Electromagnetic waves, Faraday's law

[_Andreas]

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

\nabla\times \textbf{E}=-\frac{\partial \textbf{B}}{\partial t}

gives that

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

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

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

\begin{vmatrix}<br /> \textbf{i} &amp; \textbf{j} &amp; \textbf{k}\\<br /> ik_x &amp; 0 &amp; ik_z\\<br /> E_x &amp; 0 &amp; E_z<br /> \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}

Taking the modulus of H doesn't yield the correct answer, so I'm out of clues.

 

[tiny-tim]

Hi _Andreas!

That'll only work if k and E are perpendicular

 

[_Andreas]

Hi _Andreas!

That'll only work if k and E are perpendicular

Hi!

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

BTW, the k in the last expression above (that for H) shouldn't be there.

 

[tiny-tim]

Hi!

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

(wot picture? )

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

 

[_Andreas]

Where'd my picture go?

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

Ooops! I deserve a facepalm.

Thanks.