6.2.3 Intrinsic impedance and power flow

In the lossless medium the magnetic field of a plane wave travelling along $+z$ is $H=\frac{1}{\eta }{a}_z\times E$ with $\eta =\sqrt{\mu /\epsilon }$ (Section 6.1). Replacing $\epsilon$ by ${\epsilon }_c$ gives the complex intrinsic impedance of the lossy medium,

The angle follows from $\sqrt{1-\mathrm{j}p}=(1+p^2{)}^{1/4}{e}^{-\mathrm{j}\frac{1}{2}\arctan p}$ with $p=\sigma /\omega \epsilon$, and since $p$ runs from $0$ to $\infty$, ${\theta }_{\eta }$ runs from $0^{\circ }$ (a perfect dielectric) to $45^{\circ }$ (a perfect conductor). Two compact relations between $\gamma$ and ${\eta }_c$ follow directly from Eq. (6.2.6) and Eq. (6.2.12), and they are often the fastest route in a problem:

Since $\mathrm{j}\omega \mu$ has the angle $90^{\circ }$, the first relation says that the angle of $\gamma$ is $90^{\circ }-{\theta }_{\eta }$, hence $\alpha /\beta =\tan {\theta }_{\eta }$.

With $E=a_x{E}_0{e}^{-\gamma z}$ the magnetic field is $H=a_y(E_0/{\eta }_c)e^{-\gamma z}$, and in the time domain

The magnetic field lags the electric field by ${\theta }_{\eta }$. The time-average Poynting vector (Section 6.1) is

Where does the power go? In one attenuation length $1/\alpha$ the field falls to $e^{-1}=36.8\%$ and the power to $e^{-2}=13.5\%$. The power that leaves the beam heats the medium. With the identities of Eq. (6.2.13) one can show (Exercise 6.2.3) that at every depth

the time-average Joule power per unit volume of Section 4.2.7. This is Poynting's theorem for the plane wave: the power that disappears from the wave in a slab $dz$ is exactly the heat produced in it. The proof uses only $\gamma {\eta }_c=\mathrm{j}\omega \mu$, whose real part gives $\alpha \cos {\theta }_{\eta }=\beta \sin {\theta }_{\eta }$, and the real part of $\gamma /{\eta }_c=\sigma +\mathrm{j}\omega \epsilon$, which then gives $\sigma =2\alpha \cos {\theta }_{\eta }/|{\eta }_c|$.