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Author Topic: Professor Walter Lewin's Non-conservative Fields Experiment  (Read 253073 times)

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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.



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"As far as the laws of mathematics refer to reality, they are not certain; as far as they are certain, they do not refer to reality." - Einstein

"What we observe is not nature itself, but nature exposed to our method of questioning." - Werner Heisenberg
   

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It's not as complicated as it may seem...
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.
   

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It's not as complicated as it may seem...
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
   

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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.



---------------------------
"As far as the laws of mathematics refer to reality, they are not certain; as far as they are certain, they do not refer to reality." - Einstein

"What we observe is not nature itself, but nature exposed to our method of questioning." - Werner Heisenberg
   

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It's not as complicated as it may seem...
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?
   

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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.


---------------------------
"As far as the laws of mathematics refer to reality, they are not certain; as far as they are certain, they do not refer to reality." - Einstein

"What we observe is not nature itself, but nature exposed to our method of questioning." - Werner Heisenberg
   

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It's not as complicated as it may seem...
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.
   

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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.

 


---------------------------
"As far as the laws of mathematics refer to reality, they are not certain; as far as they are certain, they do not refer to reality." - Einstein

"What we observe is not nature itself, but nature exposed to our method of questioning." - Werner Heisenberg
   
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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 ) that we are able to extract energy from them.

 8)
   

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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  ;)


---------------------------
"As far as the laws of mathematics refer to reality, they are not certain; as far as they are certain, they do not refer to reality." - Einstein

"What we observe is not nature itself, but nature exposed to our method of questioning." - Werner Heisenberg
   

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It's not as complicated as it may seem...
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?

Quote
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.
By all means check it yourself if you have doubts about my results.
« Last Edit: 2012-02-27, 22:54:32 by poynt99 »
   

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It's not as complicated as it may seem...
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?
« Last Edit: 2012-02-28, 00:23:34 by poynt99 »
   
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...

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. 

   
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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.
   

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It's not as complicated as it may seem...
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?
   
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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 from about 24 minutes onward and then consider an apology.  O0
   

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It's not as complicated as it may seem...
No, your comment is not correct - actually. You need to think about it, review Lecture 16 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..
« Last Edit: 2012-02-28, 19:31:22 by poynt99 »
   
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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...

 :)
   

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It's not as complicated as it may seem...

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.
   
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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).


   

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.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)
« Last Edit: 2012-02-28, 11:05:19 by WaveWatcher »


---------------------------
"As far as the laws of mathematics refer to reality, they are not certain; as far as they are certain, they do not refer to reality." - Einstein

"What we observe is not nature itself, but nature exposed to our method of questioning." - Werner Heisenberg
   

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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.


---------------------------
"As far as the laws of mathematics refer to reality, they are not certain; as far as they are certain, they do not refer to reality." - Einstein

"What we observe is not nature itself, but nature exposed to our method of questioning." - Werner Heisenberg
   

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It's not as complicated as it may seem...
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.

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?
   

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It's not as complicated as it may seem...
The experiment tells more about the experimenter than the physics. (Not pointed only at .99)

Please do tell.
   

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It's not as complicated as it may seem...
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
   
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