility in my mind. Either way, I'm excited to hear.
Please, speak your mind Gibbs. What possibility do you have in mind....and in what regard?
Quote from: poynt99 on 2012.04.07, 01:29:25
Please, speak your mind Gibbs. What possibility do you have in mind....and in what regard?
I don't think it is right, that's why I rather hear you. We may have measured a fixed resistance, that's all.
Not quite sure what you mean Gibbs. We use fixed resistors, yes.
Why could a fixed resistance be a problem here?
Let's say we have 1000 Ohms . The total induced voltage is 1 V , so .1 ma go through. If we measured 100Ohms, the voltage is .1V . If we makes the loop larger, more wire resistance is added. The total loop amp go down. If we measured 100Ohms as before, the voltage would indicate less. However, this scenario doesn't fit well.
I just thought of another possibility and the calculation matches. The pic below shown how you made your measurement.
I've have found that my earlier statement about sections of the rectangular loop effecting induced emf, while still true, isn't enough to explain the observations stated by others.
Rather, the act of measuring 123-3, just like the measurements on the circular secondary, splits the surface of induction into two surfaces. These two induction areas will be of opposite polarity. The induced emf into the latest secondary should remain 1V. The problem is that the .4V (123-2) is only .4V when it is being measured.
The act of measuring between any two points on similar secondary loops always has an effect upon the reading. Even in the described de-coupled state the metering circuits acts to split the induction area into two.
Gibbs,
The wire resistance is miniscule in comparison to the 1000 Ohms total we have. Even moving to a loop of twice the radius (2x the length of wire) will not skew the results by any significant amount.
With reference to your diagram, I didn't measure the emf that way.
Edit: Corrected "100" Ohms to "1000" Ohms.
:o
WW, I'm not sure what you're getting at. But what I could understand (I think), doesn't sound quite right.
Stay tuned, the explanation for all is forthcoming.
.99
Staying tuned ;D
Basically, I'm still agreeing with Ex.
Were All Tunered 8)
Glad we're all "tuned-up". ;D
Well, it seems I am going to have to do another experiment with a much longer solenoid to prove my hypothesis.
It was all going well until I became aware that my hypothesis is in conflict with Faraday's law; emf is predicted to be the same, regardless of loop radius. I find that difficult to swallow, since I know that outside the solenoid the electric field drops off as a function of 1/r. And we know that the electric field is what is responsible for the induced emf. The emf being equal to the line integral of the electric field around the conductive loop of interest.
Something doesn't add up. In fact, it would seem that the accepted theory is in conflict with itself. C.C
Don't take this all by yourself Poynt. Let us in... We can help.
You did in fact measured a drop in voltage and said to be 1/r. Can you show me your measured config. for that? I think Ex was on the right track but I also sense another element that can effect induced voltage, but I can't be sure unless I have more information.
Coming up soon Gibbs. I'm working on a detailed diagram which will help explain my hypothesis.
I understand now after studying my diagram. It's a simple relationship.
Yes the E field weakens as a function of 1/r from the solenoid, but that is to a single point at the given radius "r" from the axis. As r increases, so does the circumference at that give radius, so when we do the line integral around that larger circumference, the result is the same net E field, which should produce the same net emf.
However, things don't look so well for that rectangular loop, as you will see shortly. I'm fairly certain this explains the weaker than expected emf readings.
.99
Poynt,
From the perspective of the solenoid even the rectangular loop is still circular. The result for the entire loop should still be 1V.
Your thoughts on increased radius being compensated by increased circumference is very logical. That thought has made perfect sense to me in the past but I found it to be the wrong idea.
The only idea that seems to work is consideration of the loop surface area. Not the loop's total surface area... Rather, the loop's surface area covered by the magnetic field only from the inside of the solenoid. So, as long as the secondary is a simple loop (no resistors, no measurements of voltage across segments), the induced voltage will remain at 1V because the surface area effected by the solenoid doesn't change.
I don't think the reduced measured voltage has anything to do with the shape or size of the measured loop having a change in induction. It is the result of the size and shape of the secondary loop causing confusion in locating previously measured points.
Unfortunately, none of the above will make sense unless you understand that any attempt at measurement changes the one secondary loop into two secondary loops.
For my argument, I offer the following I posted in another forum some time ago:
Quote
When Lewin made a statement about the metering circuit resistance needing to be a certain level......
It is clear to me he only suggests a high enough resistance so the metering circuit doesn't drag the measured voltage down (overload the loop).
Except for the possibility of a very low resistance meter overloading a poorly coupled secondary, the resistance of the meter has nothing to do with the induced emf on the loop.
If the resistors were of equal value... A to D would have been .5V.
Why?
Because the act of connecting a meter loop across the diameter creates two loops ( with the meter circuit normal to the loop and internal solenoid magnetic field) .
The emf induced into EACH loop would be 1V. BUT, the meter now being connected to BOTH loops causes some confusion. No matter where you connect the meter, you will create a second loop. Doing so, ALWAYS means the meter is measuring both loops at the same time.
The relative connection of the meter (polarity of the meter) is reversed for one loop relative to the other loop.
Therefore, the meter will display a summation of the two applied emf's - .5V .
The whole Lewin exercise was simply a good demonstration of the prime component of non-conservative fields - Path Dependence.
Quote from: WaveWatcher on 2012.04.08, 10:38:57
Poynt,
From the perspective of the solenoid even the rectangular loop is still circular. The result for the entire loop should still be 1V.
Perhaps, but I am not yet convinced. The E field is a circular vector. Any loop shape other than circular is non-ideal.
Quote
Your thoughts on increased radius being compensated by increased circumference is very logical. That thought has made perfect sense to me in the past but I found it to be the wrong idea.
The math supports the idea, and it makes sense to me. You will see the equations in my next posts.
Quote
The only idea that seems to work is consideration of the loop surface area. Not the loop's total surface area... Rather, the loop's surface area covered by the magnetic field only from the inside of the solenoid. So, as long as the secondary is a simple loop (no resistors, no measurements of voltage across segments), the induced voltage will remain at 1V because the surface area effected by the solenoid doesn't change.
The induced voltage is not affected by series resistors nor decoupled measurement leads. That's been proven.
Quote
I don't think the reduced measured voltage has anything to do with the shape or size of the measured loop having a change in induction. It is the result of the size and shape of the secondary loop causing confusion in locating previously measured points.
I disagree. Again, a non-circular loop isn't optimal for induction. My diagram will show why.
Quote
Unfortunately, none of the above will make sense unless you understand that any attempt at measurement changes the one secondary loop into two secondary loops.
I don't understand this concept. Can you think of a test to prove this? The tests I've done to specifically address this issue clearly indicate that the measurement loop is not affected by the experiment. If there is no induction in the measurement loop (which I've proven for myself), and the measurement loop is high impedance (10M), how can it affect the induction loop?
Attached is the diagram illustrating a top-down view of the gradient E-field, and other salient points of interest.
I plan on doing some tests with the rectangular loop. One test will be to see if the measured emf between points 0a and 0b is equivalent to that measured between points 0b and 1a. According to the accepted theory, they should be (or not?), but I want to see it myself.
The white circle in the diagram denotes the solenoid perimeter. You see also that the E-field at that area is black (the most intense) and how it diminishes in intensity in both directions away from, and towards the axis.
The red circle is a sample point for calculating the E-field intensity, and for noting the orientation of the E-field wrt the rectangular loop. Towards the right side, it's clear that the diminishing E-field is not inducing much into the loop, as it is almost normal to the loop in that area. It is not until the extreme right of the loop that the E-field is inducing emf, and at that point the E-field is already diminished to 1/2 intensity.
Notice the equation for emf assumes a circular loop.
.99
First, you made me feel bad on art work. lol
Second, it's a good test to perform. I would add suggestion that we vary the de-coupled length to see what effect it may have on measurement.
The proposed test may be wrong.
Each measurement should encompass the same angle rather than the same length of wire, correct?
Quote from: poynt99 on 2012.04.08, 13:58:13
The proposed test may be wrong.
Each measurement should encompass the same angle rather than the same length of wire, correct?
The proposed test can't be wrong. It either confirms or denies the theory.
Edit: When I say vary the de-coupled length, I meant the length connecting to probes normal to the loop. I know it's very tall already, but the relative tall/shortness maybe a factor affecting measurement.
Quote from: poynt99 on 2012.04.08, 12:50:14
The induced voltage is not affected by series resistors nor decoupled measurement leads. That's been proven.
Nor is it affected by the resistances which are part of the loop.
The measurement leads (the meter circuit) presents a change in loop area, for one side of the -now- double loop.
Quote
I disagree. Again, a non-circular loop isn't optimal for induction. My diagram will show why.
Here is where our disagreement of solenoid position within the loop affecting or not affecting induced voltage comes into play.
In practice, performing current monitoring of a conductor via a current transformer does not change whether the cable is centered within the CT or not. CT's come in round or rectangular. You can attach a rectangular CT to a cable with the cable at one extreme inside corner. It does not have an effect upon the induced voltage. I have done this on numerous occasions. There has never been any value in centering the monitored cable within a surrounding current sensor for purposes of accuracy.
Quote
I don't understand this concept. Can you think of a test to prove this? The tests I've done to specifically address this issue clearly indicate that the measurement loop is not affected by the experiment. If there is no induction in the measurement loop (which I've proven for myself), and the measurement loop is high impedance (10M), how can it affect the induction loop?
There would be no induction within the meter circuit (as you have performed). The loop impedance has no effect upon the measured induced emf unless the secondary load is too high. Even then, the measured induced voltage will remain at 1V until Lenz starts to affect the amount of power being transformed.
The loop impedance isn't part of the formulas for induction. Even if the meter circuit was zero Ohms or infinite resistance, the emf induced for that loop will be 1V. The meter may show any value but the accuracy or ability to show the voltage decreases as resistance of the meter loop drops.
As testing should conclude, even when there is a diametrically connected short, the total induced voltage around the outside loop will still be 1V.
It should also be found that the induced emf for the loop circuit including the diameter short (say, on the left of the diameter short) will be 1V. The same should be found for the complete half circle on the right of the diameter short.
The confusing part is that the diameter short now has two induced currents opposing each other.
I'll have to think about a test to prove the idea. This becomes a relativity issue.
Quote from: GibbsHelmholtz on 2012.04.08, 14:07:23
The proposed test can't be wrong. It either confirms or denies the theory.
Edit: When I say vary the de-coupled length, I meant the length connecting to probes normal to the loop. I know it's very tall already, but the relative tall/shortness maybe a factor affecting measurement.
Gibbs,
The height should make no difference as long as wire R/L/C doesn't skew the results. You must look at the test from above, in your mind's eye.
The metering circuit, with leads, simply crosses from one point of the measured loop to another point. If you could place the scope within the measured loop with sensing leads straight to each of the two previous points (and the scope had no effect upon the magnetic fields), the same values would be recorded.
BTW:
These aren't my theories. The above was part of my education relating to transformer theory.
Quote from: poynt99 on 2012.04.08, 00:48:55
...
I know that outside the solenoid the electric field drops off as a function of 1/r. And we know that the electric field is what is responsible for the induced emf. The emf being equal to the line integral of the electric field around the conductive loop of interest.
...
Only an electric field deriving from a potential drops off as a function of 1/r, not an induced emf.
An induced electric field is conservative because ∇xE = -∂B/∂t and B is conservative.
From the flux variation ∮E.dl=-dΦ/dt and Φ=∮B.dA. B is conservative and null outside the solenoid (provided that it is long enough), so Φ is the same whatever the surface A of the loop around the solenoid.
All equations gives the same result. The first one straightforwardly explains why the emf doesn't depend on the loop diameter, provided that the same flux crosses the surface (which is experimentally not obvious to realize).
Note that the above applies only in the quasi-static fields approximation: no radiation by em waves, i.e. the circuit size must remain negligible compared to the wave length of the involved signals.
Quote from: WaveWatcher on 2012.04.08, 14:59:07
Gibbs,
The height should make no difference as long as wire R/L/C doesn't skew the results. You must look at the test from above, in your mind's eye.
The metering circuit, with leads, simply crosses from one point of the measured loop to another point. If you could place the scope within the measured loop with sensing leads straight to each of the two previous points (and the scope had no effect upon the magnetic fields), the same values would be recorded.
BTW:
These aren't my theories. The above was part of my education relating to transformer theory.
WW,
Maybe it's true. However, imagine my rectangular loop voltage V3 (V123-3), if we reduce the height to 0, voltage would be zero.
Quote from: GibbsHelmholtz on 2012.04.08, 14:07:23
The proposed test can't be wrong. It either confirms or denies the theory.
Sure it can be. I believe measurements with comparable angle are what's called for. But I will do both.
Quote from: WaveWatcher on 2012.04.08, 14:53:02
Nor is it affected by the resistances which are part of the loop.
Implied, agreed.
Quote
The measurement leads (the meter circuit) presents a change in loop area, for one side of the -now- double loop.
A change in loop area with respect to what?
Quote
In practice, performing current monitoring of a conductor via a current transformer does not change whether the cable is centered within the CT or not. CT's come in round or rectangular. You can attach a rectangular CT to a cable with the cable at one extreme inside corner. It does not have an effect upon the induced voltage. I have done this on numerous occasions. There has never been any value in centering the monitored cable within a surrounding current sensor for purposes of accuracy.
I know this.
Quote
As testing should conclude, even when there is a diametrically connected short, the total induced voltage around the outside loop will still be 1V.
It should also be found that the induced emf for the loop circuit including the diameter short (say, on the left of the diameter short) will be 1V. The same should be found for the complete half circle on the right of the diameter short.
The confusing part is that the diameter short now has two induced currents opposing each other.
I am interested only in the Lewin experiment, and the test Gibbs proposed. What you have proposed, I have not done, nor is it all that relevant to these tests.
Quote
I'll have to think about a test to prove the idea. This becomes a relativity issue.
I am asking for a proposed test to prove your notion that introduction of the decoupled meter leads somehow skews the results, or enters into the equation.
Quote from: exnihiloest on 2012.04.08, 15:27:49
Only an electric field deriving from a potential drops off as a function of 1/r, not an induced emf.
Not sure what you mean Ex. Let me restate it again:
The induced electric field at a point outside the solenoid diminishes as a function of 1/r.
The line integral of the E-field (which equals induced emf) at radius r does not diminish as a function of 1/r.
Better? ;)
Quote from: poynt99 on 2012.04.08, 16:18:38
A change in loop area with respect to what?
The other loop now created by insertion of the metering circuit. Please continue with the route you are taking.
Quote
I am interested only in the Lewin experiment, and the test Gibbs proposed. What you have proposed, I have not done, nor is it all that relevant to these tests.
There should be a point where you see my statement is at the heart of the Lewin experiment.
Quote
I am asking for a proposed test to prove your notion that introduction of the decoupled meter leads somehow skews the results, or enters into the equation.
Whatever I suggest must first show that insertion of the meter circuit does indeed segment the secondary loop into two loops. Since an additional meter would do exactly the same - again - ..... Well, I don't think it is that simple.
Not that the experiment is beyond you. Just that the experiment is beyond me, right now.
Quote from: WaveWatcher on 2012.04.08, 17:10:59
The other loop now created by insertion of the metering circuit. Please continue with the route you are taking.
I don't understand what affect that has on anything? We are adding a non-loading (essentially) non-inducing loop of wire that sits in a plane parallel to the changing flux. Two ends of that loop happen to be connected somewhere to two points on the loop that is situated normal to the changing flux. What impact does the measurement loop have on anything at all, other than the miniscule differential voltage it presents to its high impedance input amplifier/buffer at the other end of the wire loop?
Surely you're not suggesting that the tiny current in the measurement leads is skewing the experiment? Let's look at an example: Say we are measuring 400mV across two points on the loop. Let's also say the input impedance of the scope is 10M. This represents a current in the measurement leads and the loop of 40nA. This is 0.004% of the loop current of 1mA. I don't think 0.004% is anything to worry about.
Quote
There should be a point where you see my statement is at the heart of the Lewin experiment.
The Lewin experiment has been explained as far as I am concerned. The flaws in the presentation have been brought to light, and the resulting erroneous conclusions corrected. Have you not been paying attention? >:-)
Quote
Whatever I suggest must first show that insertion of the meter circuit does indeed segment the secondary loop into two loops. Since an additional meter would do exactly the same - again - ..... Well, I don't think it is that simple.
Again, so what if there is a second loop which is parallel to the changing flux and connected to the first? You mention this, but do not explain what impact you believe it has on anything. Apologies if I've missed it.
Here is an updated diagram with two points added; "00" and "11".
As promised, here are the 3 measured voltages around the rectangular loop:
V1-0 = -0.09V
V2-0 = +0.14V
V3-0 = +0.075V
Back to my proposed tests, here are the results:
V1-0 = -0.09V |Same angle (90º) &
V11-00 = -0.09V |same wire length for both measurements
V0b-0a = -0.03V |Probably 30º angle
V1a-V0b = -0.06V |45º angle, same wire length for both measurements
Conclusion? Appears it IS the angle which is relevant, vs. wire length. In the first pair of measurements, the measured emfs were the same, despite one measurement taken across a 90º corner (V1-0). It also appears that the loop shape doesn't matter for the same induce emf. I won't be able to prove this beyond a doubt until I make a solenoid with a much higher length to diameter ratio.
