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Homework Help: Uniform plane wave (help with a short walkthrough for a worked problem)

  1. Nov 26, 2006 #1
    I don't understand something on this problem:

    Q: A uniform plane wave with [itex] \vec E = \hat x E_x [/itex] propagates in a lossless simple medium ([itex] \epsilon_r = 4 \,\,\,\, \mu_r = 1 \,\,\,\, \sigma = 0 [/itex]) in the +z-direction. Assume that [itex] E_x [/itex] is sinusoidal with a frequency of 100 (MHz) and has a maximum value of [itex]+10^{-4} [/itex] (V/m) at [itex] t=0 [/itex] and [itex] z = \frac{1}{8} [/itex] (m)

    Write the instantaneous expression for [itex] \vec E [/itex] for any t and z.

    A:
    Using [itex] \cos \omega t [/itex] as a reference, we find the instantaneous expression for [itex] \vec E [/itex] to be:

    [tex] \vec E(z,t) = \hat x E_x = \hat x 10^{-4} \cos ( 2 \pi 10^8 t - kz + \psi) [/tex]

    Since [itex] E_x [/itex] equals [itex] +10^{-4} [/itex] when the argument of the cosine function equals zero, that is when:

    [tex] 2 \pi 10^8 - kz + \psi = 0 [/tex]

    we have at [itex] t = 0 [/itex], [itex] z = \frac{1}{8} [/itex]

    [tex] \psi = kz = \frac{4 \pi}{3} \frac{1}{8} = \frac{\pi}{6} [/tex]

    My question follows from this:
    Why do we introduce the phase unknown phase [itex] \psi [/itex] and set [itex] E_x [/itex] according to this. Why couldn't we do this instead,

    A:
    Using [itex] \cos \omega t [/itex] as a reference, we find the instantaneous expression for [itex] \vec E [/itex] to be:

    [tex] \vec E(z,t) = \hat x E_x = \hat x E_0 \cos ( 2 \pi 10^8 t - kz) [/tex]

    Set the according t=0, z=1/8 so that,
    [tex] \hat x E_0 \cos ( 2 \pi 10^8 (0) - k(\frac{1}{8}) = 10^{-4}[/tex] and meet the initial condition of a max of 10^-4 with [itex] E_0 [/itex] instead of setting E_0 originally and changing the phase.
     
  2. jcsd
  3. Nov 26, 2006 #2

    berkeman

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    Staff: Mentor

    I forget -- what is k in terms of lambda? It seems like the phi and the kz can influence the final equation under different constraints -- I could be wrong of course.
     
  4. Nov 27, 2006 #3

    OlderDan

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    Science Advisor
    Homework Helper

    If you used

    [tex] \hat x E_0 \cos ( 2 \pi 10^8 (0) - k(\frac{1}{8}) = 10^{-4}[/tex]

    the maximum would not be 10^-4 and at t = 0 the maximum would be at z = 0.
     
    Last edited: Nov 27, 2006
  5. Nov 27, 2006 #4
    oooooooooooooooooohhhh :)

    Thanks man, that definitely makes sense! I really appreciate it.

    berkeman:

    [tex] k = \omega \sqrt{\mu \epsilon} [/tex]
     
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