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EM wave propagation direction.

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    Given the following EM wave propagating in vacuum, find the direction along which the electric field oscillates and the direction of propagation of the wave:

    [tex] \vec{E} = (-3\hat{i} + 3\sqrt{3}\hat{j}) 10^4 e^{i[\frac{\pi}{3} (\sqrt{5}x + \sqrt{5/3} y10^7 - 8.1246 *10^{15} t]} [/tex]

    Btw how can I prevent the thread template from reappearing every time I click preview post? It is quite annoying.


    3. The attempt at a solution

    I understand the electric field oscillates along the direction [itex]-1\hat{i} \sqrt{3}\hat{j}[/itex], by looking at the amplitide. But how do I determine the direction of propagation from the phase argument? Do I just treat the x and y in the phase as if they were unit vectors of [itex]\vec{k}[/itex]?
     
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  3. Oct 2, 2011 #2

    vela

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    Pretty much. You know a wave propagating in the k direction has a phase given by [itex]i(\vec{k}\cdot\vec{r}-\omega t)[/itex], so just pick out the components of k.
     
  4. Oct 2, 2011 #3
    Thanks, I see it now. One more question, if I am asked to find the associated [itex] \vec{H}[/itex] field by calculating [itex] \frac{1}{\mu_0 c} \hat{u}\times \vec{E} [/itex], where û is the normalized wave vector, how would I split up the components of E for the determinant? Would it be like:


    [tex] E_x = -3\cdot10^4 e^{i[\frac{\pi}{3} (\sqrt{5}x + \sqrt{5/3} y10^7 - 8.1246 *10^{15} t]}[/tex]


    [tex]E_y = 3\sqrt{3}\cdot10^4 e^{i[\frac{\pi}{3} (\sqrt{5}x + \sqrt{5/3} y10^7 - 8.1246 *10^{15} t]} [/tex]

    for example? Or do I also have to split up the phase/exponent for each component of E?
     
  5. Oct 2, 2011 #4

    vela

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    You can keep the exponential part separate; it's part of the amplitude of the electric field. I would write the electric field as the product of a magnitude and a unit vector:
    [tex]\vec{E} = \left(-\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}\right)(3\times10^4~\mathrm{V/m})e^{i(\vec{k}\cdot\vec{r}-\omega t)}[/tex]The only part you need for the determinant is the unit vector. The rest just scales the result of the cross product.
     
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