I also re-ran the emf test (slightly differently)with loops of R and 2R, and the same result: with R, emf= 0.4V, with 2R, emf=0.3V. As Ex said, this could be due to the solenoid length. If it was much longer, the results might be more consistent. Will have to try it with the new solenoid.
I don't know what to say. Thank you.
Gibbs
.99,
It doesn't matter if your scope has a 100M input impedance the same emf is induced into the loop and 'any' connection of two points to that loop segments it into more than one loop.
I t has been a very long weekend. I'll check back as soon as I can.
Good Experimenting to all!
Quote from: WaveWatcher on 2012.04.09, 00:13:33
.99,
It doesn't matter if your scope has a 100M input impedance the same emf is induced into the loop
Of course the same emf is induced. Where did I ever state that the measurement loop would cause otherwise? And you are referring to emf in the secondary loop I trust, NOT the measurement loop?
Quote
and 'any' connection of two points to that loop segments it into more than one loop.
Again, so what? Please elaborate on what you believe is the significance of this.
Quote from: WaveWatcher on 2012.04.09, 00:13:33
It doesn't matter if your scope has a 100M input impedance the same emf is induced into the loop and 'any' connection of two points to that loop segments it into more than one loop.
I'm going to take a guess at what you are trying to indirectly say:
I think you are implying that the induced emf is always the same, and when it is measured as otherwise (as I have going from R to 2R), the change is due to the use of a decoupled measurement device and its leads being connected to the loop. Is that a fair conjecture?
If I've guessed right, then I can't support that notion. The measurement device and its leads and connections are having absolutely no effect on the measurements, as proven on the bench.
What I believe IS making the difference in the measurement, is the non-ideal solenoid in terms of its length to diameter ratio. My current solenoid has a Length to Diameter ratio of about 3:1 (as does Lewin's). For loops that hug the solenoid (R), this is not a problem; the induced emf is as it should be. For loops in the 2R area and larger, this is most likely a problem. The lower emf in the larger loops has nothing to do with the measurement leads at all. It simply can't. The solenoid
really is inducing a lower emf in the larger loops, because the loop radius is relatively large compared to the solenoid length.
A hypothesis of what happens when rloop > rsolenoid, and the ratio Lsolenoid:rsolenoid is relatively low:
In this case I believe the B-field in the region of the larger loops is non-zero. It not only has magnitude, but the opposite direction as it curls back to the other end of the solenoid. A such, a weaker and opposite electric field is produced near the loop and this cancels the effect of the primary electric field to a degree, hence the lower than expected induced emf.
Quote from: poynt99 on 2012.04.09, 13:42:12
Of course the same emf is induced. Where did I ever state that the measurement loop would cause otherwise? And you are referring to emf in the secondary loop I trust, NOT the measurement loop?
Yes, I try to make it a point to call the secondary 'loop' and the metering circuit - the metering circuit.
Quote
Again, so what? Please elaborate on what you believe is the significance of this.
Regardless if the metering circuit has emf induced into it or not the metering circuit either segments or sectors the loop (if on the inside) or becomes another loop in addition to the secondary loop(if on the outside).
Yes, solenoid length will have pronounced effects and there is a point where an extended loop may see 'up' and 'down'. I feel those issues are going beyond the initial experiment.
I keep saying 'view from above' - 'in your mind's eye'. If nothing else, when you have the metering circuit crossing over the solenoid (decoupled, if you choose) stand on the table and look directly down at the top of the solenoid and loop. Forget the height differences for a moment and imagine the whole experiment is in 2D. The metering leads cut across in some fashion. Even though they are not the subject of induction they change the Gaussian surface area. They don't need to be on the same plane.
As you now see, the angle of the metering circuit does matter. Why? Because changing the angle changes the Gaussian surface area of both sections of the loop now sectioned by the metering leads.
This is why I say there is no way to connect a metering circuit without changing something in the experiment.
The voltages you are seeing are the voltage induced on one loop minus the voltage of the other loop (because the original loop is now segmented or sectored by the metering circuit. The same voltages you are measuring have very little to do with the resistors.
If the resistors were equal in resistance, even down to zero Ohms, and you measured from original points A to D you will see .5V. The reason, both sectors of the original loop now have 1V induced into each of them.... At the joining conductor (the metering circuit) what little voltage is seen is the result of two opposite polarity 1V induced into each segment.
The only reason .4V was seen between A-D was because the resistors skewed the angle of the metering leads. Your 'normal to the loop' metering circuit was normal to the loop only in a mechanical sense.
Since my above description is exactly how angular position sensors work (as explained in the Maxim Application sheet brought up early in this thread) I really don't feel I need to argue the point. It would be a great thing to share the understanding but that didn't work for me when it was 'shared'. So, I can't blame you if you choose another path.
In these cases, the chosen path is almost always the best. Experimental results rule but observations must also be questioned until they can no longer be questioned.
WW,
Quote from: WaveWatcher on 2012.04.09, 23:04:22
Regardless if the metering circuit has emf induced into it or not the metering circuit either segments or sectors the loop (if on the inside) or becomes another loop in addition to the secondary loop(if on the outside).
I don't know what you mean by "inside" and "outside".
Quote
Yes, solenoid length will have pronounced effects and there is a point where an extended loop may see 'up' and 'down'. I feel those issues are going beyond the initial experiment.
They go beyond the original experiment, yes. But the issue you have seems to be with the decoupled measurement technique I use.
Quote
I keep saying 'view from above' - 'in your mind's eye'. If nothing else, when you have the metering circuit crossing over the solenoid (decoupled, if you choose) stand on the table and look directly down at the top of the solenoid and loop. Forget the height differences for a moment and imagine the whole experiment is in 2D. The metering leads cut across in some fashion. Even though they are not the subject of induction they change the Gaussian surface area. They don't need to be on the same plane.
You really believe that a very slight angle (3º max) is going to skew the measurements and/or induced emfs that much?
You are saying that it is truly an amazing coincidence that the decoupled measurements of all the summed emfs equals exactly and oppositely the summed voltages across the resistors? Is it also an amazing coincidence that the decoupled measurements across the resistors equals exactly the measurements across the resistors taken in-plane? Is it yet another amazing coincidence that my simulation of this experiment also predicts exactly what I measure between
all points on the loop using the decoupled measurement technique?
Quote
As you now see, the angle of the metering circuit does matter. Why? Because changing the angle changes the Gaussian surface area of both sections of the loop now sectioned by the metering leads.
I'm sorry, I don't see a significant effect on the experiment from the 3º angle at all.
Quote
This is why I say there is no way to connect a metering circuit without changing something in the experiment.
Then I challenge you to explain those amazing coincidences I outlined above. I say the detrimental effect (if any) is undetectable and/or insignificant to the results.
Does your above statement apply to ALL measurements? What is that
change you speak of? How big is that change (i.e. quantify it)? Do the scopes and leads in Lewin's experiment invalidate his results too?
Quote
The voltages you are seeing are the voltage induced on one loop minus the voltage of the other loop (because the original loop is now segmented or sectored by the metering circuit.
Assuming you are referring to a decoupled measurement, what the probes are measuring is representative of precisely what voltage (both emfs and potential differences) difference there is between any two points on the loop. It exists whether the measurement probes are there or not.
Quote
The same voltages you are measuring have very little to do with the resistors.
I don't know what you mean here.
Quote
If the resistors were equal in resistance, even down to zero Ohms, and you measured from original points A to D you will see .5V. The reason, both sectors of the original loop now have 1V induced into each of them.... At the joining conductor (the metering circuit) what little voltage is seen is the result of two opposite polarity 1V induced into each segment.
Equal value resistors (reasonably larger R than Rwire) would not yield a decoupled measurement of 0.5V between D and A. My calculation indicates that 0V will be measured in such case. The reason is very simple; we know the total emf is 1V, we know that emf = PD, and we know that half the PD is across each resistor. So when you do your KVL around half the loop from point D to point A (same answer for BOTH directions), we have: -0.25V + 0.5V - 0.25V = 0V.
Now for the case with no resistors at all, i.e. a closed loop with a uniform wire, I'll let you guys ponder that one (as I have) as to what voltage would be measured half way across the loop. Show your work. >:-)
Quote
The only reason .4V was seen between A-D was because the resistors skewed the angle of the metering leads. Your 'normal to the loop' metering circuit was normal to the loop only in a mechanical sense.
0.4V is seen between points D and A with the original value resistors, for the same reason we see 0V with equal value resistors. The KVL sum again works for BOTH directions. Starting from A going up the left side across the 100 Ohm we have: 0.25V -0.1V + 0.25V = 0.4V
Quote
Since my above description is exactly how angular position sensors work (as explained in the Maxim Application sheet brought up early in this thread) I really don't feel I need to argue the point.
I would caution against relying on MAXIM's theory in that paper of theirs, they made errors as well. For one eg. they claim that all the emf is induced in the resistors. Of course that is completely opposite to the truth. They don't fully understand the physics behind this apparatus even though they can get it to work.
QuoteIt would be a great thing to share the understanding but that didn't work for me when it was 'shared'. So, I can't blame you if you choose another path.
The only path that interests me is the one that leads to the truth. I would hope that would be the case for all. :)
I look forward to your responses. :)
This post is a reminder to myself that I have a possible means to measure, non-intrusively, the actual induced emf in any of these loops. This could then be compared against the decoupled measurement technique to observe if there are any significant detrimental effects to the induced emf (or any parameter) by attaching the decoupled measurement leads to the loop at any two locations.
Quote from: poynt99 on 2012.04.11, 03:02:10
This post is a reminder to myself that I have a possible means to measure, non-intrusively, the actual induced emf in any of these loops. This could then be compared against the decoupled measurement technique to observe if there are any significant detrimental effects to the induced emf (or any parameter) by attaching the decoupled measurement leads to the loop at any two locations.
Basically you have a better technique that gives the same result as decoupled ? lol
Gibbs,
Only to determine the true emf.
Decoupled is still the way forward. O0
Quote from: poynt99 on 2012.04.08, 16:24:26
The induced electric field at a point outside the solenoid diminishes as a function of 1/r.
The problem is that an "induced electric field at a point" has no effect onto a test charge at this point. A single electron outside a solenoid carrying a varying current, feels no force. You must have a closed path.
I was the first surprised. Outside a long solenoid, B is zero. But the vector potential is not null and we have E=-∂A/∂t. So we would be expecting a test charge to be submitted to a force F=q*E=-q*∂A/∂t. But we don't observe such a force.
I have led several experiments some years ago in order to transmit an information signal with the vector potential only (E and B being null at the receiver location). Not one has worked, a loop is always needed.
Quote
The line integral of the E-field (which equals induced emf) at radius r does not diminish as a function of 1/r.
Better? ;)
Yes, better :).
Now we agree that the line integral of the E-field does not diminish as a function of 1/r, and if you agree that there is no effect of this e-field on charges outside a closed path, I don't see how you could experimentally verify the 1/r effect of the "not looped e-field".
Quote from: exnihiloest on 2012.04.11, 09:17:07
Now we agree that the line integral of the E-field does not diminish as a function of 1/r, and if you agree that there is no effect of this e-field on charges outside a closed path, I don't see how you could experimentally verify the 1/r effect of the "not looped e-field".
I don't know if I agree that the induced E field has no effect on a point charge. If it did not, induction would not work.
As far as verifying experimentally the 1/r effect at a point, I don't need to, do I?
Quote from: exnihiloest on 2012.04.11, 09:17:07
Now we agree that the line integral of the E-field does not diminish as a function of 1/r, and if you agree that there is no effect of this e-field on charges outside a closed path, I don't see how you could experimentally verify the 1/r effect of the "not looped e-field".
Ex,
In regards to your statement in bold, I would suggest you read up on the betatron device (http://=http://www.britannica.com/EBchecked/topic/63352/betatron). It uses an induced electric field to accelerate electrons in a circular orbit. I think we would agree that electrons are charges?
Also, from "Matter and Interactions Vol. 2", p.872:
QuoteNo matter how an electric field is produced, it has the same effect on a charge q (that is F = qE)...
Gibbs,
Here is the idea of my test:
If we all agree that in the inductive loop, induced emf = pd (potential drops), then I propose we measure the pd's, with and without the decoupled measurement leads connected at points D and A.
So we use the in-plane lead orientation to measure the pd across both resistors, once without the decoupled leads connected between points D and A, and once with. We could do this 3 times each to get an average reading for both scenarios.
If the total pd is the same or within a few percent between each scenario, one with the decoupled leads, and one without, then we can say that the decoupled lead measurement DOES NOT have an appreciable effect on the total induced emf. Therefore, the decoupled measurement technique can be approved as a valid method to measure across any two points on the loop.
What do y'all think?
Of course this will only go forward if WW agrees that using the in-plane measurement isn't going to skew or throw off the actual induced emf in the loop.
What do you say WaveWatcher?
Quote from: poynt99 on 2012.04.11, 16:23:58
Gibbs,
Here is the idea of my test:
If we all agree that in the inductive loop, induced emf = pd (potential drops), then I propose we measure the pd's, with and without the decoupled measurement leads connected at points D and A.
So we use the in-plane lead orientation to measure the pd across both resistors, once without the decoupled leads connected between points D and A, and once with. We could do this 3 times each to get an average reading for both scenarios.
If the total pd is the same or within a few percent between each scenario, one with the decoupled leads, and one without, then we can say that the decoupled lead measurement DOES NOT have an appreciable effect on the total induced emf. Therefore, the decoupled measurement technique can be approved as a valid method to measure across any two points on the loop.
What do y'all think?
Of course this will only go forward if WW agrees that using the in-plane measurement isn't going to skew or throw off the actual induced emf in the loop.
What do you say WaveWatcher?
There is restriction to this method.
The resistor length must be as close to zero as possible. Otherwise, you will subject to the sector measurement as we have already demonstrated.
Quote from: GibbsHelmholtz on 2012.04.11, 17:04:29
There is restriction to this method.
The resistor length must be as close to zero as possible. Otherwise, you will subject to the sector measurement as we have already demonstrated.
Sorry, I don't follow your reasoning. I don't know what a "sector measurement" is.
The in-plane measurement is what Lewin uses. Are you forgetting that when using the in-plane measurement, the length of the resistor isn't critical? Remember that this measurement produces virtually identical results when the probes are directly across the resistor OR across points D and A.
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=7698;image)
Let's look at this picture. The measurement decoupled at point 0b and 1a is .06V . What would the in plane measurement for these two points? The in plane measurement would give the voltage across the resistance of that wire segment... which is about 0V (we'll ignore the loop size effect). Do you agree? If you have 100 Ohms with the length of that segment, the in plane would gives say .1V, but the decoupled would reads .1V+.06V. So we ask what is causing the .06V? Hope I didn't lost you.
Quote from: GibbsHelmholtz on 2012.04.11, 18:23:02
Let's look at this picture. The measurement decoupled at point 0b and 1a is .06V . What would the in plane measurement for these two points? The in plane measurement would give the voltage across the resistance of that wire segment... which is about 0V (we'll ignore the loop size effect). Do you agree?
Yes.
Quote
If you have 100 Ohms with the length of that segment, the in plane would gives say .1V, but the decoupled would reads .1V+.06V. So we ask what is causing the .06V? Hope I didn't lost you.
The in-plane measurement would give 0.1V (assuming the measurement is taken from the left side), and the decoupled measurement would ALSO give 0.1V, assuming the measurement is taken directly across the resistor element, with no wire segment in between the measurement probes.
You're trying to re-write Ohms law too I see. >:-)
Quote from: poynt99 on 2012.04.11, 18:37:53
The in-plane measurement would give 0.1V (assuming the measurement is taken from the left side), and the decoupled measurement would ALSO give 0.1V, assuming the measurement is taken directly across the resistor element, with no wire segment in between.
So you think even if the resistor is as long as the segment, the measurement is still .1V? The length of the resistor is also a wire. So the decoupled measurement would be resistance + length.
Quote from: GibbsHelmholtz on 2012.04.11, 18:46:01
So you think even if the resistor is as long as the segment, the measurement is still .1V?
Yes. You can't break Ohms law this way.
Quote
The length of the resistor is also a wire. So the decoupled measurement would be resistance + length.
The resistor is not equivalent to a wire, and much less emf will be induced in them as a result; the ratio being:
emf(wire)/emf(resistor) = R(resistor)/R(wire) (for approx. same length resistor & wire)
There are far fewer free electrons in a resistor compared to a wire, and it is the free electrons that become "induced" into moving due to the electric field. So, fewer free electrons, less induced emf. The vast majority of the emf is induced in the wire segments.
There is an easy way to test this.
