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Simple faraday's law problem (phi = what?)

  1. May 8, 2014 #1
    1. The problem statement, all variables and given/known data

    A toroid with N windings and radius R has cross sectional radius x (x<<R).
    The current running through the wires is given by ##I = I_0\sin (\omega t)##.
    There is a magnetic field in the center of the toroid.

    A loop of wire of radius three times that of the toroid is placed around the cross section of the toroid
    (like this, but it's circular, not a rectangle).

    Find the induced emf in the circular loop of wire.

    2. Relevant equations

    E = -dphi/dt
    3. The attempt at a solution

    ##\int B\cdot ds = \mu_0 I\implies B(2\pi R) = N\mu_0 I \implies B = \dfrac{\mu_0 N (I_0\sin\omega t)}{2\pi R}## (*)

    then ##\mathcal{E} = \dfrac{- d\phi}{dt} = - \dfrac{d(BA)}{dt} = -\pi x^2 \dfrac{dB}{dt}##, where dB/dt is just the derivative of (*).

    My question is, that should it be ##-\pi x^2## (the cross section of the toroid) or should it be the area of the loop ( pi*(3x)^2) ?
    My reasoning for pi*x^2 is that since there is no flux at parts of the circular loop, it's just the area of the cross section (no flux since no magnetic field!).
  2. jcsd
  3. May 8, 2014 #2

    rude man

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    You need to integrate B across the toroid's cross-section to the the total flux. B is a function of the radial distance from the toroid's center (not the cross-section center; I mean the center of the hole). You have oversimplified the problem by using the value of B at the outermost radial distance R.
  4. May 9, 2014 #3
    EDIT: I don't see why that is true, isn't R fixed?
    Last edited: May 9, 2014
  5. May 9, 2014 #4

    rude man

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    Sure is, but the B field isn't confined to r = R.
  6. May 9, 2014 #5
    can you specify where it is incorrect? I still don't see what you are referring to.
  7. May 9, 2014 #6

    rude man

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    As I said, you have determined the B field for r=R but have done nothing to determine what it is for r<R. R is the OUTER radius of the toroid. Ir also has an inner radius, which is R-2x. So we have R-2x < r < R.

    B is a function of the radius r: B = B(r). You have computed B(R) only. Figure out why B(r) > B(R), r < R, just the way you found B(R).
  8. May 10, 2014 #7
    Is there an elementary way to solve this then? (AP Physics C level)?

    If the cross section was a square, it would be easy, but here it's harder.
  9. May 10, 2014 #8

    rude man

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    In either case it requires integrating strips of the cross-section, from r = R-2x to r=R, to get the total flux which is what you need.
    EDIT: ON SECOND THOUGHTS, since you were given x << R your original post solution is close to correct and probably what was intended.

    I don't understand how you came up with your alternative area pi(3x)^2. What has a radius 3x???
    Last edited: May 10, 2014
  10. May 10, 2014 #9
    The larger loop of wire.
  11. May 10, 2014 #10

    rude man

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    Oh, right. But the area you want is, as you said, where the flux is, which is in the toroid cross-section only. Good work and sorry if I sidetracked you. If R >> x had not been given you would have had to integrate.
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