Problem 6.2.4: How thick must a shield be?

$\sigma /\omega {\epsilon }_0=3.5\times 10^7/(2\pi \times 10^6\times 8.84\times 10^{-12})\approx 6\times 10^{11}$, so

A reduction of $60\mathrm{dB}$ is a factor $10^{60/20}=1000$ in amplitude:

$\delta \propto 1/\sqrt{f}$ grows by $\sqrt{10^6/50}=141$: $\delta =12.0\mathrm{mm}$ and $t=6.91\delta =83\mathrm{mm}$.

An $8\mathrm{cm}$ aluminium wall is absurd. At power frequencies the magnetic field is shielded with a high-permeability alloy instead. Mu-metal, with ${\mu }_r\approx 2\times 10^4$ and $\sigma \approx 1.7\times 10^{6}S/m$, has

$t/\delta =6.91$ skin depths give $6.91\times 8.686=60.0\mathrm{dB}$. At $10\mathrm{kHz}$ the same rule gives $5.9\mathrm{mm}$: every factor $100$ in frequency divides $\delta$, and the thickness needed, by $10$ (Figure 6.2.20).