Question
(II) The electric field of a plane EM wave is given by $E_{x}=$ $E_{0} \cos (k z+\omega t), \quad E_{y}=E_{z}=0,$ Determine(a) the direction of propagation and $(b)$ the magnitude and direction of $\overrightarrow{\mathbf{B}}$.
Step 1
The argument of the cosine function is $kz + \omega t$. We can rewrite this as $k(z + \frac{\omega}{k}t)$. Show more…
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