= Solution
In a vacuum with no charge or current, the <Maxwell equations> are
$$
\nabla\cdot E=0,
\qquad
\nabla\cdot B=0,
\qquad
\nabla\times E=-\frac{\partial B}{\partial t},
\qquad
\nabla\times B=\frac1{c^2}\frac{\partial E}{\partial t}.
$$
For the proposed <plane electromagnetic wave>, $\nabla\cdot B=0$ gives
$$
k\cdot B_0=0.
$$
The <Ampère-Maxwell equation> gives
$$
E_0=-\frac{c^2}{\omega}\,k\times B_0,
$$
and substituting this into <Faraday's law> gives the <dispersion relation>
$$
\omega^2=c^2|k|^2.
$$
Thus both fields have <transverse polarization> relative to $k$, and the corresponding real electric field is
$$
\boxed{E(x,t)=\operatorname{Re}\left[-\frac{c^2}{\omega}
(k\times B_0)e^{i(k\cdot x-\omega t)}\right]}.
$$
For incidence in the positive $x$-direction, choose the incident fields
$$
B_i=B_0\cos(kx-\omega t)e_z,
\qquad
E_i=cB_0\cos(kx-\omega t)e_y.
$$
The <perfect conductor> requires the tangential electric field to vanish at $x=0$. The reflected wave therefore has
$$
B_r=B_0\cos(kx+\omega t)e_z,
\qquad
E_r=-cB_0\cos(kx+\omega t)e_y.
$$
This is <normal reflection of an electromagnetic wave from a perfect conductor>. The magnetic field in $x\leq0$ is
$$
B=B_i+B_r
=2B_0\cos(kx)\cos(\omega t)e_z,
$$
so its tangential value at the surface is
$$
\boxed{B_{\rm tangential}(0,t)=2B_0\cos(\omega t)e_z}.
$$
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