(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

At time t = 0, the vectors [itex] \textbf{E} [/itex] and [itex] \textbf{B} [/itex] are given by [itex] \textbf{E} = \textbf{E}_0 [/itex] and [itex] \textbf{B} = \textbf{B}_0 [/itex], where te unit vectors, [itex] \textbf{E}_0 [/itex] and [itex] \textbf{B}_0 [/itex] are fixed and orthogonal. The equations of motion are

[tex] \frac{\mathrm{d}\textbf{E}}{\mathrm{d}t} = \textbf{E}_0 + \textbf{B}\times\textbf{E}_0[/tex]

[tex] \frac{\mathrm{d}\textbf{B}}{\mathrm{d}t} = \textbf{B}_0 + \textbf{E}\times\textbf{B}_0[/tex]

Find [itex] \textbf{E} [/itex] and [itex] \textbf{B} [/itex] at a general time t, showing that after a long time the directions of [itex] \textbf{E} [/itex] and [itex] \textbf{B} [/itex] have almost interchanged.

3. The attempt at a solution

Now this looks like a "simple" coupled differential equations, instead of using the formal method to solve te system, I differentiate with respect to time and get

[tex] \ddot{\textbf{E}} = \dot{\textbf{B}}\times\textbf{E} = \textbf{B}_0(\textbf{E}_0\cdot\textbf{E})+\textbf{B}_0\times\textbf{E}_0[/tex]

And similar equation for [itex] \ddot{\textbf{B}} [/itex] (replace E's with B's). But I don't know what to use this result for, this is again a coupled differential equations. Is there a better way to solve this problem? I want to find E and B as functions of time.

Any hint would be appreciated.

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# Homework Help: Vector Calculus

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