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Induced emf

  1. Jan 29, 2006 #1
    If we have a loop with radius 12 cm that lies in a uniform magnetic field of 1.5T.

    The magnetic field is in the positive z direction and the loop lies in the horizontal xy-plane. Now if the loop is removed from the field in a time interval of 2.0x10^-3 s what will the average induced emf be.

    Right now first I thaught it might be zero but then seeing as the b field will be constant and A will change.

    therefore it has to be emf= - -b x dA/dt

    but how can i describe the rate of change of A?

    ps:am I even on the right track? =)
    Last edited: Jan 29, 2006
  2. jcsd
  3. Jan 29, 2006 #2
    If I understood well the magnetic field is perpendicular to the loop:
    the A will decrease from it's maximum value (the area of the loop: 4,52 x 10^-2 m^2) to zero.
  4. Jan 29, 2006 #3
    yeah so the average emf induced would be d(A/2)/dt ?
  5. Jan 29, 2006 #4
    In my opinion, we could simplify the formula:
    emf induced = B.A/dt

    B= 1,5 T
    A (of the loop) = 4,52 x 10^-2 m^2
    dt = 2,0 x 10^-3 s
    emf = 33,9 V

    But wait for the experts, it's possible that I am mistaken.
    Last edited by a moderator: Jan 29, 2006
  6. Jan 29, 2006 #5
    I thinks it is

    [tex] \frac{\Delta \Phi_B}{\Delta t} [/tex]

    since it asks average value not instantanious one, I think 33.912 V is a correct answer.
  7. Jan 29, 2006 #6
    but surely the average A confined in the magentic field during dt must be half A.. therefore the average Induced emf must be -d(A/2)/dt

  8. Jan 29, 2006 #7
    It is not an average A but a variation of A.


    efm = dPHi/dt

    PS- Sorry, but I don't know to latex.
    Last edited by a moderator: Jan 29, 2006
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