Magnetic field of a circular wave

  • Thread starter kediss
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  • #1
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Main Question or Discussion Point

Hi !

Does [itex]\vec{H}=\frac{1}{\eta}\hat{k}\times \vec{E}[/itex] always suit an electromagnetic wave ?

Because i'm getting some inconsistency for a circular wave :confused:

For instance:
[itex]\vec{E}=E_0(\hat{e}_x+e^{j\frac{\pi}{2}}\hat{e}_y)e^{-jkz}[/itex] (propagating [itex]\hat{e}_z[/itex])
[itex]\Rightarrow\vec{H}=\frac{E_0}{\eta}(\hat{e}_y-e^{j\frac{\pi}{2}}\hat{e}_x)e^{-jkz}[/itex]

The poynting vector results:
[itex]\vec{S}=\vec{E}\times\vec{H}=\frac{E^2_0}{\eta}e^{-2jkz}(1+e^{j\pi})\hat{e}_z=\vec{0}[/itex] which is quite disconcerting :/

Thx a lot for your help !
 

Answers and Replies

  • #2
34,278
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Your expression is not circular wave. Actually, it is not a wave at all, it is not time-dependent.

However, neglecting prefactors and with fixed orthogonal E and H:
[itex]\vec{E} \propto (e_x - e_y)[/itex]
[itex]\vec{H} \propto (e_x + e_y)[/itex]

=> [itex]S_z = E_x H_y - E_y H_x \propto 1-(-1) = 2 \neq 0[/itex]
 
  • #3
3,740
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I think that if you use complex representation, the Poynting vector is calculated as
S=1/2(E x H*) where the star means complex conjugate.
With this definition, the real part of S is the average power density.
 

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