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Induced EMF through variable-resistance-wire due to Faraday's Law

  1. Jan 21, 2012 #1
    Hello,

    While considering Faraday's Law of Induction, I tried to think of a situation which would expose some difficulties I have with the notion that there is an induced EMF without clear regions of relatively high and low voltage (as in a battery). Here is what I thought would get me started:

    Consider a circular loop of wire. The top half is made of wire twice as resistant as the bottom half. A uniform magnetic field changes with time through the surface bounded by the loop. Is the (induced) electric field at a point at the top of the loop equal to the electric field at a point diametrically opposed (and therefore in a part of the loop half as resistant)?

    Faraday's Law seems to make a statement about the EMF induced around the whole loop, but not how it would vary within that loop due to changes in resistance. Or does it?

    edit: Remove a mistaken consideration.

    Thanks for any help!
     
    Last edited: Jan 21, 2012
  2. jcsd
  3. Jan 21, 2012 #2

    Philip Wood

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    It doesn't. The emf is independent of the resistance of the loop, and you could perfectly well talk about the emf around an imaginary path in a vacuum - which also has the value dphi/dt. The cross-section and material of the conductor determine the current that flows as a result. This must be the same all round the loop, even if the top half has a different resistance from the bottom half. Their combined resistance (in series), together with the emf, determines the current.

    [I've ignored self-inductive effects, that is emfs arising from changing magnetic flux due to changing current in the loop itself. These emfs DO depend, for obvious but indirect reasons, on the resistance of the loop.]
     
  4. Jan 21, 2012 #3
    So to put some figures to it (ignoring self inductive effects): Let's say the top half of the loop has resistance 2R and the bottom half has resistance R. If the induced EMF is V, then is it safe to say that the current around the loop will be constant throughout and equal to (V/3R)?
     
  5. Jan 21, 2012 #4

    Philip Wood

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  6. Jan 21, 2012 #5
    Actually the emf does indeed vary with resistance. A low resistance loop results in lower emf than a high R loop. I posted equations a few years ago detailing this. I'll find it and repost. When current exists in the loop it has a magnetic flux that opposes the external flux per law of Lenz.

    With high R, the emf is independent of the R value as long as the induced current;s magnetic flux is too small to oppose the external flux. But with low R values, voltages varies with R due to cancellation of external flux by the internal flux.

    Dr. Walter Lewin at MIT has a lecture note regarding a loop with 2 different R values. The emf along the high R section is higher. I'll dig that up as well. Cheers.

    Claude
     
  7. Jan 21, 2012 #6
  8. Jan 22, 2012 #7

    Philip Wood

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    Cabraham: the effect to which you are drawing attention is none other than the self-induction which I mentioned in the second paragraph of my original post.
     
  9. Jan 22, 2012 #8
    But when the loop R is low, self-induction cannot be ignored. Your mention in the 2nd paragraph is valid only with the condition that R is a high value.

    A classic example is the induction motor. The R value of the squirrel cage rotor is extremely low. Without considering self-induction, the induction motor cannot be discussed. One cannot completely explain induction motor behavior w/o self-induction.

    Ignoring self-induction and only considering mutual induction is only conditionally valid. The squirrel cage induction motor is a clear case of where self-induction cannot be ignored.

    That is all I'm saying. I agree with you that if the R value is high enough so that self-induction is very small relative to mutual induction, then the value of R can vary with negligible influence on induced emf.

    Claude
     
  10. Jan 22, 2012 #9

    Philip Wood

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    We are completely in agreement. I didn't mean to imply in my original post that it was always fine to ignore self-induction, just that I had done so in my first paragraph! But re-reading the post, I understand that this could have been misleading.
     
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