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Author Topic: Professor Walter Lewin's Non-conservative Fields Experiment  (Read 253081 times)

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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.


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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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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.
 



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

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


   

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

   

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


   

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


   

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

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It's not as complicated as it may seem...
What will be the indicated voltage on CH2?
   

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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  :)


---------------------------
"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 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.
   
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What will be the indicated voltage on CH2?




I suppose .1V .
   

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It's not as complicated as it may seem...
Let's wait for Harvey and WW to post their answers before I respond.
   

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

   
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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.


   

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

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

   

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It's not as complicated as it may seem...
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. ;)
   
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The positive probe is on top, so the answer, if correct, would be -0.1V. ;)





Fair enough, I'll buy that.


   
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