Problem 6.2.5: A lossy magnetic dielectric

$\omega \sqrt{\mu \epsilon }=\frac{10^8\sqrt{2\times 8}}{3\times 10^8}=\frac{4}{3}rad/m$, so

Dividing the two relations,

and $\sigma =0.515\times 10^8\times 8{\epsilon }_0=3.65mS/m$: a lossy medium, neither low-loss nor a good conductor.

With ${\epsilon }_c=\epsilon (1-\mathrm{j}\tan {\delta }_c)$,

$v_p=\omega /\beta =10^8/1.374=7.28\times 10^{7}m/s$.

The direction of $H$ is $a_k\times a_E=a_z\times a_x=a_y$, its amplitude $0.5/177.7=2.81mA/m$, and $H$ lags $E$ by ${\theta }_{\eta }$:

With $\sigma =3.646mS/m$ the exact $\gamma =\sqrt{\mathrm{j}\omega \mu (\sigma +\mathrm{j}\omega \epsilon )}=0.3333+\mathrm{j}1.3744{\mathrm{m}}^{-1}$. The angle of $\gamma$ is $90^{\circ }-{\theta }_{\eta }=76.37^{\circ }$, and indeed $\arctan (1.374\times 3)=76.37^{\circ }$.