= Solution
Write $\psi=kx-\omega t$ and $v_p=\omega/k$. The <electric field> is
$$
\mathbf E=A\sin\psi\,(\hat{\mathbf y}+6\hat{\mathbf z}).
$$
It has <linear polarization>, along the fixed line $E_z=6E_y$, at angle $\tan^{-1}6\simeq80.54^\circ$ from $+y$ toward $+z$. For the associated <plane electromagnetic wave>, <Faraday's law> gives $\mathbf B=v_p^{-1}\hat{\mathbf x}\times\mathbf E$. Hence
$$
\boxed{\mathbf B=\frac{A}{v_p}\sin\psi\,(-6\hat{\mathbf y}+\hat{\mathbf z}),\qquad B_x=0.}
$$
In vacuum $v_p=c$. Both fields are transverse, and $\mathbf E\times\mathbf B$ points along $+x$.
Back to article page
