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Induced EMF and a circular loop of wire

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data
    A 25-turn circular loop of wire has a diameter of 1m. In 0.2 seconds it is flipped 180 degrees at a location where the magnitude of the Earth's magnetic field is 50 micro T. What is the emf generated in the loop?

    2. Relevant equations

    3. The attempt at a solution
    When I use NABwsin(wt)=V, I get zero. When I use -(dB/dt)(A)=V, I get a non-zero number. I believe the 2nd equation works but why doesn't the first work?

    Thanks for any help.
  2. jcsd
  3. Apr 25, 2010 #2
    Where does the first equation come from? If you used d/dt[BA] = V is B changing?
  4. Apr 25, 2010 #3
    Thanks for the reply. NABwsin(wt)=V comes from taking the derivative of NABcos(wt)=magnetic flux. The 2nd equation should be taking into account the change in angle. Shouldn't both equations lead to the same answer?
  5. Apr 25, 2010 #4
    I believe they should. How did you get 0 for the first equation? I do not get 0.
  6. Apr 25, 2010 #5
    Show the work on both because when I use the 2nd equation all that happens is I derive the first equation.
  7. Apr 25, 2010 #6
    sin(wt) when t=(.2) equals 0.

    Hm..I think I'm not using the equation correctly?
  8. Apr 25, 2010 #7
    Doh! I see what you mean :/.
  9. Apr 25, 2010 #8
    Here's my thoughts: By plugging in wt only for the final time, you only are calculating the emf produced at that instant. So maybe you should use:

    V = -d/dt(BA) = -[BA(Final) - BA(Initial)]/[t(Final) - t(Initial)]

    sorry cannot get fraction in latex to work, I am a newb.
    Last edited: Apr 25, 2010
  10. Apr 25, 2010 #9

    V = -\frac{d}{dt}[BA] = -[\frac{BA_f - BA_i}{t_f - t_i}]


    I did it in latex :).
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