Let's insert 100Ohms into segment 0b-1a. Let's say the segment is 4 inches long. A typical resistor is .5 inch long after clipping the leads off. We can join 8 resistors together with 12.5 Ohms each resistor. Now we have 100 Ohms resistor with 4 inches long (no leads). Is this fair? Will the measurement the same in plane and decoupled?
Quote from: poynt99 on 2012.04.11, 16:23:58
Of course this will only go forward if WW agrees that using the in-plane measurement isn't going to skew or throw off the actual induced emf in the loop.
What do you say WaveWatcher?
I don't know how another statement from me should matter. I was waiting to see what method you would invent that didn't change the Gaussian surface area and also not change the quantity of such areas. The method you describe will do one, the other or both. It doubt it will change the actual emf induced in the loop under test. That isn't the point. If that was understood you probably wouldn't bother the proposed test.
Please continue, anyway. I'm hoping that at some point you will discover what I'm talking about on your own. It will make more sense that way.
Quote from: GibbsHelmholtz on 2012.04.11, 19:54:35
There is an easy way to test this.
Let's insert 100Ohms into segment 0b-1a. Let's say the segment is 4 inches long. A typical resistor is .5 inch long after clipping the leads off. We can join 8 resistors together with 12.5 Ohms each resistor. Now we have 100 Ohms resistor with 4 inches long (no leads). Is this fair? Will the measurement the same in plane and decoupled?
Yes, this could be done, but it would not likely work as you envision. A segment of wire even 0.1mm long is most likely going to produce more emf than the resistor material.
Perhaps we can simply find a long non-inductive resistor, and use that? Even one inch would be sufficient I believe. I will look and see what I have.
Quote from: poynt99 on 2012.04.11, 21:58:35
Yes, this could be done, but it would not likely work as you envision. A segment of wire even 0.1mm long is most likely going to produce more emf than the resistor material.
Perhaps we can simply find a long non-inductive resistor, and use that? Even one inch would be sufficient I believe. I will look and see what I have.
Poynt,
The result over the entire length (say 100mm) just wire alone is .06V. What would .1mm length contribute?
.06V/100mm = ?V/.1mm
V = .006/100 = .00006V
So .1mm length contribute .00006V to the measurement. It's negligible.
Quote from: WaveWatcher on 2012.04.11, 21:09:04
I don't know how another statement from me should matter. I was waiting to see what method you would invent that didn't change the Gaussian surface area and also not change the quantity of such areas. The method you describe will do one, the other or both. It doubt it will change the actual emf induced in the loop under test. That isn't the point. If that was understood you probably wouldn't bother the proposed test.
Please continue, anyway. I'm hoping that at some point you will discover what I'm talking about on your own. It will make more sense that way.
WW, unless you can produce a test that accomplishes what's needed, then it's relatively easy to rest on the notion that the decoupled leads are tainting the experiment, isn't it? What do you suggest then as a means to determine this?
Anyone????
Are you resting "your case" on this WaveWatcher? I'd guess you're pulling out of the debate then. How unfortunate if you're content leaving it inconclusive... for you that is.
Let's summarize where we are at this point, regarding the Lewin experiment according to WW:
1) All the measurements I've taken on my apparatus are invalid, notwithstanding the non-ideal solenoid. This, despite close correlation with a simulation of the experiment.
2) Lewin's in-plane measurements are invalid.
3) An in-plane measurement of pd's in the experiment invalidates the emf measurement.
4) It is beyond practical means to design and implement a test to quantify the true induced emf in Lewin's experiment.
5) Regardless if induction does not occur in the measurement leads, the measurement taints the experiment nonetheless.
6) We are to take WW's word for the above, even though he has not shown any proof, nor confirmed anything by experiment.
C.C
Quote from: GibbsHelmholtz on 2012.04.11, 22:15:57
Poynt,
The result over the entire length (say 100mm) just wire alone is .06V. What would .1mm length contribute?
.06V/100mm = ?V/.1mm
V = .006/100 = .00006V
So .1mm length contribute .00006V to the measurement. It's negligible.
Gibbs, either you're not following along, or you didn't understand.
It has little to do with length. It has ALL to do with free electrons. This should hopefully make it clear:
Imagine part of a loop is composed of nichrome wire that is 9 inches long. That 9 inches of nichrome wire has a resistance of 100 Ohms. Attached to this to complete the loop is one inch of solid copper wire. Guess where 99.9999999999999999999999999% of the induced emf is going to appear? Across that one inch of copper wire.
Okay,
Let's approach this from another angle. Again, measurement is being decoupled at point 0b-1a. It is being read .06V. Now we decrease the height of decoupling. Voltage starts to drop down and as the height reach 0, voltage reads zero. Decoupled is decoupled, why does this happens?
Unfortunately Gibbs, I won't be able to perform either of these two experiments (I know you want me to prove the 9 inch nichrome, 1 inch copper thing) for a number of days, as I'm going in for a hernia repair surgery in the morning. :'(
Quote from: poynt99 on 2012.04.11, 22:43:30
Unfortunately Gibbs, I won't be able to perform either of these two experiments (I know you want me to prove the 9 inch nichrome, 1 inch copper thing) for a number of days, as I'm going in for a hernia repair surgery in the morning. :'(
I'm sorry Poynt. I did not want you to perform those experiments. I rather experiment is the last thing we do. Hernia is bad enough, let's not get a stomach ulcer because of worry. We all hope you well.
The way this device/system works is conclusive for me and I urge anyone in doubt to perform the experiments themselves. Don't take my word for it
I could be and probably am wrong on some points. In the end, the opinions of those I repeat, from those with far more expertise than I could claim, mean nothing when you can only relate your observations to what you know.
Faraday was a genius at visualizing what happened in his experiments. He went overboard on more than one occasion. Thanks to him we think we have wave-particle duality, spooky action at a distance, Einstein spending his last years trying to convince people he had some of his ideas wrong and who knows what else.
Number 1 is incorrect. There doesn't seem to be anything wrong with the measurements.
Number 2 is incorrect. Lewin's in-plane resistor measurements weren't resistor measurements and I'm sure he knew it. His purpose was to demonstrate non-conservative fields and related issues. He succeeded.
Number 3 is not precise. Any measurement will display the view of the observer (the meter or scope) not the loop.
Number 6 is incorrect. I almost never take the word of another when it comes to experimental observations. Why should anyone take the word of Faraday, Lorentz, Gauss, other well known names or even little 'ol me?
I'm sure the functions demonstrated have made some good money for a few. Others just have to work with it every day. The majority will never see an example in front of them or when they do, they will think it is something else.
BTW:
The resistor length doesn't matter. The same amount of emf is induced on the closed loop no matter the resistor value or length. Since potentials across sections of the loop are meaningless I suggest not wasting much time on inventing a zero length resistor.
I'm still working on 'my version' when time permits.
.........Re: The hernia stuff
OUCH! I just read that and felt a sharp pain where I went through that.
Take the good drugs while you can get them O0
Quote from: GibbsHelmholtz on 2012.04.11, 22:40:58
Okay,
Let's approach this from another angle. Again, measurement is being decoupled at point 0b-1a. It is being read .06V. Now we decrease the height of decoupling. Voltage starts to drop down and as the height reach 0, voltage reads zero. Decoupled is decoupled, why does this happens?
Gibbs,
Why do you say this happens?
If all you are doing is shortening the leads down to practically zero height above the loop, the leads are still normal to the loop, AND they are not in-plane. The indicated voltage would still be 0.1V. In order to measure 0V, the leads would have to approach from the side, in-plane with the loop.
Quote from: WaveWatcher on 2012.04.11, 22:54:47
The way this device/system works is conclusive for me and I urge anyone in doubt to perform the experiments themselves. Don't take my word for it
I've done them. You seem to refute my results.
Then you say there isn't anything wrong with my measurements:
Quote
Number 1 is incorrect. There doesn't seem to be anything wrong with the measurements.
It's difficult to understand where you're coming from WW. I would very much like if you could try to clarify your main point, as it seems to have gone unnoticed, or been misunderstood.
Quote
.........Re: The hernia stuff
OUCH! I just read that and felt a sharp pain where I went through that.
Take the good drugs while you can get them O0
Thanks guys :) Hopefully they give me some good stuff.
Quote from: poynt99 on 2012.04.11, 22:57:28
Gibbs,
Why do you say this happens?
If all you are doing is shortening the leads down to practically zero height above the loop, the leads are still normal to the loop, AND they are not in-plane. The indicated voltage would still be 0.1V. In order to measure 0V, the leads would have to approach from the side, in-plane with the loop.
Okay, I think this approach is good. Actually, the voltage is not constant .1V as you travel the height. But let's not use that. Use the .06V in your experiment. It's the wire length from 0b-1a. As you travel to 0 height, it measured just the wire length resistance which is virtually 0 V. I hope you're not arguing it's still .06V.
.99,
I must agree with your measurements. I disagree with your conclusions.
My main point? Wow! Maybe a drawing will help.
Gibbs,
If I understand your method, zero volts should be expected when the meter and probes are on the inside and in-plane with the loop while connected across the diameter of that loop. Also, the measured emf should increase as the meter and probe leads extend above the loop - up to a point.
Quote from: WaveWatcher on 2012.04.11, 23:14:31
Gibbs,
If I understand your method, zero volts should be expected when the meter and probes are on the inside and in-plane with the loop while connected across the diameter of that loop. Also, the measured emf should increase as the meter and probe leads extend above the loop - up to a point.
WaveW,
I can't really answer your comment because I do not know exactly the setup you describe. If the meter and probes are across the diameter and the resistance is equal on both sides or zeros, then the voltage is zero. Yes,the measurement should increase as decoupling go higher up (except in the case of directly across the diameter, it does nothing).
Quote from: GibbsHelmholtz on 2012.04.11, 23:13:10
Okay, I think this approach is good. Actually, the voltage is not constant .1V as you travel the height. But let's not use that. Use the .06V in your experiment. It's the wire length from 0b-1a. As you travel to 0 height, it measured just the wire length resistance which is virtually 0 V. I hope you're not arguing it's still .06V.
Sounds confusing Gibbs. I'm not sure what you mean.
If all you are doing is lengthening or shortening the decoupled leads, then the measurement will remain constant, no matter what length they are. This holds for a measurement across a resistor and across a wire segment. Maybe you need to make a diagram to make it more clear?
Quote from: WaveWatcher on 2012.04.11, 23:14:31
.99,
I must agree with your measurements. I disagree with your conclusions.
Yes, the measurements will be correct, but you disagree that the measurement leads are non-intrusive to the experiment, correct? I hope that makes sense. In other words, "garbage in, garbage out", is what you mean right? Kinda like Rosemary's measurements. :D
I'm looking forward to your conclusions and explanations that refute my own.
Quote from: poynt99 on 2012.04.12, 01:59:39
If all you are doing is lengthening or shortening the decoupled leads, then the measurement will remain constant, no matter what length they are. This holds for a measurement across a resistor and across a wire segment. Maybe you need to make a diagram to make it more clear?
Ahah, yes, all I'm doing is lengthening or shortening the decoupled leads. However, .06V won't be constant across that wire segment. If you see it carefully, when you shorten decoupled leads to zero height for segment 0b-1a, it will indeed reads 0V instead of .06V.
.99,
In no way are you or your experiment a comparison to Rosemary. I can only imagine what her results would be. Perhaps our disagreement is due to zipons? >:-)
Good Luck with the Op Darren O0
Quote from: Peterae on 2012.04.12, 09:27:01
Good Luck with the Op Darren O0
Ditto!
Make sure all holes that should be there - are still there after you wake up.
Be sure no freely moving parts are secured with an accidental stitch! I'm not speaking of them tying your shoes together :D
Gibbs,
Is it possible that the proximity of the meter to the loop is having an effect on meter performance?
I'm not saying the readings are other than I would expect. Simply, all possible effects should be ruled out.
Quote from: WaveWatcher on 2012.04.12, 10:51:40
Gibbs,
Is it possible that the proximity of the meter to the loop is having an effect on meter performance?
I'm not saying the readings are other than I would expect. Simply, all possible effects should be ruled out.
WaveW,
I show you a picture below. Hope you can see why I think the height matter.
-----------
Get better soon Poynt
Gibbs,
Yes, I do. You must consider the amount/density and angle of the field lines entering the metering circuit loop and the secondary loop.
Raising the meter increases the number of possible lines entering the metering loop. The value seen will be the sum of the two segments (created at the secondary loop by inserting the measuring circuit) plus any possible induction now occurring in the metering circuit.
Lowering the height to almost zero simply decreases the amount induced into the metering loop. At or near zero height, the sum will be zero because each of the two segments of the secondary loop (again, attaching two meter probes and meter segments the secondary loop into two loops) always have 1V induced, each. Since the current path taken by both segments is the same direction, those current paths meet in the metering circuit and cancel each other to zero.
In your above drawing, there is only one secondary loop segment because the meter loop is at an extreme edge of the secondary rectangle. If that rectangle was a circle there will always be two segments created by attaching the meter circuit.
The height does matter to the observer but doesn't matter to the measured loop. Please be aware that the height will 'seem' to matter to the measured loop but only because the metering loop has a direct effect upon what it displays depending upon the meter's relation to the measured loop and the percent of field lines seen from the source solenoid.
Once the meter is high enough to register a voltage you should be able to change the perpendicular angle to a lower number and see further measured voltage change, both plus and minus.
I hope that is clear enough.
Quote from: WaveWatcher on 2012.04.12, 13:45:04
Gibbs,
Yes, I do. You must consider the amount/density and angle of the field lines entering the metering circuit loop and the secondary loop.
Raising the meter increases the number of possible lines entering the metering loop. The value seen will be the sum of the two segments (created at the secondary loop by inserting the measuring circuit) plus any possible induction now occurring in the metering circuit.
Lowering the height to almost zero simply decreases the amount induced into the metering loop. At or near zero height, the sum will be zero because each of the two segments of the secondary loop (again, attaching two meter probes and meter segments the secondary loop into two loops) always have 1V induced, each. Since the current path taken by both segments is the same direction, those current paths meet in the metering circuit and cancel each other to zero.
In your above drawing, there is only one secondary loop segment because the meter loop is at an extreme edge of the secondary rectangle. If that rectangle was a circle there will always be two segments created by attaching the meter circuit.
The height does matter to the observer but doesn't matter to the measured loop. Please be aware that the height will 'seem' to matter to the measured loop but only because the metering loop has a direct effect upon what it displays depending upon the meter's relation to the measured loop and the percent of field lines seen from the source solenoid.
Once the meter is high enough to register a voltage you should be able to change the perpendicular angle to a lower number and see further measured voltage change, both plus and minus.
I hope that is clear enough.
WaveW,
I find it is easier to apply KVL to get the meter reading. Let's pick the metering loop through that segment .
dB/dt(meter) + V (meter) + V(segment) = 0
dB/dt(meter) is the induction due to field lines. This depends on the height of the decoupling.
V(meter) is the voltage register on the meter
V(segment) is just the current x resistance of the segment
Simple, don't you think?
Yes :D
In some cases KVL provides the same answers. There is where I disagree with using KVL. The potentials you measure only exist when the meter is connected.
The only accurate way to measure between any two points on this loop is to place the meter in the center with probes extending to the inside of the diameter and across the diameter.
Don't be surprised if the results are all zero. There are no real potentials on the loop.
Quote from: poynt99 on 2012.04.11, 13:17:49
I don't know if I agree that the induced E field has no effect on a point charge. If it did not, induction would not work.
As far as verifying experimentally the 1/r effect at a point, I don't need to, do I?
"If it did not, induction would not work": this is the main point to verify. This objection is logical and I have already tried to verify it.
If the induced E field has an effect on a point charge, then it has also an effect on charges in a small circuit not encircling the varying flux.
To test it, I have used a toroid coil. On one side, near the toroid, I put two large metallic cylinders acting as capacitors.
They were placed at some distance from each other and a wire connected the two capacitors with a voltmeter in series. All was placed in order that there is the presumed emf in the wire. But no voltage was measured.
I have put an inductance in series with the wire to make the circuit more sensitive because resonant at the frequency used to power the toroid coil. Same null result.
A voltage was measured only when the wire was placed through the toroid hole and the two capacitors at each end were near enough from each other. In this case I consider that the circuit was looped thanks to the displacement currents between the two capacitors.
Note that if this effect of E=-dA/dt on a single charge was real, its corollary is that the reversed effect would also be possible, and so, we could induce an emf in a toroid coil from outside, without any conductor through the hole. I'm not aware that any experiment of this kind gave positive results.
What I observe is the experimental impossibility to show any effect of E=-dA/dt on elements of circuit not encircling the flux and consequently the effect on a single charge is doubtful or too weak to be measured. I don't know (yet) why. This effect is indispensable to test the 1/r effect, otherwise with a closed circuit around the flux, we can only measure the well known trivial emf not depending on r.
WW,
As you don't approve of the last proposed test, I have another to run past you. The goal here of course is to introduce a measurement technique that can non-intrusively determine if the addition of two decoupled test leads skews the actual induced emf.
The idea; use a current probe (Rogowski coil, toroid, etc) to determine the induced current, with and without the decoupled leads attached.
What say you?
