1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook