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I Walter Lewin paradox

  1. May 24, 2018 #1
    Right answer for the Walter Lewin demo/paradox about electromagnetic induction in Lecture 16 ?
     
  2. jcsd
  3. May 24, 2018 #2

    Charles Link

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    For the Professor Lewin video that I previously saw on the subject, I think professor Lewin had most of the information correct. In any case, the additional measurements that are made in this video are very interesting. See post 52 of this thread for the Professor Lewin video that I saw: https://www.physicsforums.com/threa...asure-a-voltage-across-inductor.880100/page-3 And yes, he does call that one "Lecture 16".## \\ ## Edit: To the OP @mabilde I see you chose the same name as the author of this video. Are you the author? ## \\ ## Additional editing: In a conductor, even though there is an EMF, where EMF ## \mathcal{E}=\int \vec{E}_{induced} \cdot d \vec{s} ##, a reverse electrostatic field will necessarily develop=otherwise the current would necessarily be very large by the equation ## \vec{J}=\sigma \, \vec{E} ## where ## \sigma ## is very large for a conductor. In that sense, I don't know that obtaining a voltage measurement inside the coil necessarily proves anything. In any case, an interesting set of experiments, but I think further discussion is in order. ## \\ ## And my input on this: A voltmeter and/or an oscilloscope consists of connecting wires plus some resistance and measures the current in the resistor circuit. It becomes part of the circuit loop, and there is nothing magical about it. It really can't measure the electric field or an EMF between two points in the circuit. It becomes part of the circuit, and if the resistance is chosen properly and the leads are connected properly, ideally it doesn't disrupt the circuit being studied, but that does not mean that (changing) external magnetic fields won't effect the results that are obtained. The EMF that is used to analyze the circuit containing the voltmeter can usually be obtained by Faraday's law. IMO, there is no need to try to localize the source of the EMF's as is done in this video by Professor Cyriel Mabilde.
     
    Last edited: May 24, 2018
  4. May 25, 2018 #3

    sophiecentaur

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    I often find this sort of thing confusing but it seems that it works like this. Every elemental length of wire will have an elemental emf induced in it.
    Your measurement leads have an emf induced in them which will depend on their path through the field from the primary and this adds (vectorially) to the PDs across the two resistors and the emfs induced in the two sides of the secondary.
    Trivial situation: If the measurement leads are short enough and connected straight across each resistor then the emfs will be the 'share' of 1V that each resistor gets - as a normal potential divider. This is because there are no other emfs involved and you could think of the circuit as a 1V (AC) battery (or a 0.5V battery in each leg of the loop.
    More complicated situation: Connect the voltmeter diametrically across the mid points of the two semicircles. In my diagram, I have split the induced emf into four 0.25V contributions. The dashed line splits the circuit into two symmetrical parts and, treating it as a simple DC circuit, the total PD measured will be 0.4 X 2 = 0.8V, because you have, effectively a potential divider. (I hope I have my + signs right in the diagram)
    Even more complicated: Move the measurement leads off centre or make them longer and tilt them. They will have their own emf induced into them, which can vary from 0.5V to -0.5V so the measured value can be anything between 0.1V and 0.9V, depending.
    Walter Leven 1.jpg
     
  5. May 25, 2018 #4
    Walking clockwise through the loop, then the emf's have the right polarity, but the resistor drops are opposed to this direction. The sum will be ZERO
    Using two voltmeters (in series) as in your "split" circuit, will give 0.2 Volts on each voltmeter.
    For example, considering the upper left : 0.25 Vemf -0.05 Vdrop = 0.2V
    and for upper right: - 0.25Vemf +0.45 Vdrop = 0.2 V etc...So the sum of both voltmeters will be 0.4 Volts and not 0.8 Volts!
    Note that in the first (left) case your voltmeter Hi is connected to the positive side of the emf, but in the right case the voltmeter Hi is pointing to the negative side of the emf.
    Hope this will help to understand also the rest of my video and thanks for watching it.
    Best greetings from Belgium
    Cyriel Mabilde
     
  6. May 25, 2018 #5

    Charles Link

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    Now here is how I would solve this voltmeter problem: Let the EMF from the changing magnetic field be clockwise and 1.0 volt. Let the resistor on the left be ## R_1=100 \, \Omega ## and the resistor on the right be ## R_2=900 \, \Omega ##. Let the voltmeter across the middle be ## R_3= 1 \, M \Omega ## . Let the clockwise current through ## R_1 ## be ## i_1 ##, and the clockwise current through ## R_2 ## be ## i_2 ##. The downward current through ## R_3 ## is ## i_3 ##. ## \\ ## The 3 loop equations are: ## \\ ## 1) ## .5=i_1R_1+i_3 R_3 ## ## \\ ## 2) ## .5=i_2 R_2-i_3 R_3 ## ## \\ ## 3) ## i_1=i_2+i_3 ##. ## \\ ## A little algebra gives (eliminating ## i_2 ##) ## \\ ## ## 1.0=i_1R_1+(i_1-i_3)R_2 ## ## \\ ## and ## \\ ## ## 0=i_1(R_1-R_2)+i_3(R_2+2R_3)##. ## \\ ## The ## i_3 R_2 ## term can be considered small in two places, so that ## i_1=\frac{1.0}{R_1+R_2} ## and ## \\ ## the voltmeter reading is ## i_3 R_3=-\frac{1}{2}\frac{(1.0)(R_1-R_2)}{R_1+R_2}=+.4 \, volts ##. ## \\ ## @mabilde Thank you very much for your video. It was very interesting and instructive.
     
    Last edited: May 25, 2018
  7. May 26, 2018 #6

    sophiecentaur

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    And greetings to you from sunny Essex UK.
    Thanks for your comment. (The demo on the video is great, btw) I obviously got the signs in my calculation wrong and, because I am familiar with Potential Dividers, I can do it in an easy way . Take the top left corner as reference zero for the first half of the calculation. The two resistors are a potential divider with 10:1 ratio. The upper voltmeter should be showing the unbalanced volts between the two nodes of 0.2V (= 0.25-(0.5/10)) Likewise the other voltmeter will see -0.2V, giving 0.4V overall.
    It's been a long time since I did this stuff and I'v looked at the problem again. I should have written some reference letters on the diagram and it would be easier to discuss but @Charles Link has solved the problem in yet another way (with Kirchoff 1) - and got the same answer!!!.
     
  8. May 29, 2018 #7

    Charles Link

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    @mabilde I am very glad you posted this video. The experimental apparatus and results is very interesting. I am also very glad that it showed the results that confirms a combination of Faraday's law, and Kirchhoff's voltage law.
     
  9. May 29, 2018 #8

    rude man

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    Mr Mabilde's voltage measurements inside his 1-turn coil are accurate as shown by the fact that when he connects the two rotatable probes together he gets 0 volts. His methodolgy avoids the sort of lead pickup problems to which you're referring.

    The virtue of Mr Mabilde's internal rotating probes is that he zeros out any B field parallel to his probe wires' loop normal inside & outside the large coil. The fringe field is sizable as he showed with his B plots inside & outside that coil. There is no way a voltmeter lead arrangement like Prof Lewins' can avoid this large fringe field's effect on voltmeter readings.
     
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