I like the idea of a no-core Rogowski coil.
I would expect the current measured to be the same anywhere on the loop. I am quite sure KCL holds unless there is a fair amount of inductance somewhere on the loop. Well, KCL will hold regardless. If such an inductance value existed their would be signs of phase shift.
I'm using a 2-section solenoid so metering leads may be in-plane but perpendicular to the magnetic field. So far, any two measured points result in zero, just as I think it should. We'll see when I don't need my daughters to hold the experiment during the test :-X
I take that as an approval then.
So, two tests: measure the loop current with and without the decoupled leads attached. Either the current will be the same in both cases, or not. If they are the same, do you agree the decoupled measurement method is valid?
Valid?
As in reflecting the voltage between the probe tips -or- voltage measurement created just by connecting the meter?
If is the latter then, yes.
They aren't the same.
1. Reducing or eliminating inductive coupling on the metering circuit is only part of the problem.
2. Using current readings and Ohm's law to determine potential drop between two points is only straightforward with conservative fields.
With the Rogowski coil (with no core) some of what I'm saying should be clarified.
I think it would be interesting to measure current at each resistor, separately, with the metering circuit in the so-called 'decoupled' position but with probes at opposite sides of the diameter (original points A & D).
Quote from: WaveWatcher on 2012.04.16, 20:27:10
Valid?
As in reflecting the voltage between the probe tips -or- voltage measurement created just by connecting the meter?
Neither.
Perhaps it will be helpful if I try to restate the problem/challenge:
First, I am going to assume that you agree that an emf will be induced in the loop, agreed?
Second, it is my understanding that you assert that the TRUE induced level of loop emf is being skewed by the application of two decoupled measurement leads placed anywhere on the loop.
Do you concur with my understanding of your objection?
As current probes might be just as susceptible to "influence" by the experiment, I have yet another option for you to consider WW:
Make the induction loop, a second measurement loop. Or put another way, change from two resistors to one, and make it a 1k value. So we have a wire loop that goes 357º around the solenoid, and the loop is loaded with a single 1k resistor (which makes up the last 3º of rotation). This second measurement loop does the integration essentially around the entire solenoid, giving us a pretty accurate reading of the induced emf.
Now, after we note what the total induced emf is by this method, we introduce our two decoupled leads once again across any two random points on the loop and observe if the emf value measured by the "second measurement loop" as described above has changed any appreciable amount.
If there is little or no change, then can we say that the decoupled measurement method is valid, i.e. non-intrusive to the experiment?
Quote from: poynt99 on 2012.04.16, 22:23:10
Neither.
Perhaps it will be helpful if I try to restate the problem/challenge:
First, I am going to assume that you agree that an emf will be induced in the loop, agreed?
Yes.
Quote
Second, it is my understanding that you assert that the TRUE induced level of loop emf is being skewed by the application of two decoupled measurement leads placed anywhere on the loop.
Do you concur with my understanding of your objection?
I can't concur with that statement. The actual emf induced into each of the loop segments created by connecting your metering circuit will still equal 1V.
.....
Wait a minute... (Just saw your follow-up post)
What is wrong with leaving the loop under test as is and measuring current at both resistors while the voltage metering circuit segments the loop under test?
Have you already done so and found that the resistor currents aren't equal?
KCl will hold but only for the circuit being observed.
357
o or 360
o doesn't matter. The same emf is induced into the loop because it is closed by a resistor. The skewing I mentioned is related to current balance between two segments of the same loop.
The emf induced into a second separate loop is determined by the surface area bounded by that loop, not by the surface area bounded by the other loop.
I think we are still speaking separate languages.
Gentlemen,
Correct me if I'm wrong but is the issue about the meter when connect to the loop change the amount of current flowing through the resistors?
If I understand correctly, the only way it could "stripped" some current from the main loop is current flowing through the probes, KCL... right? How about insert a low value cap in series with the probes?
-----------------
Poynt, hope you're doing better. Welcome back.
Gibbs,
The input impedance of the meter means the meter isn't dragging current of any appreciable amount.
I don't know how to state the real issue any more plainly than - connecting the metering circuit segments the area of induction of the secondary loop into two areas of induction.
Where these two areas of induction meet, in the meter circuit, the current paths are opposing one another.
It is as simple as looking down upon the loop and the metering loop while connected. Decoupling means absolutely nothing except to minimize errors caused by unwanted induction.
The area of induction is what matters with the discussion, not the loop except that it defines the area of induction.
Quote from: WaveWatcher on 2012.04.17, 02:21:08
I can't concur with that statement. The actual emf induced into each of the loop segments created by connecting your metering circuit will still equal 1V.
Then I think you are still most likely the only one that understands what you really mean WW. I would suggest you succinctly state what the problem is,
and what the effect of that problem is.
Quote
Wait a minute... (Just saw your follow-up post)
What is wrong with leaving the loop under test as is and measuring current at both resistors while the voltage metering circuit segments the loop under test?
Have you already done so and found that the resistor currents aren't equal?
I think it's rather unlikely that the current at each resistor will or even possibly can be different. With practically zero induction and infinite impedance in the measurements leads, they will not alter the induced emf, nor the induced current in the secondary loop. So no, I have not separately measured the resistor currents. I don't even know how you would propose one do that ???
Quote
The skewing I mentioned is related to current balance between two segments of the same loop.
I want some of what you've been smokin' 8) (esp right now)
Quote
The emf induced into a second separate loop is determined by the surface area bounded by that loop, not by the surface area bounded by the other loop.
Therefore, the measurement loop emf = 0V.
Quote
I think we are still speaking separate languages.
I think the real problem is that we just don't understand what you're getting at.
Quote from: GibbsHelmholtz on 2012.04.17, 02:51:51
Gentlemen,
Correct me if I'm wrong but is the issue about the meter when connect to the loop change the amount of current flowing through the resistors?
You and I both are struggling to understand. Hopefully WW can clarify his point of view.
Quote
Poynt, hope you're doing better. Welcome back.
Feeling better each day Gibbs, thanks. ;)
Quote from: WaveWatcher on 2012.04.17, 03:04:40
I don't know how to state the real issue any more plainly than - connecting the metering circuit segments the area of induction of the secondary loop into two areas of induction.
Where these two areas of induction meet, in the meter circuit, the current paths are opposing one another.
What? :o
Please explain. ???
Quote from: poynt99 on 2012.04.17, 20:05:10
What? :o
Please explain. ???
This is the crux of the matter. I do intend to explain. It is going to take more than words.
At this point, I'm thinking the only understandable explanation must include rotatable 3D graphics and a video conference with a live demonstration.
If you find something to smoke that cuts my working hours I'll buy a kilo :D
Quote from: WaveWatcher on 2012.04.17, 20:17:44
This is the crux of the matter. I do intend to explain. It is going to take more than words.
At this point, I'm thinking the only understandable explanation must include rotatable 3D graphics and a video conference with a live demonstration.
If you find something to smoke that cuts my working hours I'll buy a kilo :D
A well-annotated diagram usually does the trick WW. I'd encourage you to give it a shot.
I think the only way to settle this is to see who understand how the system works.
I posted a problem below.
The total induced emf is 1V
Scope reading position in-plane reads .2V
R1 = 1 Ohm
Find R2.
Quote from: GibbsHelmholtz on 2012.04.18, 00:22:04
I think the only way to settle this is to see who understand how the system works.
I posted a problem below.
The total induced emf is 1V
Scope reading position in-plane reads .2V
R1 = 1 Ohm
Find R2.
;) :)
I'll think about it. However, anyone offering an answer should think about this:
Electric potential (emf) only has meaning for electric fields produced by static charges (battery, capacitor stored energy, etc.). It has no meaning for electric fields produced by induction.
...
I started to build a set of graphics to show what is happening and had to break for dinner.
Quote from: GibbsHelmholtz on 2012.04.18, 00:22:04
...
The total induced emf is 1V
Scope reading position in-plane reads .2V
R1 = 1 Ohm
Find R2.
Impossible to find, without knowing the surface topology between the wires connecting the voltmeter and R2. There is a part of the magnetic flux generated by the current in the circular circuit that loops outside around, in a reversed direction. It is clear that the flux crossing this surface is not null and therefore generates an emf that disturbs the measurement.
If this surface is null because the wires would tightly follow the circle and then would be kept side by side, from R2 to the voltmeter, then R2=0.25 Ohm.
Quote from: exnihiloest on 2012.04.18, 10:57:12
If this surface is null because the wires would tightly follow the circle and then would be kept side by side, from R2 to the voltmeter, then R2=0.25 Ohm.
Yes, very possible.
I must question how the total emf induced on the loop is determined.
Measurement?
Using amper-turns-ratio and applied voltage to the solenoid?
Then, there is another problem that hasn't been accepted as fact; what are the measured potentials of the same two points at different secondary loop emf levels?
The ratio of I
R1:I
R2 will change with secondary loop emf and current levels and meter circuit topology and relative orientation.
The question cannot be answered without making several assumptions.
Quote from: exnihiloest on 2012.04.18, 10:57:12
Impossible to find, without knowing the surface topology between the wires connecting the voltmeter and R2. There is a part of the magnetic flux generated by the current in the circular circuit that loops outside around, in a reversed direction. It is clear that the flux crossing this surface is not null and therefore generates an emf that disturbs the measurement.
If this surface is null because the wires would tightly follow the circle and then would be kept side by side, from R2 to the voltmeter, then R2=0.25 Ohm.
There is a flux outside in reverse direction, but it's minimal so we'll ignore those.
R2 = 0.25 Ohms is correct. You should show us how you derived your answer.
Here is another problem. I think if you can solve these problems, you can solve any problem in decoupled and in plane measurement.
R1 = 6 Ohms
R2 = 18 Ohms
In plane reading .2V
Find the loop's induced EMF
Quote from: exnihiloest on 2012.04.13, 10:17:20
"If it did not, induction would not work": this is the main point to verify. This objection is logical and I have already tried to verify it.
If the induced E field has an effect on a point charge, then it has also an effect on charges in a small circuit not encircling the varying flux.
You didn't measure anything because 1) you're making an incorrect assumption, and 2) you are going about "measuring" it incorrectly.
Any loop/circuit not including the solenoid will not produce an emf.
If you want to measure the induced electric field in its non-integral form, you should use two small metallic balls spaced about 3/4" apart. They have magnet wire leads twisted together, which connect to a scope. Position the balls so that one is closer to the solenoid axis, and the other is radial outward. You will measure a difference in potential across the two balls.
Quote from: GibbsHelmholtz on 2012.04.18, 13:08:01
There is a flux outside in reverse direction, but it's minimal so we'll ignore those.
R2 = 0.25 Ohms is correct. You should show us how you derived your answer.
Here is another problem. I think if you can solve these problems, you can solve any problem in decoupled and in plane measurement.
R1 = 6 Ohms
R2 = 18 Ohms
In plane reading .2V
Find the loop's induced EMF
On the run here and still wondering how you determine the emf induced but, I'll say .8V is what others would calculate. If things are as they should be, it will be closer to .9V and that still isn't what may actually be induced into the loop with no metering connected.
Quote from: poynt99 on 2012.04.18, 17:18:40
If you want to measure the induced electric field in its non-integral form, you should use two small metallic balls spaced about 3/4" apart. They have magnet wire leads twisted together, which connect to a scope. Position the balls so that one is closer to the solenoid axis, and the other is radial outward. You will measure a difference in potential across the two balls.
I hadn't thought of that method. Nice O0
Quote from: WaveWatcher on 2012.04.18, 20:36:43
I hadn't thought of that method. Nice O0
Here's a 50 pin chip that was designed to help detect E fields. 8)
Quote from: GibbsHelmholtz on 2012.04.18, 13:08:01
Here is another problem. I think if you can solve these problems, you can solve any problem in decoupled and in plane measurement.
R1 = 6 Ohms
R2 = 18 Ohms
In plane reading .2V
Find the loop's induced EMF
Quote from: WaveWatcher on 2012.04.18, 20:36:03
On the run here and still wondering how you determine the emf induced but, I'll say .8V is what others would calculate. If things are as they should be, it will be closer to .9V and that still isn't what may actually be induced into the loop with no metering connected.
Gibbs,
Are you planning on making your future tests somewhat challenging? :P :)
The answer to this last one is:
loop's induced emf=0.2666V
[/quote]
Quote from: WaveWatcher on 2012.04.18, 20:36:03
On the run here and still wondering how you determine the emf induced but, I'll say .8V is what others would calculate. If things are as they should be, it will be closer to .9V and that still isn't what may actually be induced into the loop with no metering connected.
WaveW,
There's really nothing big about this. You just need to perform KVL to find the answer.
a)
dB/dt + V(meter) + I(loop) x 6 Ohms = 0The two unknown is dB/dt which is what we need to find and I(loop). We need another equation to find I. Do another loop.
b)
dB/dt + V(meter) + I(loop) x 18 Ohms = 0In this loop, dB/dt = 0 . Everything else is known except for I(loop). It turns out to be .0111 amps. Solve for equation a).
dB/dt = .2 + .0666
It's lengthy but it's the correct format.
--------------------------------------
Whatever Poynt, I just think you got lucky. :P
Try this one. Now we got 2 identical loops, .2666V induced in each loop. Find the meter reading V. :P
Quote from: GibbsHelmholtz on 2012.04.18, 21:41:36
WaveW,
There's really nothing big about this. You just need to perform KVL to find the answer.
I have the feeling you aren't going to tell me how you determine the induced emf.
So, you cranked your drive voltage down by 75% or you decreased the area of the loop by 75% or a mix of both?
Have you determined the induced emf by any other method than using KVL during this whole story?
Quote from: WaveWatcher on 2012.04.18, 22:31:53
I have the feeling you aren't going to tell me how you determine the induced emf.
So, you cranked your drive voltage down by 75% or you decreased the area of the loop by 75% or a mix of both?
Have you determined the induced emf by any other method than using KVL during this whole story?
WaveW,
The induced emf is the unknown. It is dB/dt of the loop. It's not 1V given like Lewin's .
I think KVL is all we need.
Quote from: GibbsHelmholtz on 2012.04.18, 21:41:36
Whatever Poynt, I just think you got lucky. :P
Try this one. Now we got 2 identical loops, .2666V induced in each loop. Find the meter reading V.
Gibbs, I think we're still going to see 0.2V on that meter.
Quote from: poynt99 on 2012.04.19, 00:20:51
Gibbs, I think we're still going to see 0.2V on that meter.
Hm... looks like you did got lucky on the last one.
The answer is
.466 V :P
Forgive me Gibbs.
KVL does not and never will apply to non-conservative fields. KVL may produce the correct results depending upon the view of the observer/meter. Potentials measured between points on a closed loop are meaningless.
The only correct measurement you can obtain is zero volts unless you break the loop in one place and measure potential difference between the open ends.
So, the 'now known' emf is taken from KVL?
I submit to you that the real emf is still unknown.
Quote from: GibbsHelmholtz on 2012.04.19, 00:28:41
Hm... looks like you did got lucky on the last one.
The answer is .466 V :P
It's possible Gibbs, however I won't be fully convinced until I try it. Good one nonetheless ;)
Quote from: WaveWatcher on 2012.04.19, 00:29:36
Forgive me Gibbs.
KVL does not and never will apply to non-conservative fields. KVL may produce the correct results depending upon the view of the observer/meter. Potentials measured between points on a closed loop are meaningless.
The only correct measurement you can obtain is zero volts unless you break the loop in one place and measure potential difference between the open ends.
So, the 'now known' emf is taken from KVL?
I submit to you that the real emf is still unknown.
Hi WaveW,
I've observed your thinking for a while. I want to ask what do you consider the real value? I see three value: the induced voltage, Ohm's law voltage, and meter voltage. They are all different. The induced voltage and meter voltage change, but the Ohm's law voltage can never change between two points of interest.
About KVL and non conservative field. I think Lewin's argument is based on philosophy and not the math. Lewin's argument is this:
V1+V2+V... = dB/dt
KVL is this:
V1+V2+V... -dB/dt = 0
I mean they are different only on philosophy. I think we should come to an agreement with this. Maybe there is higher philosophy about free energy with this but what important to me is how can we determine parameters based on known parameters. Yes, even Poynt have a hard time with this. lol
Once again:
The closed loop voltage = dB/dt (Faraday)
The closed loop voltage = 0 (KVL by bring dB/dt to the other side)JUST FOLLOW THESE RULES.
Quote from: GibbsHelmholtz on 2012.04.19, 01:02:49
Hi WaveW,
I've observed your thinking for a while. I want to ask what do you consider the real value? I see three value: the induced voltage, Ohm's law voltage, and meter voltage. They are all different. The induced voltage and meter voltage change, but the Ohm's law voltage can never change between two points of interest.
