Home exercise 6.2.5: Skin depth and 20 dB

$\alpha =\beta =1/\delta =1/(5\times 10^{-6})=2\times 10^{5}Np/m$ (rad/m).

$e^{-\alpha z}=0.1$ gives $z=\delta \ln 10=5\times 2.3026=11.5\mu \mathrm{m}$.

$\lambda =2\pi \delta =31.4\mu \mathrm{m}$ and $v_p=\omega /\beta =\omega \delta =2\pi \times 10^{10}\times 5\times 10^{-6}=3.14\times 10^{5}m/s$.

$\sigma /\omega {\epsilon }_0=1.01\times 10^6/0.556=1.8\times 10^6\gg 1$: the good-conductor formulas were allowed. And $20\mathrm{dB}$ is $20/8.686=2.30\mathrm{Np}$, that is $2.30\delta$. A conductivity of about $1MS/m$ is that of stainless steel (Problem 6.2.8). Copper at $10\mathrm{GHz}$ would have $\delta =0.66\mu \mathrm{m}$ and $R_s=0.026\Omega$, almost eight times less loss: waveguide parts are made of copper, brass or aluminium, or plated with silver, but never left as bare steel.