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Prove that emf from E equals emf from dB/dt

  1. May 14, 2017 #1
    Exercise:
    I'm suppose to prove that e1 = e2
    e1 = ∫E*dl
    e2 = ∫dB/dt*dS

    where S is the surface encircled by the conture c.
    c is a box with with a length (in x axis) and b height (in y axis).
    [​IMG]

    for an electromagnetic wave:
    E = E0*sin(kx - wt) (in y axis)

    I'm ASSUMING this means that
    B = B0*sin(kx - wt) (in z axis)

    Pathetic Attempt:
    e1 calculated on the outside of c becomes
    b*E(x + a,t) - b*E(x,t) =
    E0*sin(kx + ka - wt) + E0*sin(kx - wt)
    This expression is confusing and doesn't lead to any simplification at all.

    e2 = ∫dB/dt*dS
    dB/dt = B0*-w*cos(kx - wt)

    Now heres the tricky part, how do I integrate this infinitly thing vector over a surface?
    The wave has no thickness, shouldn't the integral be zero?!
     
  2. jcsd
  3. May 14, 2017 #2
    I'm also suppose to prove this explicitly, I DONT EVEN KNOW WHAT THAT MEANS!!!
     
  4. May 14, 2017 #3

    TSny

    User Avatar
    Homework Helper
    Gold Member

    What happened to the factor of b in going to the last line. Also, what happened to the negative sign between the two terms?

    You've made a good start once you make the corrections above.

    OK

    Not sure what you mean by "this infinitly thing vector".
    ##\frac{\partial B}{\partial t}## is a function of ##x## and ##t##. ##t## is some fixed but arbitrary time. So, you can think of ##\frac{\partial B}{\partial t}## as a function of ##x## that you need to integrate over the rectangular area. What would be a good way to break up the area into infinitesimal elements ##dS##? Recall how you did similar integrals in your calculus course.
     
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