About KVL and non conservative field. I think Lewin's argument is based on philosophy and not the math. Lewin's argument is this:
V1+V2+V... = dB/dt
KVL is this:
V1+V2+V... -dB/dt = 0
I mean they are different only on philosophy. I think we should come to an agreement with this. Maybe there is higher philosophy about free energy with this but what important to me is how can we determine parameters based on known parameters. Yes, even Poynt have a hard time with this. lol
The induced voltage is always correct and fixed. Ohm's law voltage is always correct but the path must be considered. If the path includes more than one possibility your meter position will convince you the current is going either way and Ohm's law is broken. There are only two cases where the meter voltage is correct:
1. the loop is broken to insert the meter probes (Voltage)
2. the meter and the probes are made up like a single, solid and straight line. That is placed across the diameter of the loop. The reading should be zero.
Using the diagram where I 'failed', I calculate the induced emf to be .799V not considering the change from the induced emf of the previous exam. I don't truly know what that value was but assume that you have changed little and the actual induced emf is almost exactly the same as before.
KVL and Faraday are not differing philosophies. KVL is a round peg and Faraday a square hole. There are ways to make KVL fit. That doesn't mean it is the correct method.
KVL has never included dB/dt. Only the later adaptations of KVL include that. From that point on it is no longer KVL but Maxwell's flavor of Faraday.
Quote
Once again:
The closed loop voltage = dB/dt (Faraday)
The closed loop voltage = 0 (KVL by bring dB/dt to the other side)
JUST FOLLOW THESE RULES.
Thanks, but if I followed those rules I would not be able to understand much of the machinery I design and diagnose on a daily basis.
All resolvers would cease to function. Older navigational gyroscope systems would instantly fail. Certain types of adjustable transformers and inductive couplers would fail.
This all has little to do with Lewin.
once again:
He was presenting a very old script from as far back as the 70's and WAS taught in all transformer theory classes, as far as I assumed.I am dismayed with people I thought I knew and have worked side by side with for years. When I mentioned the subject of this discussion his ears perked. Within two minutes of explaining the discussion he shook his head and said he didn't know I was such an idiot.
I grabbed an old resolver from inventory and placed it on his desk. He ask, why? I said I'll answer that question when you can tell me how it works.
His answer was, "Oh, sorry 'bout that".... DUH!
First this: Much of what we've been discussing lately doesn't directly relate to the Lewin experiment, but it most certainly is related and helpful to investigate. I have a pretty good grip on the dynamics involved in this experiment, but I am learning some new things here as well.
Quote from: WaveWatcher on 2012.04.19, 02:21:27
The induced voltage is always correct and fixed. Ohm's law voltage is always correct but the path must be considered. If the path includes more than one possibility your meter position will convince you the current is going either way and Ohm's law is broken.
Fellows: The PATH is important only when our measurement leads are in-plane with the loop. Ohm's law is NEVER broken when the measurement is taken decoupled.
The loop is a translator; it performs the integral of the E-field at radius r. That integrated E-field results in a loop emf. No emf can be induced if there is no conductive loop around the solenoid.
Now, once the loop is present, it is a real-time translator/integrator, and we must then treat the loop as any "normal" circuit, with an emf and loads. The only difference now between this circuit and one with a DC battery, is the emf source is localized with the battery. With an induced emf, the emf will be spread over any length (or lengths) of wire in the loop, and it will vary in amplitude with time.
The only way to measure all the points on a loop with an induced emf, is to decoupled the leads, otherwise the leads become influenced by the induced E-field. Once decoupled, all measurements and calculations we normally use on DC circuits apply to this one as well. Ohm's Law, and KVL are completely valid.
Where KVL breaks down, is when we are observing or measuring the E-field, or its integral. We inadvertently do this when our measurement leads are in-plane with the induction loop. When the measurement leads are directly in-plane with the induced E-field, they are at its mercy. What ever the E-field is doing, the meter will reflect. The meter can not measure the induced emf in the wire segments this way.
As emf is induced in the wire segments, there is an excess and shortage of electrons created at each end, due to restrictions caused by the resistors. This bunching of charges at each end of the wire segments and resistors "distorts" the induced E-field so that it too is concentrated at the junctions of the resistor elements and the wire segments. This is why moving the in-plane measurement lead probes anywhere in-between the wire segments in the Lewin experiment, makes no discernible difference in what is indicated on the scope.
Now, decouple the measurement leads (raise them normal to the loop) and the induced E-field no longer has any influence on them. You are now able to measure the true emf and potential drops anywhere on the loop, without interfering with the induced emf.
Gibbs,
Did you decide why one of your earlier measurements was zero when the meter circuit was very close to the plane of the loop?
What reason did you choose?
Quote from: poynt99 on 2012.04.19, 03:15:53
Fellows: The PATH is important only when our measurement leads are in-plane with the loop. Ohm's law is NEVER broken when the measurement is taken decoupled.
Fellows: The PATH is always important when working with non-conservative fields. Ohm's law is NEVER broken.
I think everyone reading this thread must make there own decision.
This is no theory of mine so I cannot concede.
The decoupling methods depicted simply work. There is no argument that the methods fail to prevent unwanted errors by induction.
When you must modify a circuit law to suite your findings you are no longer using that circuit law.
Quote from: WaveWatcher on 2012.04.19, 09:39:19
Gibbs,
Did you decide why one of your earlier measurements was zero when the meter circuit was very close to the plane of the loop?
What reason did you choose?
WaveW,
I still use only 1 method.
dB/dt + V(meter) + I x R segment = 0
Since the meter is close, there is no flux going through the meter loop. Meter reading has to be the same as I x R segment. It's not zero, it is just extremely small.
-----------------
Hey, I got some more setup for you guys. Determine the meter reading on this one. If possible, show some KVL. lol
Quote from: WaveWatcher on 2012.04.19, 10:04:41
Fellows: The PATH is always important when working with non-conservative fields. Ohm's law is NEVER broken.
I think everyone reading this thread must make there own decision.
WW, when your measurement leads are in-plane,
what are you measuring, and what are you "working with"?
And yes, Ohm's Law CAN APPEAR to be broken when working with non-conservative fields, let's make that distinction. I have shown that, and I can prove it. Have you shown and proved the contrary?
My challenge to you is to prove/show that it CANNOT APPEAR to be broken.
WW,
What is the voltage indicated on CH2 of the scope? Assume 1V emf.
Quote from: WaveWatcher on 2012.04.18, 12:06:42
Yes, very possible.
I must question how the total emf induced on the loop is determined.
Measurement?
You just need to open the loop and connect its two ends to a voltmeter. Or easier: connect the two voltmeter probes together, each one passing each side of the solenoid.
Quote
Using amper-turns-ratio and applied voltage to the solenoid?
It works also.
Quote
Then, there is another problem that hasn't been accepted as fact; what are the measured potentials of the same two points at different secondary loop emf levels?
The ratio of IR1:IR2 will change with secondary loop emf and current levels and meter circuit topology and relative orientation.
The question cannot be answered without making several assumptions.
There is
no potentials. The induced emf doesn't derive from a potential.
At the position of the circuit, the circuit being removed, there is a circular electric field but no potential differences. The potential differences are created when you place a circuit with "obstacles" like an open circuit or resistances. The electric field move the charges which are prevented to move freely and so, positive or negative charges tend to accumulate on each side of the "obstacle" ("positive charges" meaning lack of electrons), creating the potential difference.
We logically assume that we have the maximum potential difference when the "obstacle" is total, i.e. when the circuit is open and we measure the voltage at its two ends. And this gives us the circular electric field (E=U/circuit length). Potential differences can be simply measured with a voltmeter, provided that there is no flux crossing the surface of the measurement circuit.
Quote from: poynt99 on 2012.04.18, 17:18:40
You didn't measure anything because 1) you're making an incorrect assumption, and 2) you are going about "measuring" it incorrectly.
Any loop/circuit not including the solenoid will not produce an emf.
...
This is true but not related to my reply which applied to the following image:
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=7753;image)
I suppose that there is a flux crossing the surface limited by the circle, isn't it?! If so, of course one can correctly measure a potential difference at R2, provided that there is no flux crossing the surface of the measurement circuit (limited by the 2 wires from the voltmeter and r2) which is not the case here. As already said, there is a flux disturbing the measurement if the wires are put as on the image. I have said that my reply applies only if this is corrected by a careful positionning of the voltmeter wires, eliminating the disturbing flux by reducing the surface of the measurement circuit to zero (this can be easily done by connecting the wires directly to R2 and by keeping them as near from each other that is possible.
So your points 1 and 2 are wrong because you didn't take account of the domain of validity of my answer, indeed given ("If this surface is null because the wires would tightly follow the circle and then would be kept side by side, from R2 to the voltmeter, then R2=0.25 Ohm").
Quote from: poynt99 on 2012.04.19, 15:15:47
WW,
What is the voltage indicated on CH2 of the scope? Assume 1V emf.
Ex, are you WW?
Is this question asked by you, poynt99?
Are we on a forum or not?
If so, questions and replies are public, every message is open and every one can reply about everything. It's the first time that I have a message canceled because I gave an answer to a question intended preferably to someone else!
For personal messages not addressed to the group, for games between two participants only, the mail should be prefered.
ETA: If I don't specify who I'm directing the question to, then yes, it's open to all. I specifically asked WW, because I wanted his answer before anyone else's. Simple enough?
Quote from: exnihiloest on 2012.04.19, 17:07:08
This is true but not related to my reply which applied to the following image:
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=7753;image)
You mentioned your own tests in an attempt to prove whether or not an induced E-field can influence, or be influenced by a point charge. THAT is what the following referred to:
Quote from: poynt99 on 2012.04.18, 17:18:40
You didn't measure anything because 1) you're making an incorrect assumption, and 2) you are going about "measuring" it incorrectly.
Any loop/circuit not including the solenoid will not produce an emf.
...
I wasn't referring to Gibbs' test question and diagram.
Quote from: poynt99 on 2012.04.19, 13:43:11
WW, when your measurement leads are in-plane, what are you measuring, and what are you "working with"?
And yes, Ohm's Law CAN APPEAR to be broken when working with non-conservative fields, let's make that distinction. I have shown that, and I can prove it. Have you shown and proved the contrary?
My challenge to you is to prove/show that it CANNOT APPEAR to be broken.
Huh?
I'm agreeing with you on that. It is never broken but may appear to be broken when working with non-conservative fields.
Quote from: poynt99 on 2012.04.19, 13:43:11
WW, when your measurement leads are in-plane, what are you measuring, and what are you "working with"?
Then, the loop being measured is that one made up with the metering loop and the side of the loop under test, including that one resistor and the counter voltage created at the junction of those two loops.
I've said this before so I just typed it again to answer your question. Not that you will see it as an answer.
Quote from: exnihiloest on 2012.04.19, 16:45:32
There is no potentials.
Notice is used the word 'measured' not actual. I see no disagreement between us on this matter.
Quote from: GibbsHelmholtz on 2012.04.19, 13:34:45
WaveW,
I still use only 1 method.
dB/dt + V(meter) + I x R segment = 0
Since the meter is close, there is no flux going through the meter loop. Meter reading has to be the same as I x R segment. It's not zero, it is just extremely small.
Ok. So, you are using KVL to determine the induced emf? If true, this explains everything.
I'm am saying that the meter reading is close to zero because the meter circuit divides the induction area into two sections. Those two sections have the same direction around the induction area. Since they are the same direction the meter sees opposite current direction of equal value and results as zero, the joining point for those two sections is the meter circuit. The slight variation from zero is the meter circuit not quite on the same plane as the loop under test.
Quote from: poynt99 on 2012.04.19, 15:15:47
WW,
What is the voltage indicated on CH2 of the scope? Assume 1V emf.
Thanks for the test but I can no longer assume the induced emf is actually 1V.
the bell rang - back to work....
Quote from: WaveWatcher on 2012.04.19, 18:16:29
Ok. So, you are using KVL to determine the induced emf? If true, this explains everything.
I'm am saying that the meter reading is close to zero because the meter circuit divides the induction area into two sections. Those two sections have the same direction around the induction area. Since they are the same direction the meter sees opposite current direction of equal value and results as zero, the joining point for those two sections is the meter circuit. The slight variation from zero is the meter circuit not quite on the same plane as the loop under test.
I must say you have a different way looking at things. I can see it a little bit. My advice is find every situation possible that could break this logic. I think only you can do that.
I don't say your thinking is wrong, I just think it has limitation.
Quote from: WaveWatcher on 2012.04.19, 18:17:53
Thanks for the test but I can no longer assume the induced emf is actually 1V.
When you're out for recess again, consider that the stated induced emf is a "known". That is, accept that "emf" is one given piece of information on the test, then find the unknown as requested.
So,
if the induced emf is 1V, what will CH2 of the scope indicate?
Quote from: poynt99 on 2012.04.19, 17:32:46
You mentioned your own tests in an attempt to prove whether or not an induced E-field can influence, or be influenced by a point charge. THAT is what the following referred to:
I wasn't referring to Gibbs' test question and diagram.
Ok, sorry!
Quote
If you want to measure the induced electric field in its non-integral form, you should use two small metallic balls spaced about 3/4" apart. If you want to measure the induced electric field in its non-integral form, you should use two small metallic balls spaced about 3/4" apart. They have magnet wire leads twisted together, which connect to a scope. Position the balls so that one is closer to the solenoid axis, and the other is radial outward. You will measure a difference in potential across the two balls.
Thuesday I asked the question of the effect of an induced current onto a point charge on 4 physics forums because I was tired of not mastering this subject. The threads are not yet ended but there is some progress.
When the balls are not far from one another, we can possibly show an effect. If we place the balls each side of the hole of the toroid, the effect is much stronger.
The reason that emerges from these experiments is that, in reality, this apparently open circuit is closed between the balls due to the displacement currents. Such circuits with open capacitors are in fact closed circuits. When the displacement currents cross totally or partially the hole of the toroid, the circuit is looped around the magnetic flux, therefore a current can be detected.
In my first experiment, the displacement currents through the toroid hole should have been very weak, question of solid angle that represents the hole viewed from the capacitors, comparing to the whole space where the displacement currents can travel from one capacitor to the other one.
In the experiment that you propose, the position of the balls is better than in mine for the propagation of the displacement currents through the hole because one ball is placed on the axis.
But the conclusion is still that there is strictly no effect when the circuit is looped without flux crossing it. This is a direct consequence of theory: according to Stockes theorem, the surface integral of the curl of a vector field over a surface is linked to the line integral of the vector field over its boundary. As B=rot(A) and B=0, the line integral of the vector E-field on any closed circuit where B=0 is null.
Now that we know that
- we can't expect for an induced current in a closed circuit placed outside a toroid, and
- a "dipole circuits" with terminal capacities is closed thanks to the displacement currents and thus doesn't provide more emf than the part due to the flux through the "leakage circuit" created by the displacement currents through the toroid hole,
well, we have to reinvent a completly new kind of experiments . The ideal would be to observe the effect of a toroid coil on single charges in vacuum. Not easy for a small lab...
Quote from: GibbsHelmholtz on 2012.04.19, 18:40:41
I must say you have a different way looking at things. I can see it a little bit. My advice is find every situation possible that could break this logic. I think only you can do that.
I don't say your thinking is wrong, I just think it has limitation.
I can't take the blame for the thinking. These aren't my theories.
Quote from: WaveWatcher on 2012.04.19, 20:02:14
I can't take the blame for the thinking. These aren't my theories.
Maybe you ought to consider that option....thinking on your own that is.
btw, what are these resolvers you mentioned? Got a link?
Quote from: poynt99 on 2012.04.19, 20:08:56
Maybe you ought to consider that option....thinking on your own that is.
btw, what are these resolvers you mentioned? Got a link?
Forget the resolver. I have doubts you would perceive the connection. You are still on in-plane and normal positions.
Quote from: WaveWatcher on 2012.04.19, 20:26:34
Forget the resolver. I have doubts you would perceive the connection. You are still on in-plane and normal positions.
??? Really? Okiiiiiii. ;)
Looked into "resolvers" and found out they're what I suspected, an angular position sensor.
Yes angle affects induction, but we know that. Not sure WW what you feel I wouldn't or don't understand about this technology that has been used since WW2 and is still being used? ???
Here are a few references on "resolvers" for those that may be interested.
I have one for you Gibbs.
What voltage will be indicated on CH1 and CH2? (Assume as always that the emf=1V)
Quote from: poynt99 on 2012.04.20, 02:25:57
I have one for you Gibbs.
What voltage will be indicated on CH1 and CH2? (Assume as always that the emf=1V)
Well, It should be
CH1 .9V
Ch2 .1V
To be specific, one of them is positive and one of them is negative (sorry, forgot which one is positive probe again).
However, I sense something sinister going on here. lol
Quote from: GibbsHelmholtz on 2012.04.20, 04:18:42
Well, It should be
CH1 .9V
Ch2 .1V
...
Well done!
In fact we just need to add the potential difference at the resistance terminals + the voltage induced in the measurement circuit.
CH1 is directly connected to R900, there is no flux crossing the connecting wires, so V=R900*Vtot/(R100+R900)=0.9v where Vtot=1v.
CH2 is also connected to R900 but there is a complete loop around the flux providing 1v in series and opposition to the 0.9v at R900 terminals. So it remains 1-0.9=0.1v.
