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Optics problem: electromagnetic theory

  1. Nov 21, 2004 #1
    If anyone who's good at optics/physics can help me with this electromagnetic part of optics...it'll be great.

    so here's the problem:

    imagine an electromagnetic wave with it's E-field in the y-direction. Show that (dE/dx)=-(dB/dt) applied to the harmonic wave B

    E=Eo cos(kx - wt) B=Bo cos(kx - wt)

    yields to the fact that Eo=cBo
    in agreement with Ey=cBz

    thanx in advance :blushing:
  2. jcsd
  3. Nov 21, 2004 #2


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    Hi cuti_pie75.

    Just work out (dE/dx)=-(dB/dt) with what you're given.
    I`ll do dE/dx:

    [tex]\frac{dE}{dx}=\frac{d}{dx} E_0 \cos(kx-\omega t) = -kE_o\sin(kx-\omega t)[/tex]
    Differentiate B with respect to t and use the equation (dE/dx)=-(dB/dt).

    You also need that [itex]\vec E_0=E_y \hat y[/itex] (this was given).
    And assuming the direction of the wave is in the +x-direction. [itex]\vec B_0= B_z \hat z[/itex].
    Last edited: Nov 21, 2004
  4. Nov 21, 2004 #3
    Thank you very much for your fast reply Galileo, everything's much more clearer to me now...but i just have one more question:

    i did the derivative for B in respect to t and comes to: kEo = -wBo
    so my question is -w/k = c?

    anyway, sorry for the bother and thanx again!
  5. Nov 21, 2004 #4


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    Almost, remember the equation is (dE/dx)=-(dB/dt), so you have a minus sign. There is another minus sign coming from the derivative of the cosine and another one from the chain rule (derivative of (kx-wt) with respect to t is -w).
    So you get kEo=wBo
    w/k indeed equals the speed of the wave (c in this case).
    the way I always remember it, is by knowing that a traveling wave always has the form f(x-vt) with v the speed of the wave. kx-wt=k(x-w/kt), so w/k is the speed,
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