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Calculating an expression for work in a magnetic field

  1. Jun 17, 2013 #1
    As shown, I must derive a formula to find work.

    http://i.imgur.com/0U0eLqm.jpg

    I proceeded as follows.

    W= F * D
    = (B * I * L) * D

    V = BA/T

    I = V/R
    = (BA/t)/R
    = BA/tR

    F = B * (BA/tR) * A
    W = ((B^2*A^2)/tR) * A
    W = (B^2 * A^3)/tR

    Does that seem right? IIRC, the B*I*L formula is only for solonoids?

    Thanks
     
  2. jcsd
  3. Jun 17, 2013 #2
    I don't understand this

    "F = B * (BA/tR) * A
    W = ((B^2*A^2)/tR) * A
    W = (B^2 * A^3)/tR

    However consider Power = VI and work = power X time
    I think you will get a similar ansawer
     
  4. Jun 17, 2013 #3
    I solved L in terms of B, A, R and T, as the question doesn't allow me to use I. Only thing I'm confused about is if I can use V = B * I * L in this situation, as I thought that was for solonoids only.
     
  5. Jun 17, 2013 #4
    What is L? I recall that Force = B X current X length of wire.
    Think about it this way. V = dphi/dt and this is what you have.
    Power = V^2/r

    Work = power X Time, yes?
     
  6. Jun 17, 2013 #5
    ((BA/t)^2 / r) * t gives me

    (BA)^2 / tR, which is a different equation then the other one I got. So is mine wrong?
     
  7. Jun 18, 2013 #6
    I got (BA)^2 / tR. In your post #1 you have an extra A. There might be a math error. Do you have the correct answer?
     
  8. Jun 18, 2013 #7
    Unfortuantetly, I don't have an answer. I'll check it today. Thanks :)
     
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