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Homework Help: Faraday's Law

  1. Mar 4, 2005 #1
    Induced emf around a closed path in a time-varying magnetic field.
    A magnetic field is given in the xz-plane by B=Bo*cos(pi)(x-Uot)ay Wb/M^2. Consider a rigid square loop situated in the xz-plane with its vertices at (x,0,1), (x,0,2),(x+1,0,2) and (x+1,0,1).
    1.What is the expression for the emf induced around the loop in the sense defined by connecting the above points.
    2.If the loop is moving with the velocity [tex]V = U_{o}a_{x}[/tex] m/s instead of being stationary what is the induced emf

    This is what I got for flux. Can someone check me if I’m doing this right? If it is ok then I can go to next step. I hope, the latex code comes out right. Thanks for help.

    Sorry, I should be more specific. This is the exact expression:
    [tex]\ B = B_{o}cos{\Pi}(x-U_{0}t)a_{y}[/tex]
    So:
    [tex]\psi=\int_{s}B\cdot\,ds=\int_{0}^{2} \int_{0}^{1}B_{0}cos{\Pi}(x-U_{0}t)a_{y}\cdot\, dx\,dz\,a_{y}[/tex]

    My problem is that I'm not sure that integral limits are correct. Thanks
     
    Last edited: Mar 4, 2005
  2. jcsd
  3. Mar 4, 2005 #2
    Can you double check your expression for B. For the expression you have given,

    when y=0 (xz plane) , B=0. Therefore flux=0 since the loop is also on the xz plane.
     
  4. Mar 4, 2005 #3
    I am assuming ay is the unit vector along y direction. Magnetic field B is independant of z. It is a function of only x. Therefore, integrate only with restpect to x.

    [tex]\psi=-\frac{d}{dt}\int_{s}B\cdot\,ds=-\frac{d}{dt}\int_{x}^{x+1}B_{0}cos{\Pi}(x-U_{0}t) dx[/tex]

    Induced emf is a function of x.
     
    Last edited: Mar 4, 2005
  5. Mar 5, 2005 #4
    Thanks Gamma, I got the answer.

    [tex]-2B_{0}U_{o}cos{\Pi}(x-U_{0}t)[/tex]

    So for part 2, emf would be the same because the loop is perpendicular to B? Is that right?
     
  6. Mar 5, 2005 #5

    Not really. Now x is a function of t.

    [tex]\psi=\int_{s}B\cdot\,ds=\int_{x}^{x+1}B_{0}cos{\Pi}(x-U_{0}t) dx[/tex]

    First evaluate the above integral. Then differentiate wrt to t. When you differentiate the above to find the emf, don't forget x=x(t). Use the fact that dx/dt = Uo
     
  7. Mar 5, 2005 #6
    I did integrate and differentiate and induced emf is equal to:

    [tex]-2B_{0}U_{o}cos{\Pi}(x-U_{0}t)[/tex]

    But in second part of problem "If the loop is moving with the velocity [tex]V = U_{o}a_{x}[/tex] m/s instead of being stationary what is the induced emf"

    I think that emf would be 0 becuase moving loop will produce a positive current. But I don't know how to proved it.
     
  8. Mar 5, 2005 #7
    I have explained how to go about it. First find [tex]\psi = \psi (x)[/tex].
    See my post #5. Yes, you will get zero.

    regards.
     
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