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Electromagnetic waves, Faraday's law

  • Thread starter _Andreas
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  • #1
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Homework Statement


Homework 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.

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:

Answers and Replies

  • #2
tiny-tim
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Hi _Andreas! :smile:

That'll only work if k and E are perpendicular :wink:
 
  • #3
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Hi _Andreas! :smile:

That'll only work if k and E are perpendicular :wink:
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.
 
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  • #4
tiny-tim
Science Advisor
Homework Helper
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Hi!

But they are. Take a look at the picture I drew.
(wot picture? :confused:)

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

Exactly … so if k is perpendicular to E, then |k x E| = |k| |E| :wink:


Ooops! I deserve a facepalm.

Thanks.
 

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