Problem 6.2.8: Decibel attenuation of a microwave oven wall

$\omega {\epsilon }_0=2\pi (2.45\times 10^9)\frac{10^{-9}}{36\pi }=\frac{2.45}{18}=0.136S/m$, so $\sigma /\omega {\epsilon }_0=8.8\times 10^6\gg 1$: a good conductor, and $k=\frac{\omega }{c}\sqrt{1-\mathrm{j}\sigma /\omega {\epsilon }_0}\approx (1-\mathrm{j})\sqrt{\pi f{\mu }_0\sigma }$.

It is $d/\delta =107.7$ skin depths thick: $e^{-\alpha d}=e^{-107.7}=1.6\times 10^{-47}$, and

One per cent of the field ($40\mathrm{dB}$) is left at the depth $\delta \ln 100=42.7\mu \mathrm{m}$.

Absorption alone gives $936\mathrm{dB}$. The reflection at the surface between air and steel adds more: $|{\eta }_c|=\sqrt{\omega {\mu }_0/\sigma }=0.127\Omega \ll {\eta }_0$, and the usual estimate $20{\log }_{10}({\eta }_0/4|{\eta }_c|)$ (Section 6.3) gives about $57\mathrm{dB}$. The wall is, for all purposes, a perfect shield.

The exact $k$ has $-\mathrm{Im}k=1.0773\times 10^5{\mathrm{m}}^{-1}$, equal to the good-conductor $\alpha$ to five digits. Rounding $\alpha$ to $1.077\times 10^{5}Np/m$ before exponentiating gives $e^{-107.7}=1.68\times 10^{-47}$ and $935.5\mathrm{dB}$, unrounded $1.63\times 10^{-47}$ and $935.8\mathrm{dB}$. The cited source's version of this wall has ${\mu }_r=500$, which gives $\delta =0.42\mu \mathrm{m}$ and $A_{\mathrm{dB}}\approx 2\times 10^4\mathrm{dB}$. Austenitic stainless steel (type 304, for example) is practically nonmagnetic, ${\mu }_r\simeq 1$, as assumed here.