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Homework Help: Faraday's law

  1. Dec 13, 2008 #1
    1. The problem statement, all variables and given/known data

    Actually this is not exactly a homework. I am trying to understand the following situation only.

    Consider a ring shaped coil of N turns and area A. Connect it to an external circuit with a twisted pair of leads ( this info is trivial). The resistance of the circuit along with the coil itself is R. Now the coil in a magnetic field.

    Suppose the flux through the coil is somehow altered from its initial steady state value (A) to final value (B).

    The author claims that the total charge Q that flows through the circuit as a result is independent of the rate of change of the flux. I am having hard time understanding this. Can anyone help me understand it.

    2. Relevant equations

    [tex]\oint[/tex] E.dl = -d[tex]\Phi[/tex]/dt

    3. The attempt at a solution

    faraday's law is the most relevant law here, according to the book. But I am just not getting what the author is saying.
  2. jcsd
  3. Dec 13, 2008 #2


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    I think this lecture covers a lot of what you want to know.
    http://ocw.mit.edu/ans7870/8/8.02/videolectures/wl-802-lec16-220k.ram [Broken]

    (Requires Real Player, if that's not a problem for you.)

    About 11 minutes in if you are in a rush.
    Last edited by a moderator: May 3, 2017
  4. Dec 13, 2008 #3

    i followed the video. It was helpful.

    However, i am still not sure about independence of charge with the rate of change of flux. On applying faraday's law

    EMF = -Nd[tex]\Phi[/tex]/dt

    In the condition mention in the question above, B is the final M_flux A and the initial M_flux. We are trying to derive Q such that it is independent of d[tex]\Phi[/tex]/dt.

    I am confused with initial and final magentic flux. On just using d[tex]\Phi[/tex]/dt, here is what i got

    I = Nd[tex]\Phi[/tex] cos(theta)/dt*(R)

    and I = dQ/dt

    But still Q is dependent on d[tex]\Phi[/tex]/dt.

    Any clue ?
    Last edited: Dec 14, 2008
  5. Dec 15, 2008 #4


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    What is cos theta here? And try equating the expression for I with V/R where V is as given by Faraday's law.
  6. Dec 15, 2008 #5
    cosine theta is a mistake here. It has to be omitted.

    Yeah, i did use Ohm's law there.

    But my confusion at this point is, since the final and initial fluxes are given, in Faraday's formula, should emf be

    emf = -N d(B-A)/dt or just -N d(flux)/dt ?

    The final expression is supposed to show that Q is independent of rate of change of flux
    Last edited: Dec 15, 2008
  7. Dec 15, 2008 #6


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    It should be [tex]emf = -\frac{B-A}{\delta t}[/tex].
  8. Dec 15, 2008 #7
    did you forget N ?
  9. Dec 15, 2008 #8


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    No I didn't. N was already included in both B and A. Remember that B, A are themselves the flux through the coil. Anyway it should make no difference in the solution.
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