Problem 5.6.11: \vec H(x;t) from a phasor \vec E

The factor $e^{+j2x}$ gives $\beta =2rad/m$ and a wave travelling to $-x$. $16j=16e^{j\pi /2}$: amplitude $16V/m$, phase $90^{\circ }$. The medium is lossless, so

and with $\mu =4{\mu }_0$, $\omega \mu =10^8\times 4\times 4\pi \times 10^{-7}=160\pi \Omega /\mathrm{m}$.

$\nabla \times E=-j\omega \mu H$, so $H=\frac{j}{\omega \mu }\nabla \times E$:

The intrinsic impedance of the medium is $\eta ={\eta }_0\sqrt{{\mu }_r/{\epsilon }_r}=120\pi \cdot \frac{2}{3}=80\pi \Omega$, and $|E|/|H|=16\times 5\pi =80\pi$. $E\times H\propto a_z\times a_y=-a_x$: the power flows to $-x$, as $e^{+j2x}$ says.