= Solution
The <source-free Maxwell equations in a linear medium> are
$$
\nabla\cdot D=0,\qquad
\nabla\cdot B=0,
$$
$$
\nabla\times E=-\frac{\partial B}{\partial t},
\qquad
\nabla\times H=\frac{\partial D}{\partial t},
$$
with $D=\epsilon E$ and $B=\mu H$.
Take complex plane waves
$$
E=E_0e^{i(k\cdot x-\omega t)},
\qquad
B=B_0e^{i(k\cdot x-\omega t)}.
$$
The divergence equations give
$$
k\cdot E_0=k\cdot B_0=0,
$$
and <Faraday's law> gives
$$
\boxed{B_0=\frac1\omega k\times E_0}.
$$
Ampère's law gives
$$
k\times\frac{B_0}{\mu}=-\omega\epsilon E_0.
$$
Substituting the expression for $B_0$ and using
$k\times(k\times E_0)=-|k|^2E_0$ yields
$$
\frac{|k|^2}{\mu\omega}E_0
=\omega\epsilon E_0.
$$
Hence the <plane electromagnetic wave in a linear medium> has
$$
\boxed{\omega^2=\frac{|k|^2}{\epsilon\mu}},
\qquad
\boxed{v=\frac{\omega}{|k|}
=\frac1{\sqrt{\epsilon\mu}}},
$$
and equivalently
$$
\boxed{B_0=\sqrt{\epsilon\mu}\,\widehat k\times E_0}.
$$
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