I think this method is easy. It is straigforward to get the voltage at the terminals resistance. Once we get it, we add the voltage induced in the measurement circuit, due to the flux crossing its surface.
Quote from: GibbsHelmholtz on 2012.04.20, 04:18:42
Well, It should be
CH1 .9V
Ch2 .1V
To be specific, one of them is positive and one of them is negative (sorry, forgot which one is positive probe again).
However, I sense something sinister going on here. lol
Nothing sinister, just Ohm's law and an understanding of what's going on. ;)
CH1 = 0.9V
CH2 = -0.1V
O0
Quote from: exnihiloest on 2012.04.20, 09:40:09
Well done!
In fact we just need to add the potential difference at the resistance terminals + the voltage induced in the measurement circuit.
CH1 is directly connected to R900, there is no flux crossing the connecting wires, so V=R900*Vtot/(R100+R900)=0.9v where Vtot=1v.
CH2 is also connected to R900 but there is a complete loop around the flux providing 1v in series and opposition to the 0.9v at R900 terminals. So it remains 1-0.9=0.1v.
I think this method is easy. It is straigforward to get the voltage at the terminals resistance. Once we get it, we add the voltage induced in the measurement circuit, due to the flux crossing its surface.
Thanks Exn. and Poynt,
Now I see it. It's not you that make that sinister, it's myself. lol I found a limitation to my method and assumption. I posted another picture here. This one I have to be careful whether I subtract or add voltage. In another word, direction is relevant. Before I just subtract it.
Here's one for you Ex.
What voltage will be indicated on CH1 and CH2?
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=7782;image)
There is no flux crossing between A-R900-D and the wires from CH1, so the situation is the same as if CH1 was directly connected to R900: V1=V*R900/(R100+R900)=0.9v.
There is no flux crossing between A-R100-D and the wires from CH2, so the situation is the same as if CH2 was directly connected to R100: V2=V*R100/(R100+R900)=0.1v.
There is no potential difference between point 1 and 9, or between point 1' and 9'. The potential differences are created only by the resistances. So the point positions of the connexion of the wires between 1 and 9 or between 1' and 9' doesn't a matter when the wires follow the circle (no flux crossing in between).
This is very understandable if we had a superconductor loop: there would be no potential difference anywhere along the circle, in despite the varying flux crossing the circuit, and no voltage could be measured even if we connected the scope wires at diametrically opposed points or anywhere else, provided that the wires follow the circle and are never crossing the disk surface.
You are correct Ex, but the 0.1V is of course -0.1V.
What do you mean by "flux crossing"?
I posed a question a number of posts back, but no one responded; what would be measured across D and A if the loop was made up of a uniform resistance wire, and no resistors? (provide an answer for the in-plane and decoupled cases).
Quote from: poynt99 on 2012.04.21, 13:02:13
You are correct Ex, but the 0.1V is of course -0.1V.
I suppose you mean that the two signals are in opposition. I agree. I don't like the negative notation because we deal with ac currents.
Quote
What do you mean by "flux crossing"?
I posed a question a number of posts back, but no one responded;
Sorry I didn't remark the question. I mean any varying magnetic flux passing through the surface of a circuit.
For example, "no flux crossing between A-R900-D and the wires from CH1" means no flux through the green surface, "no flux crossing between A-R100-D and the wires from CH2" means no flux through the blue surface (see the image). It is implicit only if it is presumed that there is no flux outside the circle.
Quote
what would be measured across D and A if the loop was made up of a uniform resistance wire, and no resistors? (provide an answer for the in-plane and decoupled cases).
Same thing as R100=R900: v=0.5v.
Here is another way to see the problem. We keep the same image and we are interested only in CH1 measurement.
Remove the right half resistive circle between A and D. So the left half resistive circle is connected to a right half circle which now consists of the wires connected to the scope. We still have the same disk surface crossed by the same flux, so the scope measures the full potential difference of 1v, which is the emf because no current flows in the circuit (the scope has a high impedance). Note that all is happening as if the voltage came from a pseudo-generator having an internal resistance R which is that of the left half resistive circle.
Now reconnect the right half resistive circle. The previous emf conditions still apply (same surface crossed by the same flux). So from the viewpoint of the scope, our right half resistive circle appears only as a load. Our pseudo-generator of internal resistance R provides now a current to a load of same value R because the two half circuit have the same resistivity.
Therefore the scope see now half the previous voltage.
Quote from: exnihiloest on 2012.04.21, 15:30:13
I suppose you mean that the two signals are in opposition. I agree. I don't like the negative notation because we deal with ac currents.
Actually, it's pulsed DC, and it produces mainly a pulsed DC response on the loop. The meters are "polarity sensitive" as the good professor said, and therefore there is a positive and a negative response wrt the meters.
Quote
For example, "no flux crossing between A-R900-D and the wires from CH1" means no flux through the green surface, "no flux crossing between A-R100-D and the wires from CH2" means no flux through the blue surface (see the image). It is implicit only if it is presumed that there is no flux outside the circle.
This doesn't make sense.
By this reasoning, if I understand it correctly, means if we take a measurement directly across R100, both from the left side and right side, the two readings should be the same, correct?
Accordingly, flux
never crosses through the measurement wires as long as they are close to the loop. Also, are you assuming a non-infinite length solenoid?
Ex,
I wish you Good Luck. Perhaps your flare for details and formulae will clarify these concepts for us. O0
Quote from: poynt99 on 2012.04.21, 15:53:30
Actually, it's pulsed DC, and it produces mainly a pulsed DC response on the loop. The meters are "polarity sensitive" as the good professor said, and therefore there is a positive and a negative response wrt the meters.
This doesn't make sense.
Induced emf from a pulsed DC is not a pulsed DC. The dc component is lost because only the time variation of the flux generates emf. For instance, a 1µs pulse from 0 to 10v, or from -5v to 5v or from -1000v to -990v and so on will induce exactly the same emf pulse. Only the signal transitions have an effect. It doesn't mean that the induced signal will have positive and negative alternance of equal amplitude (it's true only if the pulse has the same rise and fall time), but that the time integral of the induced pulse will give zero (the sum of the products time duration x voltage =0) although the time integral of the source signal is not null.
Quote
By this reasoning, if I understand it correctly, means if we take a measurement directly across R100, both from the left side and right side, the two readings should be the same, correct?
Accordingly, flux never crosses through the measurement wires as long as they are close to the loop.
Also, are you assuming a non-infinite length solenoid?
I'm not sure to understand what you mean.
a) It is right to assume that there is no flux through the blue area which is between R100 and the CH2 wires, even if CH2 is connected to R100 instead of A and D. So the scope really measures R100 voltage.
b) It is right to assume that there is no flux through the green area which is between R900 and the CH1 wires, if CH1 is connected to R900 instead of A and D. So the scope really measures R900 voltage.
But c) it is wrong to assume that there is no flux between R100 and the CH1 wires, CH1 being connected either to A/D or to R100. Between R100 and the CH1 wires, there is all the flux crossing the disk, which is also the surface of the measurement circuit. So CH1 measures R100 voltage minus 1v which is induced in the measurement circuit. Of course we retrieve the same result 0.9v or -0.9v depending on the sign convention, but it is easier to consider case b) to get the solution.
In other words, we can consider either that CH1 measures directly the voltage at R900 because there is no flux crossing the area between R900 and the scope wires, or that it measures R100 but with a correction, because we must take account of the induced emf in the circuit measurement which in this view, is now enclosing all the flux.
Quote from: exnihiloest on 2012.04.21, 20:34:16
Induced emf from a pulsed DC is not a pulsed DC. The dc component is lost because only the time variation of the flux generates emf. For instance, a 1µs pulse from 0 to 10v, or from -5v to 5v or from -1000v to -990v and so on will induce exactly the same emf pulse. Only the signal transitions have an effect. It doesn't mean that the induced signal will have positive and negative alternance of equal amplitude (it's true only if the pulse has the same rise and fall time), but that the time integral of the induced pulse will give zero (the sum of the products time duration x voltage =0) although the time integral of the source signal is not null.
You're splitting hairs here Ex. Reference the Lewin experiment, a DC source is used to pulse the coil. Consequently, there is a corresponding unipolar response on the loop (polarity dependent on meter connection). i.e. the current does not change direction, it rises, then falls back to zero.
Quote
I'm not sure to understand what you mean.
a) It is right to assume that there is no flux through the blue area which is between R100 and the CH2 wires, even if CH2 is connected to R100 instead of A and D. So the scope really measures R100 voltage.
b) It is right to assume that there is no flux through the green area which is between R900 and the CH1 wires, if CH1 is connected to R900 instead of A and D. So the scope really measures R900 voltage.
But c) it is wrong to assume that there is no flux between R100 and the CH1 wires, CH1 being connected either to A/D or to R100. Between R100 and the CH1 wires, there is all the flux crossing the disk, which is also the surface of the measurement circuit. So CH1 measures R100 voltage minus 1v which is induced in the measurement circuit. Of course we retrieve the same result 0.9v or -0.9v depending on the sign convention, but it is easier to consider case b) to get the solution.
In other words, we can consider either that CH1 measures directly the voltage at R900 because there is no flux crossing the area between R900 and the scope wires, or that it measures R100 but with a correction, because we must take account of the induced emf in the circuit measurement which in this view, is now enclosing all the flux.
Interesting approach. It happens to work, but it is not the correct way to analyze this in-plane scenario.
For some odd reason, many here seem obsessed with the solenoid "flux". The flux outside the solenoid is zero, so according to you Ex, the measurement leads will
never be influenced by the solenoid "flux" (and you would be correct). The "correction factor" you propose is nonsense. The scope indicates what is there across the two points on the loop. We never ask or ponder whether the measurement is relative to the R100 or R900 resistor ???, the measurement simply IS what it IS. The scope certainly doesn't care.
I've stressed this a hundred times it seems, but it does not sink in;
forget about the flux, it is simple a means to an end, and that END is the resulting induced electric field. The E-field is what counts. No E-field, no induced emf. The E-field is all around the solenoid, inside and outside, the flux is internal only to the solenoid. The flux has no direct influence on the loop or the measurement leads, ever.
Quote from: poynt99 on 2012.04.21, 21:25:55
...
I've stressed this a hundred times it seems, but it does not sink in; forget about the flux, it is simple a means to an end, and that END is the resulting induced electric field. The E-field is what counts. No E-field, no induced emf. The E-field is all around the solenoid, inside and outside, the flux is internal only to the solenoid. The flux has no direct influence on the loop or the measurement leads, ever.
...
But the flux is what creating the E field (for this set up). I don't think it just happens to work. It would works with many more complex in-plane case. I think the E field you're speaking of is the same as the flux method. Can you use your method to solve this problem I posted a while ago? Perhaps we can find a relationship.
Quote from: poynt99 on 2012.04.21, 21:25:55
You're splitting hairs here Ex. Reference the Lewin experiment, a DC source is used to pulse the coil. Consequently, there is a corresponding unipolar response on the loop (polarity dependent on meter connection). i.e. the current does not change direction, it rises, then falls back to zero.
I'm not splitting hairs, I justified what I said. I say that "+" or "-" are not relevant for signals with a null dc component and I explain why it is the case here.
An induced current never rises and fall back to zero. It rises and fall back below zero at some time, otherwise there would be a dc component. If transformers could pass DC, it would be known for a long time!
Quote
For some odd reason, many here seem obsessed with the solenoid "flux". The flux outside the solenoid is zero, so according to you Ex, the measurement leads will never be influenced by the solenoid "flux" (and you would be correct).
This idea is not according to me, but to your interpretation of what others say, including me, which is different.
The Faraday's law doesn't imply that the circuit be in the flux. The Faraday's law stipulates only that the surface of the circuit must be crossed somewhere by a varying flux. The flux can cross the surface even through only 1% of the surface and anywhere. If your loop is at 1km from the solenoid, still around, the same emf will be induced.
Quote
The "correction factor" you propose is nonsense. The scope indicates what is there across the two points on the loop. We never ask or ponder whether the measurement is relative to the R100 or R900 resistor ???, the measurement simply IS what it IS. The scope certainly doesn't care.
"The measurement simply IS what it IS": yes, I agree. The measurement that you imagine by connecting CH1 to R100 while the wires are enclosing the disk surface is the measure of the potential difference at the resistance terminals + the emf induced in your measurement circuit, because the total flux of the solenoid is crossing your measurement circuit.
It is what it is: 0.9v, which is a real measurement that is what it is, but naturally not the potentiel difference felt by the R100 resistance.
Of course you will not get the same measurement by connecting CH2 to R100 by the other side, which gives the real potential difference at R100.
The nonsense is only to imagine that both measurements should give the same result.
My "correction factor" was only to obtain the potential difference at R100 terminal from the measurement of CH1 (in the same way that when you measure the potential difference at the terminals of two identical resistances in series, you can use a "correction factor" to obtain the potential difference at only one: by dividing by two).
Quote
I've stressed this a hundred times it seems, but it does not sink in; forget about the flux, it is simple a means to an end, and that END is the resulting induced electric field. The E-field is what counts. No E-field, no induced emf. The E-field is all around the solenoid, inside and outside, the flux is internal only to the solenoid.
You can get the same results by using the E-field, as by using the flux. But you can't oppose one method to the other, because the two methods are strictly equivalent due to the Stokes theorem. If one failed, the other too. I said it somewhere else but you missed the point:
"According to Stockes theorem,
the surface integral of the curl of a vector field over a surface is linked to the line integral of the vector field over its boundary.".
Demonstration that they are the same:
Emf = -dΦ/dt (Faraday's law = flux crossing method)
Φ = ∫∫B.n.dS (definition of the flux, B and n are vectors, n is the unit vector, normal to the circuit surface S).
(1) Emf = -dΦ/dt = -∂(∫∫B.n.dS)/∂t = -∫∫∂B/∂t .n.dSEmf = ∮E.dl (integral of the electric field over the circuit = E-field method, l is refering to the circuit path).
∮E.dl =∫∫(∇xE).n.dS (Stokes theorem applied to the E-field, S being the surface enclosed by l)
and
∇xE = -∂B/∂t (Maxwell)
so
∮E.dl = -∫∫(∂B/∂t).n.dS = equation (1) = -dΦ/dtQED
The only interest of the E-field, is that it gives the local field acting on the electrons, and therefore the force F=q.E which is the cause of their motion. On the contrary, if we supposed that the flux would be the cause of the current, the phenomenon would be a non local (therefore outside classical physics) because there is no flux at the position of the wire enclosing the solenoid. Nevertheless in our approximation of the quasi-stationary states (we suppose that there is no EM waves), the both methods are strictly equivalent to calculate the emf but the method by the flux is by far the simpler.
Quote
The flux has no direct influence on the loop or the measurement leads, ever.
I agree, but the flux variation through the circuit surface is an event concomitant with the local E-field along the circuit, and so, there is no discrepancy in using one method or the other, as proved by the equations.
Quote from: GibbsHelmholtz on 2012.04.22, 00:07:31
But the flux is what creating the E field (for this set up).
[...]
I think the E field you're speaking of is the same as the flux method.
I confirm the second affirmation. For the first one I would rather say that it is the motion of charges in the solenoid coil that creates both the flux inside and the E-field outside. This is disputable. One can oppose that a flux variation can be obtained from a moving permanent magnet. But even in this case, the magnet field coming from electron spins and orbitals, there are still moving charges.
Very smart proposition, Gibbs! I confess that I have been very surprised that we should get more than 1v, according to my previous method. We should have 1v + 0.9v = 1.9v!
And we will, the method works: it is finally easy to see it, by removing R900 and to follow a wire from the scope: the circuit has 2 turns before returning to the scope.
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=7798)
Quote from: exnihiloest on 2012.04.22, 16:34:24
Very smart proposition, Gibbs! I confess that I have been very surprised that we should get more than 1v, according to my previous method. We should have 1v + 0.9v = 1.9v!
And we will, the method works: it is finally easy to see it, by removing R900 and to follow a wire from the scope: the circuit has 2 turns before returning to the scope.
Thanks for the E field equivalent analysis, Exn.
Yes, you can remove the R900 and see it has 2 turns, but that's not the only way. You can pick the path through the R900 and it only has 1 turn. I've come up with a rule using E field.
Pick a path
Count the number of turns it covers the flux
Multiply the Emf(1V) by the number of turns
Determine the direction of emf
Add or subtract resistance voltage according to the Emf polarity
Let's say you pick the path through R100, now you have 2 turns. The total induce emf in those two turns is 2V. Now you have to add or subtract the R100 . If you follow the whole loop, you should see - + + - . This tells you to subtract. 2V - .1V = 1.9V
Let's say you pick the path through R900, now you have 1 turn. Total induced emf in the 1 turn is 1V . Now you have to add or subtract the R900. If you follow the whole loop, you see + - + - . This tells you to add. 1V + .9V = 1.9V
I agree 100%, gibbs. Now it becomes easy. O0
I would like now to go further with the question of whether or not it is possible to show an effect of an induced E-field on charges which are not in a circuit enclosing a varying flux.
I think that this thread is well intended for such a question.
