Problem 6.1.6: Material characterisation

Given: the amplitudes $E_0=1.333V/m$ and $H_0=4.243mA/m$, both fields varying as $\cos (\omega t-\beta y)$, $\mu ={\mu }_0$, $\sigma =0$. Unknown: ${\epsilon }_r$. Relations: $H=\frac{1}{\eta }{a}_k\times E$ and $\eta =\sqrt{{\mu }_0/{\epsilon }_r{\epsilon }_0}={\eta }_0/\sqrt{{\epsilon }_r}$.

The phase $\omega t-\beta y$ is constant on planes $y=$ const that move towards $+y$. With the fields, $a_E\times a_H=a_x\times (-a_z)=a_y=a_k$. In phasor form $H=\frac{1}{\eta }{a}_y\times 1.333{\mathrm{e}}^{-\mathrm{j}\beta y}{a}_x=-\frac{1.333}{\eta }{\mathrm{e}}^{-\mathrm{j}\beta y}{a}_z$: the minus sign of $H_z$ is exactly what a wave towards $+y$ needs.

The refractive index is $n=\sqrt{{\epsilon }_r}=1.2$, so $v_p=c/n=2.5\times 10^{8}m/s$, and

$E_0^2/2\eta =1.333^2/(2\times 314.2)=2.83mW/{\mathrm{m}}^2$, the same as $\frac{1}{2}{E}_0{H}_0$. Since ${\epsilon }_r=({\eta }_0/\eta {)}^2$, an error of $1\%$ in the measured $E_0/H_0$ gives $2\%$ in ${\epsilon }_r$ (Figure 6.1.29).