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Electromagnetic waves, Faraday's law

  1. Apr 18, 2009 #1
    1. The problem statement, all variables and given/known data
    2. Relevant equations

    For a plane polarized electromagnetic wave traveling along the z axis, with its E vector parallel to the x axis and its H vector parallel to the y axis, Faraday's law

    [tex]\nabla\times \textbf{E}=-\frac{\partial \textbf{B}}{\partial t}[/tex]

    gives that

    [tex]\frac{E}{H}=\frac{\omega\mu}{k}[/tex], where E and H are the moduli of E and H.

    ETA: It is assumed that [tex]\textbf{H}=\textbf{B}/\mu[/tex]

    My problem is with an em-wave traveling in the x-z plane (were E and k have both x and z components). Apparently, I'm supposed to get the same ratio between E and H as above.

    3. The attempt at a solution

    [tex]\begin{vmatrix}
    \textbf{i} & \textbf{j} & \textbf{k}\\
    ik_x & 0 & ik_z\\
    E_x & 0 & E_z
    \end{vmatrix}=i\omega\mu H\textbf{j} \Longrightarrow \textbf{H}=\frac{1}{\omega\mu}(k_zE_x-k_xE_z)\exp i(k_xx+k_zz-\omega t)\textbf{j}[/tex]

    Taking the modulus of H doesn't yield the correct answer, so I'm out of clues.
     
    Last edited: Apr 19, 2009
  2. jcsd
  3. Apr 19, 2009 #2

    tiny-tim

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    Hi _Andreas! :smile:

    That'll only work if k and E are perpendicular :wink:
     
  4. Apr 19, 2009 #3
    Hi!

    But they are. Take a look at the picture I drew.

    BTW, the [tex]k[/tex] in the last expression above (that for H) shouldn't be there.
     
    Last edited by a moderator: Apr 19, 2009
  5. Apr 19, 2009 #4

    tiny-tim

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    (wot picture? :confused:)

    Exactly … so if k is perpendicular to E, then |k x E| = |k| |E| :wink:
     
  6. Apr 19, 2009 #5
    Where'd my picture go?



    Ooops! I deserve a facepalm.

    Thanks.
     
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