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Author Topic: Professor Walter Lewin's Non-conservative Fields Experiment  (Read 251635 times)
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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:


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").


   
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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?
« Last Edit: 2012-04-21, 16:45:33 by poynt99 »
   

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It's not as complicated as it may seem...
This is true but not related to my reply which applied to the following 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:
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.
   

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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.


---------------------------
"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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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.


---------------------------
"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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There is no potentials.


Notice is used the word 'measured' not actual. I see no disagreement between us on this matter.


---------------------------
"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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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.



---------------------------
"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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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....


---------------------------
"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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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.

   

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


   

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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.


---------------------------
"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 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?
   

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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.


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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...
Forget the resolver. I have doubts you would perceive the connection. You are still on in-plane and normal positions.
??? Really? Okiiiiiii.  ;)
   

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

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It's not as complicated as it may seem...
I have one for you Gibbs.

What voltage will be indicated on CH1 and CH2? (Assume as always that the emf=1V)
   
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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
   
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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.


   

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


   

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It's not as complicated as it may seem...
Here's one for you Ex.

What voltage will be indicated on CH1 and CH2?
   
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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.

   

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


   
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