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Just a quick question if I may.

The Poynting vector, or the energy flux density, is given by:

$\mathbf{S} = \frac{1}{\mu_{0}}(\mathbf{E} \times \mathbf{B})$

So it's the cross product between the $\mathbf{E}$-field and $\mathbf{B}$-field. So depending on the direction of the fields, the Poynting vector will point in some direction. So lets say the $\mathbf{E}$-field has the direction $\mathbf{e}_{y}$ and the $\mathbf{B}$-field has the direction $\mathbf{e}_{z}$, then the resulting direction for $\mathbf{S}$ will be $\mathbf{e}_{z}$.

So my question is, is that the direction of which the energy is flowing, or is there some fancy thing I need to know, like it's the opposite or something like that ?

Thanks in advance :)

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Why didn't you look it up at Poynting_vector. –  Ali May 27 '13 at 17:56
    
You are going to keep on collecting downvotes unless you change the result. It is in the $e_x$ direction, not $e_z$ –  Eduardo Guerras Valera May 27 '13 at 23:17
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1 Answer 1

up vote 2 down vote accepted

The Poynting vector was defined as directional energy flux density. Therefore, it naturally shows the way energy flows and you do not have to switch the direction or anything. So, if you have an $\mathbf{E}$-field in the direction $\mathbf{e}_y$ and $\mathbf{B}$-field in the direction $\mathbf{e}_z$, Poynting vector is in the direction $\mathbf{e}_x$ and that is the direction in which energy flows.

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Ok, that should cover it :) Thanks you very much. –  Denver Dang May 27 '13 at 17:09
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