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Induced emf and Faraday's law

  1. Feb 11, 2008 #1
    1. The problem statement, all variables and given/known data

    A radio antenna that receives an AM radio station that emits at 800 kHz is constituted of an inductor of 120 turns with a radius of 0.6cm. The inductor is home to an iduced emf du to a magnetic field oscillating from the radio wave. The magnetic field is [tex]\vec{B}[/tex]=1.0*10[tex]^{-5}[/tex]sin(2[tex]\pi[/tex]ft)[tex]\vec{k}[/tex]. What is the induced emf in the inductor? We suppose that the magnetic field is oriented according to the inductor's axe.

    2. Relevant equations
    emf = -Nd[flux]/dt

    d[flux] = int[B*dA]

    3. The attempt at a solution

    I tried to integrate and find the flux, but Im not that great at calculus, but then what I really don't get is that Im gonna have to derive to find the emf. It seems like Im going in circles. By the way, the final answer is -0.682cos(5.03*10[tex]^{-6}[/tex]t) V.

    Thanks for the help!
  2. jcsd
  3. Feb 11, 2008 #2
    I'm checking it right now, but seems to me that B here doesn't vary with area or location or anything

    I'm assuming that the d in front of [flux] is a typo, as is the 2^pi(should be 2pi)

    So you're integrating B*dA....>_> if B doesn't depend on the area, it just comes out of the integral and you get B*A, which is pi*r^2

    then you take the time derivative of THAT(which most certainly depends on time)times -N

    Your concern would be right except you're integrating over dA, then differentiating with respect to time

    Edit: This is correct, and I checked it on windows calculator so it took like 6 start overs when I clicked the wrong thing :(
    Last edited: Feb 11, 2008
  4. Feb 11, 2008 #3
    Thank you very much for the help, it works now. I was just complicating the problem. For some odd reason I was integrating the function for the magnetic field in the flux calculation, but it all works now. Thank you!!!
    Last edited: Feb 11, 2008
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