We know for sure, provided that we accept Maxwell's equation ∇xE=-∂B/∂t, that there is no emf in a closed circuit whose the surface is not crossed by a varying flux, even if it lies in a not null induced E-field: near a toroid coil it is impossible to detect a current in a closed circuit without conductor passing through the toroid hole. This means that the contribution of the E-field at some places along the circuit cancels its contribution at some other places.
But we are not restricted to closed circuits. What with open circuits? I have tried in an already mentionned experiment with a piece of wire terminated by terminal capacitors at each end. But I couldn't detect any current going back and forth in my setup. I guess that the capacitors form a closed circuit due to the displacement currents between them. As the circuit is looped and there is no flux crossing, there is no emf.
This seems to be a problem but it opens other great prospects: could we engineer the path of displacement currents closing a circuit in order to modulate an enclosing flux and therefore to generate induced emf?
Therefore the question is double and both aspects being related to unknown domains (at my knowledge) could be promising.
Quote from: exnihiloest on 2012.04.23, 07:37:37
I agree 100%, gibbs. Now it becomes easy. O0
I would like now to go further with the question of whether or not it is possible to show an effect of an induced E-field on charges which are not in a circuit enclosing a varying flux.
I think that this thread is well intended for such a question.
We know for sure, provided that we accept Maxwell's equation ∇xE=-∂B/∂t, that there is no emf in a closed circuit whose the surface is not crossed by a varying flux, even if it lies in a not null induced E-field: near a toroid coil it is impossible to detect a current in a closed circuit without conductor passing through the toroid hole. This means that the contribution of the E-field at some places along the circuit cancels its contribution at some other places.
But we are not restricted to closed circuits. What with open circuits? I have tried in an already mentionned experiment with a piece of wire terminated by terminal capacitors at each end. But I couldn't detect any current going back and forth in my setup. I guess that the capacitors form a closed circuit due to the displacement currents between them. As the circuit is looped and there is no flux crossing, there is no emf.
This seems to be a problem but it opens other great prospects: could we engineer the path of displacement currents closing a circuit in order to modulate an enclosing flux and therefore to generate induced emf?
Therefore the question is double and both aspects being related to unknown domains (at my knowledge) could be promising.
Yes, this is a hot topic. I think part of the reason you cannot have good detection is because of the weak signal. I've done this experiment in a different way. I have a plasma ball and tried to detect an E field using my volt meter. The set up is as below. As I varies the distance between them, the voltage drop. Reverse the lead the voltage reverse polarity. There is also a video that could be similar to this experiment. He used LED to detect this E field. I believe this E field also a displacement current, but 90 degrees to the displacement current of a closed loop containing magnetic flux.
http://www.youtube.com/watch?v=7TlyO9o-fSo
I'm not sure that this kind of setup detects an induced E-field. A double-diode-led detector circuit lights strongly when approaching the hot end of a coil and switches off on the other side. We should not have this effect by detecting a real induced E-field, which is uniformly distributed around a coil and symmetrically between the two ends.
For me the detector is a capacitively coupled to the hot part of the coil which is usually resonant thus building a high voltage. So only an ordinary electric field is detected, although there are real displacements currents from the coil to the dectector.
If we make a too sensitive device, it will always detect something. However the induced E-field is strong. If we have 1v/turn in a long solenoid coil, and we consider a circuit around of circumference 10 cm, this means that the E-field is 10v/m, which should be large enough to be easily detected.
In an old experiment intended for another goal, I had realized a coaxial coil and powered it with an HF current. The coaxial cable was to shield the capacitive coupling. I could light a low energy lamp bulbs at a distance:
http://exvacuo.free.fr/div/Sciences/Experiences/Induction/Lampe%201.jpg
http://exvacuo.free.fr/div/Sciences/Experiences/Induction/Lampe%20Ensemble.jpg
But
1) only one end of the shield of the coaxial cable was connecting to the ground, otherwise the induced emf in the shield would have been shortcut, and consequently it would have acted as a shorcut transformer secundary. But by connecting only one end, the induced E-field in the shield makes it acquiring the same voltage as the inner conductor. So this way I had not eliminated the capacitive coupling.
2) I don't know the physical principle that lights the lamp. At the beginning the gas is not ionized, so a magnetic field has no action. Could it be the induced E-field that excites the electrons? Or simply the electrical field due to the HV section of my resonant coi,l capacitively coupled to the lamp?
3) The lamp could be lit whatever its position around the coil. The frequency was high (some Mhz). So the last problem is that there are likely electromagnetic waves. A power is radiated, so I don't know exactly what this experiment can really show, expect a nice effect.
Quote from: exnihiloest on 2012.04.23, 14:59:00
I'm not sure that this kind of setup detects an induced E-field. A double-diode-led detector circuit lights strongly when approaching the hot end of a coil and switches off on the other side. We should not have this effect by detecting a real induced E-field, which is uniformly distributed around a coil and symmetrically between the two ends.
For me the detector is a capacitively coupled to the hot part of the coil which is usually resonant thus building a high voltage. So only an ordinary electric field is detected, although there are real displacements currents from the coil to the dectector.
If we make a too sensitive device, it will always detect something. However the induced E-field is strong. If we have 1v/turn in a long solenoid coil, and we consider a circuit around of circumference 10 cm, this means that the E-field is 10v/m, which should be large enough to be easily detected.
In an old experiment intended for another goal, I had realized a coaxial coil and powered it with an HF current. The coaxial cable was to shield the capacitive coupling. I could light a low energy lamp bulbs at a distance:
http://exvacuo.free.fr/div/Sciences/Experiences/Induction/Lampe%201.jpg
http://exvacuo.free.fr/div/Sciences/Experiences/Induction/Lampe%20Ensemble.jpg
But
1) only one end of the shield of the coaxial cable was connecting to the ground, otherwise the induced emf in the shield would have been shortcut, and consequently it would have acted as a shorcut transformer secundary. But by connecting only one end, the induced E-field in the shield makes it acquiring the same voltage as the inner conductor. So this way I had not eliminated the capacitive coupling.
2) I don't know the physical principle that lights the lamp. At the beginning the gas is not ionized, so a magnetic field has no action. Could it be the induced E-field that excites the electrons? Or simply the electrical field due to the HV section of my resonant coi,l capacitively coupled to the lamp?
3) The lamp could be lit whatever its position around the coil. The frequency was high (some Mhz). So the last problem is that there are likely electromagnetic waves. A power is radiated, so I don't know exactly what this experiment can really show, expect a nice effect.
Actually, this is also one of my concern about this effect. Is it an induced E field or an E field from HV hot end. The E field according to our thinking would be uniformly distributed as you mentioned. I think I'll agree that it is HV hot end. If so, the capacitive coupling would link back to ground end. The experiment you performed as well as Tesla coil wireless lighting would also share the same concept. We can consider this as a closed circuit except in the form of extremely low current high voltage.
Quote from: GibbsHelmholtz on 2012.04.24, 00:28:47
Is it an induced E field or an E field from HV hot end.
I must say, this is a very good way to describe the reality of the experiments offered as 'not being from RF energy'.
We all know that charges accumulate best on physical points. While the battery and other larger surface objects provided little area for charge accumulation, either side may be a small point for accumulation. Both sides pointed is not required.
The 'AV plug' provides the pointed side of unbalanced charges so highest potential accumulates there. The circuit is closed due to capacitive coupling. I watched the whole series hoping a new proof would emerge. The only proof I saw was that the experimenter still doesn't have support for his claims.
The question I have left is:
If we induced 10kV into a uniform loop of wire... Where is the hot end? lol
In practice, it is not so difficult than it looks. It depends on which end is the most coupled to the ground and which is the most coupled to the detector. For example the cold point of a big vertical Tesla coil would be the side near the ground, even if this coil end was not connected to the ground.
Generally the primary coil that induces the current determines the cold side because it is capacitively coupled to the secondary coil and it is at a low potential. If it is wound at the middle of an HV winding, then the cold point is the middle and there are 2 hot ends with a voltage half the HV.
It must be concluded that a Faraday cage is needed if we want perform correct measurements of induced E-field.
Another flaw to consider is the magnetic field which could disturb a pure induced E-field measurement. As a Faraday cage doesn't shield magnetic fields, only a toroid coil with a high permeability core, not saturated, can guarantee the lowest leakage of magnetic field.
I'm just thinking if we have a loop with current flowing, is there a hot end since hot and cold tailing each other. What if the current increasing or decreasing? This is how induction manifest and it seems like a hot and cold end is needed for induction.
There is only a potential difference between the two ends. The coupling with the vicinity determines the potential along the line, relative to the ground, the section of highest potential becoming the hot spot.
A potential is always relative.
Quote from: GibbsHelmholtz on 2012.04.22, 00:07:31
But the flux is what creating the E field (for this set up).
So what? Irrelevant. The flux has about the same relevance as the burning coal does in a steam engine. Neither of them ultimately do the work. They are simply a step in the energy conversion process. Ultimately, it is the E-field that does the work on the loop charges, and it is only the E-field that is far-reaching outside the solenoid.
Quote
I don't think it just happens to work. It would works with many more complex in-plane case. I think the E field you're speaking of is the same as the flux method. Can you use your method to solve this problem I posted a while ago? Perhaps we can find a relationship.
If your flux method works and you are happy with it, good for you guys ;) I'm going to continue applying the E-field for any analysis, as both the loop and the measurement leads measure the line integral of the E-field alone.
Quote from: exnihiloest on 2012.04.22, 15:44:06
I'm not splitting hairs, I justified what I said. I say that "+" or "-" are not relevant for signals with a null dc component and I explain why it is the case here.
An induced current never rises and fall back to zero. It rises and fall back below zero at some time, otherwise there would be a dc component. If transformers could pass DC, it would be known for a long time!
Yeah, you were,
and you're doing it again.
I think you know very well what I mean, and there's nothing incorrect about what I said. C.C
Quote from: poynt99 on 2012.04.25, 22:55:20
...
I think you know very well what I mean,
Certainly not. I'm intellectually honest. What you say here is to judge someone on presumed intent.
Quote
and there's nothing incorrect about what I said. C.C
Maybe what you mean is correct, but I know only what you say and what you say is not correct. In physics things must be clearly said, without ambiguity. It's not a game where everyone would have to guess what others really mean.
When you assert: "the current does not change direction, it rises, then falls back to zero", this is
false for an induced current. And I say it, question of truth. An induced emf always rises above zero and then decreases under zero, perhaps several times if it oscillates, and the integral of emf over time is always zero, meaning that there is no DC component.
If you meant other thing that what you said and that I have reformulated here, then you should say it more clearly. If you don't agree with what I say, I would appreciate that you say why, in terms of electromagnetism.
Quote from: poynt99 on 2012.04.25, 22:51:55
...
it is the E-field that does the work on the loop charges, and it is only the E-field that is far-reaching outside the solenoid.
If your flux method works and you are happy with it, good for you guys ;) I'm going to continue applying the E-field for any analysis, as both the loop and the measurement leads measure the line integral of the E-field alone.
I agree with your first point: the local E-field is the causal explanation for the motion of charges, so it's the best.
But you must also admit that the flux variation is a concomitant event and that the knowledge of the flux is an easier way to calculate the emf. The most important:
the equality between the integration of the E-field along the circuit and that of the flux through the circuit surface, both giving the emf, is not a physical principle, it is a direct implication based only on logical mathematical arguments applied to Maxwell equations (Stokes theorem). Accepting Maxwell's equations implies to accept both methods to calculate the emf: one of them can't be true while the other would be false, without logical fallacy.
Quote from: poynt99 on 2012.04.25, 22:51:55
If your flux method works and you are happy with it, good for you guys ;) I'm going to continue applying the E-field for any analysis, as both the loop and the measurement leads measure the line integral of the E-field alone.
Ya know what... I like that attitude. lol Someone has to go against the flow. lol As long as you understand both sides of the argument, you are free to choose any path. I don't recommend everyone think the same thing, that's how things come to an end. ;)
Quote from: exnihiloest on 2012.04.26, 06:55:07
When you assert: "the current does not change direction, it rises, then falls back to zero", this is false for an induced current. And I say it, question of truth. An induced emf always rises above zero and then decreases under zero, perhaps several times if it oscillates, and the integral of emf over time is always zero, meaning that there is no DC component.
Ex,
Draw out what V(D-A) looks like vs. time on a graph and post it. Use the actual voltage across the two points, i.e. what would be measured with decoupled leads. Assume same conditions as Lewin's experiment; step function on input to solenoid (as I have previously prescribed), and 1V induced emf. Actual time is not important, just draw out the basic shape of V(D-A) until it reaches equilibrium.
Now, also plot out the loop current. Assume induced i(t) is 1mA peak.
We may be talking about two different things, and this exercise will sort that out.
How about a response Ex? ;)
Quote from: poynt99 on 2012.05.01, 12:27:56
How about a response Ex? ;)
Before going further, I would like to understand what is your goal. Is it to affirm that, when a constant voltage is abruptly applied to a solenoid, there is an induced emf whose the mean voltage is a not null DC?
The equation is simple:
L*di/dt + R*i(t) = V * h(t)
L is the inductance, R its resistance, V the voltage and h(t) the Heaviside step function h(t)=0 for t<0 and h(t)=1 for t>=0.
Just before t=0, i=0.
Just after: L*di/dt + R*i(t) = V whose the solution for i is well known:
i = V/R * (1 - e
-t/τ) where τ=
R/L L/R
Only the variation of i produces emf. If we suppose a perfect coupling with the solenoid: emf = k * di/dt.
di/dt =d/dt (V/R * (1 - e
-t/τ)) = V/R * d/dt (- e
-t/τ) = V/R * 1/τ * e
-t/τ = V/L * e
-t/τAs e
-t/τ is always >0, so is di/dt and emf. Your goal is correct.
I must admit that my affirmation pretending the contrary made an implicit assumption that I should have given: it is true only when the initial and final conditions are the same (for example i=0, or i=anything but the signal is periodic) which was not the case here.
Do you agree?
I just still ask me a final question about the t<0 to t=0 transition, as modelized in the equation. Is it physical? I think it is but I have some doubt.
My goal is to establish what your understanding is (and if it is the same as mine) of what would be observed on an oscilloscope with the leads placed across nodes D and A using a decoupled method. The input condition is a step function (step and hold) from 0 to some value that produces an induced emf of 1Vp in our 100/900 Ohm loop. Please plot the response across A and D and the loop current.
Read the post again for details of what I am asking if you are uncertain.
It's not ill will but I'm a bit confused. I didn't follow all the story and there are now hundreds posts. I would appreciate that you synthesize the question and put links to what is precisely on the stack for you, schematics and so on... otherwise I will be unable to reply because I could not reread the hundreds of posts of this thread for guessing which are of help.
Huh? ???
Just re-read post #575 above (your quote) and my response and question. That should be sufficient for you to draw out the wave forms. Nothing has changed, it's the same experiment we've been discussing for a while now.
Quote from: poynt99 on 2012.05.02, 21:04:24
Huh? ???
Just re-read post #575 above (your quote) and my response and question. That should be sufficient for you to draw out the wave forms. Nothing has changed, it's the same experiment we've been discussing for a while now.
Hello Houston - We have a problem.
Which schematics is "V(D-A)" referring to? Is it the first post? post #17? Which "same experiment"? You produced so many problems with variants of measurement. You would like that I follow a step by step procedure but I don't know why... ???
Communication problem, we are not on the same wave length. Hello, hello, Poynt, I don't hear you, re-tune, clarify, please.
The Lewin experiment is not much different than a standard air-core transformer, so all the same theories apply.
Have you forgotten what our argument was about over only the last several postings in this thread?
It applies to transformers too, so if this makes it easier for you, simply plot out the voltage across the terminals of a 1k Ohm terminated air-core secondary when the primary is excited by a step and hold input voltage. 1Vp is induced in the secondary.
You DO remember what our argument was about I trust? I'm asking you to plot this out to see if you have the same understanding as me, because your responses thus far (excluding your recent ones claiming you don't know what I'm referring to) indicate that you don't.
Stop playing silly-bugger Ex, and just answer the question. C.C
I will reply later if I can. ADSL here is out of order due to cables theft in my area (for re-selling the copper). The repair is announced for next Monday. I'm using now a cellular phone as modem but it's very expensive...
Quote from: poynt99 on 2012.05.03, 18:17:13
The Lewin experiment is not much different than a standard air-core transformer, so all the same theories apply.
Have you forgotten what our argument was about over only the last several postings in this thread?
It applies to transformers too, so if this makes it easier for you, simply plot out the voltage across the terminals of a 1k Ohm terminated air-core secondary when the primary is excited by a step and hold input voltage. 1Vp is induced in the secondary.
Stop playing silly-bugger Ex, and just answer the question. C.C
I don't play. I have not to "just answer the question", without knowing why I should answer. It is surprising that you refuse to give the reasons of your question. In any case, see the attached picture for the answer.
I shall not answer any other question that doesn't make sense for me.
Quote from: exnihiloest
ADSL here is out of order...
It should work only this evening. I am exasperated by the delay, so I post from my office.
Quote from: exnihiloest on 2012.05.10, 07:15:36
I don't play. I have not to "just answer the question", without knowing why I should answer. It is surprising that you refuse to give the reasons of your question. In any case, see the attached picture for the answer.
I shall not answer any other question that doesn't make sense for me.
It's strange that you're most likely the only one in this thread that did not know what the question was about or why it was being asked, especially since the pertinent discussion was between you and I. ???
It all started with my statement HERE (http://www.overunityresearch.com/index.php?topic=739.msg22890#msg22890) regarding the loop response as seen by the scope.
Then your refutation 2 posts later, HERE (http://www.overunityresearch.com/index.php?topic=739.msg22896#msg22896).
Then I elaborate further in an attempt to clarify, HERE (http://www.overunityresearch.com/index.php?topic=739.msg22897#msg22897). Quoted:
Quote from: poynt99 on 2012.04.21, 21:25:55
You're splitting hairs here Ex. Reference the Lewin experiment, a DC source is used to pulse the coil. Consequently, there is a corresponding unipolar response on the loop (polarity dependent on meter connection). i.e. the current does not change direction, it rises, then falls back to zero.
Which you then attempt to refute again, HERE (http://www.overunityresearch.com/index.php?topic=739.msg22910#msg22910). To quote you:
Quote from: exnihiloest on 2012.04.22, 15:44:06
I'm not splitting hairs, I justified what I said. I say that "+" or "-" are not relevant for signals with a null dc component and I explain why it is the case here.
An induced current never rises and fall back to zero. It rises and fall back below zero at some time, otherwise there would be a dc component. If transformers could pass DC, it would be known for a long time!
So now look at your nice graph (which is correct) you just posted. Do you see your error now and that I was correct in what I had said? It should also be clear to you now why I asked you to graph out the voltage and current.
Quote from: poynt99 on 2012.05.10, 13:44:43
...
Do you see your error now and that I was correct in what I had said?
I had already said it in post #577, with the math!
"As e-t/τ is always >0, so is di/dt and emf.
Your goal is correct [to affirm
"an induced emf whose the mean voltage is a not null DC"]. I must admit that my affirmation pretending the contrary made an implicit assumption that I should have given: it is true only when the initial and final conditions are the same (for example i=0, or i=anything but the signal is periodic)
which was not the case here."
And I asked you: "Do you agree?". No reply. I understand now that you missed the point of my answer 577 and asked again the same question until you get a picture, while both of us agreed. Waste of time.
Ex,
To put it plainly, you said: "An induced current never rises and fall back to zero".
With the step function input (as per the subject matter of this thread from the beginning), the secondary most certainly does.
AND, there ARE two polarities of voltage as measured according to the experimental setup.
I watch this thread with interest and now I see a question pertaining to applying DC to a transformer - and leaving it connected?
Surely it is known that once the current is removed the current will fall below zero and oscillate above an below for a short period? The same goes for primary and secondary.
If you apply a steady DC current to the primary the secondary will a rise in current and then a gradual fall to zero - and not beyond - until the switch is opened.
In any case, the complete cycle of on-off will produce AC. Even the least reduction of primary applied DC, after the applied peak it reached, will cause the secondary to see a reverse in current.
Why the argument?
In terms of the Lewin experiment, the pertinent interest is what happens from t=0 to the point e-induced falls to zero. Even this latter "falling" part is of limited interest, but showing the complete "half-cycle" makes things look nice and provides a sense of what is going on in terms of the input stimulus and how induction works.
Seems there was simply a misunderstanding of what I said which started this all off.
Quote from: poynt99 on 2012.05.10, 21:25:10
...
To put it plainly, you said: "An induced current never rises and fall back to zero".
Yes I did. It was in post #557.
But I just told you that
in my post #577 I admitted that it was a mistake in the general case, including our particular case of this thread, by stipulating that "an induced current never rises and fall back to zero" is true only when the start and final values of the primary current are the same. It should now be definitely clear.
It is surprising that the time constant depends only on the resistance of the primary, not on the load of the secondary provided that it is much larger. Also surprisingly, the greater the resistance, the lower the time constant.
With a primary circuit resistance of 0.001 Ohm, the time constant is on the order of s. With 1 ohm resistance, it is on the order of ms.
(The input voltage rises to 1v at t=0 and is kept constant after).
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=7950)
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=7952)
Quote from: exnihiloest on 2012.05.11, 12:46:04
It is surprising that the time constant depends only on the resistance of the primary, not on the load of the secondary provided that it is much larger. Also surprisingly, the greater the resistance, the lower the time constant.
It's not surprising if you know the equations for the time constant "tau":
For resistive/capacitive, tau=RC
For resistive/inductive, tau=L/R
Tks for the recall, Poynt. I wrongly wrote τ=R/L in reply #577 (without consequence). I have corrected.
So now it's not surprising... :(
Quote from: exnihiloest on 2012.05.11, 12:46:04
It is surprising that the time constant depends only on the resistance of the primary, not on the load of the secondary provided that it is much larger. Also surprisingly, the greater the resistance, the lower the time constant.
With a primary circuit resistance of 0.001 Ohm, the time constant is on the order of s. With 1 ohm resistance, it is on the order of ms.
(The input voltage rises to 1v at t=0 and is kept constant after).
I found that time constant depends on resistance of secondary, however, only at certain resistance. The lower the resistance, the lower the time constant. lol
frequency change sim (http://www.falstad.com/circuit/#%24+1+5.0E-6+9.78399845368213+21+5.0+50%0Ac+176+176+112+176+0+9.999999999999999E-5+2894.9461100479775%0Aw+304+208+304+176+0%0Aw+304+176+176+176+2%0Aw+304+272+304+320+0%0AT+304+208+384+272+0+4.0+1.0+-17.079398690230796+-2.8906862898772454+0.999%0Ar+112+176+112+96+0+1.0E-4%0Av+112+96+304+96+0+0+40.0+100000.0+0.0+0.0+0.5%0As+304+96+304+176+0+1+false%0As+112+176+112+320+0+0+false%0As+384+208+464+208+0+0+false%0Aw+464+272+384+272+0%0Aw+304+352+112+352+0%0Aw+112+352+112+320+0%0Ar+464+272+464+208+0+1000.0%0As+464+208+544+208+0+1+false%0Ar+544+272+544+208+0+1.0%0Aw+544+272+464+272+0%0Aw+304+352+304+320+0%0Ao+0+64+0+35+76624.77704329444+12259.964326927111+0+-1%0Ao+13+64+0+35+655360.0+819.2+1+-1%0Ao+15+64+0+35+10240.0+13107.2+2+-1%0A)
Quote from: GibbsHelmholtz on 2012.05.11, 20:37:43
I found that time constant depends on resistance of secondary, however, only at certain resistance. The lower the resistance, the lower the time constant. lol
...
It depends on the order of magnitude of the resistance of secondary compared to that of the primary, and of the mutual coupling coefficient. If the resistance of secondary is of the same order or lower than the resistance of the primary, the former can't be negligible when viewed from the primary through the coupling, and so the apparent resistance of the primary is lower. The lower the resistance, the
higher the time constant, according to the simulation and math.
In the simulation, with a coupling=0.9, τ= around 1ms for Rout=10*Rin, 2ms for Rout=Rin, 6ms for Rout=Rin/10, 10ms for Rout=Rin/100.
With a coupling=0.1, τ= around 1.1ms for Rout=10*Rin, 3ms for Rout=Rin, 8ms for Rout=Rin/10, 13ms for Rout=Rin/100.
Quote from: exnihiloest on 2012.05.12, 08:35:54
It depends on the order of magnitude of the resistance of secondary compared to that of the primary, and of the mutual coupling coefficient. If the resistance of secondary is of the same order or lower than the resistance of the primary, the former can't be negligible when viewed from the primary through the coupling, and so the apparent resistance of the primary is lower. The lower the resistance, the higher the time constant, according to the simulation and math.
In the simulation, with a coupling=0.9, τ= around 1ms for Rout=10*Rin, 2ms for Rout=Rin, 6ms for Rout=Rin/10, 10ms for Rout=Rin/100.
With a coupling=0.1, τ= around 1.1ms for Rout=10*Rin, 3ms for Rout=Rin, 8ms for Rout=Rin/10, 13ms for Rout=Rin/100.
OKay, I see what you mean now. You looking at the time constant for the secondary. If we take our view broader, you will see that the time constant for the secondary is the opposite to the time constant for the primary.
My simulation involves a bit more than primary and secondary. It also involves time constant for capacitor and condition at resonance. I've simulate the increase in frequency in primary circuit by changing secondary resistance. I believe this is the Thane's effect ...
simulated. :o
Primary and secondary being coupled, there is only one time constant because coil currents are linked to the variation of a unique magnetic flux. E=-dφ/dt and phi is the same for the two coils.
This time constant depends on both resistances.
Quote from: exnihiloest on 2012.05.13, 09:41:39
Primary and secondary being coupled, there is only one time constant because coil currents are linked to the variation of a unique magnetic flux. E=-dφ/dt and phi is the same for the two coils.
This time constant depends on both resistances.
It's difficult to express my view in this matter. Let me start with the graph below. The red curve is primary current, blue curve is secondary current.
The first case where the blue curve go up with the red is where secondary R<<L secondary. The second case where the blue curve drop as primary current max is where secondary R>>L secondary. Notice also the rise time for both cases is significantly different.
We can say that one is in phase and one is out of phase. The in phase where R is so high has no time constant.
I have not such graphs. In mines all is happening as expected. I measure the voltage, not the coils current. What do you measure? Could you give us the schematics?
Here's the sim.
Current is observed from primary and secondary resistance.
Circuit Simulator Applet (http://www.falstad.com/circuit/#%24+1+5.0E-6+27.727228452313398+21+5.0+50%0Av+144+160+256+160+0+0+40.0+5.0+0.0+0.0+0.5%0Ar+144+160+144+272+0+1.0%0AT+256+160+336+272+0+4.0+1.0+-1.169394647844953E-19+1.1705652130591094E-19+0.999%0As+336+160+432+160+0+1+false%0Ar+432+272+432+160+0+100.0%0As+432+160+512+160+0+1+false%0Ar+512+272+512+160+0+1.0%0Aw+336+272+432+272+0%0Aw+432+272+512+272+0%0As+144+272+256+272+0+1+false%0Ao+1+64+0+35+7.62939453125E-5+9.765625E-5+0+-1%0Ao+4+64+0+35+7.62939453125E-5+9.765625E-5+0+-1%0Ao+6+64+0+35+7.62939453125E-5+9.765625E-5+1+-1%0A)
The simulator applet gives two lines that evolve in unison. I don't retrieve your previous graph with the inverse slopes of the second view.
In the previous week I came across a paper about quantum mechanics which talked about an electronics simulation of coupled quantum oscillators.
When I saw the schematic, I found it very innovative in spite it is simple:
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=7958)
Here is the idea: there are two coupled LC circuits, one with a resistance that damps the oscillation while the negative resistance in the second one provides energy and compensates the attenuation from the other side. The system is said to bypass the bandwidth theorem and to transfer energy quasi-instantly between the two sides: "the latter case indicates the possibility of transforming an initial state to an orthogonal final one, or in more practical terms, transferring energy from one side to the other, in an arbitrarily short time interval."
http://arxiv.org/abs/1205.1847
My goal is not to discuss this complicated paper about QM, but only the part concerning the electronics simulation: the left side of page 2.
I don't know for sure if it can be of interest for our actual topic. I'm testing spice simulations. I have already modelized the complete schematics of a neg resistance with an op amp and shall now test the circuit response.
As the discussion is drifting, I'll move the latest posts to another new thread, if that is ok with you guys.
Quote from: poynt99 on 2012.05.14, 12:20:00
As the discussion is drifting, I'll move the latest posts to another new thread, if that is ok with you guys.
Sure Poynt, whatever you feel is necessary.
Exn,
The second graph where the inverse slope saying that the current in the secondary start out high and fall off to zero as time goes by. It takes about 5 seconds into the sim to see it fall off close to zero.
Quote from: poynt99 on 2012.05.14, 12:20:00
As the discussion is drifting, I'll move the latest posts to another new thread, if that is ok with you guys.
I have put it here because we discussed about coupling and this is an original setup not yet experimented. I imagined that there were possibilities to link it to this thread but I don't know how. So if you think it is out of topic, don't hesitate to move it.
Quote from: GibbsHelmholtz on 2012.05.14, 12:51:17
...
The second graph where the inverse slope saying that the current in the secondary start out high and fall off to zero as time goes by. It takes about 5 seconds into the sim to see it fall off close to zero.
Bizarre.
In my simulations with ltspice the currents in primary and secondary are always well correlated.
Quote from: exnihiloest on 2012.05.14, 06:47:58
In the previous week I came across a paper about quantum mechanics which talked about an electronics simulation of coupled quantum oscillators.
When I saw the schematic, I found it very innovative in spite it is simple:
(http://www.overunityresearch.com/index.php?action=dlattach;topic=739.0;attach=7958)
Here is the idea: there are two coupled LC circuits, one with a resistance that damps the oscillation while the negative resistance in the second one provides energy and compensates the attenuation from the other side. The system is said to bypass the bandwidth theorem and to transfer energy quasi-instantly between the two sides: "the latter case indicates the possibility of transforming an initial state to an orthogonal final one, or in more practical terms, transferring energy from one side to the other, in an arbitrarily short time interval."
http://arxiv.org/abs/1205.1847
My goal is not to discuss this complicated paper about QM, but only the part concerning the electronics simulation: the left side of page 2.
I don't know for sure if it can be of interest for our actual topic. I'm testing spice simulations. I have already modelized the complete schematics of a neg resistance with an op amp and shall now test the circuit response.
Hey Exn.,
I came across this guy simulation. The circuit keep going after power is disconnected. If this is a bug, someone should identify what it is and notify falstad. I think it might be related to your post of negative resistance sim. I did some modification and it gives out the same result. I monitor several point on the circuit and seems like every part is functioning.
http://www.youtube.com/watch?v=cD_Bwwvlc3I&feature=youtu.be
Crazy sim (http://www.falstad.com/circuit/#%24+1+5.0E-6+0.5999443210467817+40+7.0+50%0Aw+224+176+192+176+0%0Ar+192+176+192+240+0+10000.0%0Aw+192+240+224+240+0%0Aw+192+240+192+272+0%0Aw+192+272+224+272+0%0Ac+192+272+192+336+0+3.0E-7+0.49458523800804954%0Ar+192+176+192+112+0+10000.0%0Aw+192+112+288+112+0%0A165+224+144+240+144+2+0.8728416402528315%0Aw+288+112+352+112+0%0Aw+352+112+352+176+0%0Aw+128+64+128+112+0%0Aw+128+112+192+112+0%0As+128+112+128+192+0+1+false%0As+128+192+192+272+0+1+false%0Ar+128+272+128+192+0+100.0%0Aw+480+304+480+336+0%0Aw+352+208+480+208+0%0Av+128+336+128+272+0+0+40.0+5.0+0.0+0.0+0.5%0Aw+128+336+192+336+0%0Ag+192+336+192+368+0%0Ar+192+336+480+336+0+1.0E-4%0AT+480+208+544+304+0+4.0+2.0+34.69609383920891+-17.23929220302742+0.999%0Aw+544+304+544+336+0%0Aw+544+336+480+336+0%0Ad+544+208+544+304+1+0.805904783%0Ar+544+208+528+64+0+100.0%0As+128+64+528+64+0+0+false%0Ao+26+64+0+35+0.078125+7.8125E-4+0+-1%0Ao+5+64+0+35+5.0+1.953125E-4+1+-1%0Ao+1+64+0+35+2.5+3.90625E-4+2+-1%0Ao+24+64+0+35+0.01953125+0.0015625+3+-1%0A)
Hi Gibbs, The bug can be related to any thing, but likely to the 555 model. Generally, only a nominal functioning is modelized. A 555 is specified for a power voltage range 4.5v - 16v. If you power it with 1.5v, the model doesn't apply. Another example is transistor 2N2222A. We can use it as a negative resistance by connecting its emitter and collector in a reverse way. But this is not simulable with an ordinary 2N2222A model because it's out of a normal functioning.
The simulations are restricted by Kirchoff's equations saying that 1-1=0. If you observe "1", you can be sure there is a bug or the model is not used in its domain of validity.
I know you are convinced, otherwise you would have already built the setup with a real 555, isn't it? ;)
Quote from: GibbsHelmholtz on 2012.05.19, 00:55:48
...
I think it might be related to your post of negative resistance sim.
...
A negative resistance must be considered as a power supply. It really provides energy. Spice allows for negative resistance values. If you put one in a circuit, you get a "perpetual motion".
But in the real world, a negative resistance is negative only around an operating point obtained by biasing the dipole with a DC source, and it is this DC source that provides energy.
I'm now revisiting the Kapanadze's device with this idea that a negative resistance (may be in a spark gap?) could be the cause of energy gain in his coupled circuits. I'm not yet convinced if it is a scam or if he has really something.