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Magnetic field of a circular wave

  1. Sep 3, 2012 #1
    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 !
     
  2. jcsd
  3. Sep 4, 2012 #2

    mfb

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    2016 Award

    Staff: Mentor

    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]
     
  4. Sep 4, 2012 #3
    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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