Imagine that I have a very nice aluminium Faraday cage with some sensitive circuit inside, to not complicate the problem the cage has no holes and the aluminium is $$2\,\mathrm{mm}$$ thick.

As far as I remember from the electrodynamics course, the electromagnetic waves have two components, an oscillating electric field and an oscillating magnetic field, but Faraday cages made of aluminium, given their low permeability, protect almost only from electric fields.

My question is, given that nowadays we have radio waves coming from everywhere (GSM, WiFi, etc), will the cage also protect the circuit inside from the magnetic components of these waves? Or in other words, could interference still arise?

• Radio waves are by definition rapidly oscillating. The B field is inevitably accompanied by electric effects, which even aluminum would short out. However, the aluminum cage would be ineffective against a steady (DC) magnetic field. Commented Mar 15, 2021 at 11:28
• if youre looking to block harmful waves that are either directed at you or just happen to be there then i think wood is the best (wood probably absorbs quite a lot of the frequencies you do) Preferably lacquered wood. Especially for magnetic fields Commented Mar 15, 2021 at 18:48
• Why would wood absorb quite a lot of frequencies you do? Wood has very different composition from human body. Commented Mar 15, 2021 at 23:29

Strictly speaking, if there are no holes worth the name then it is not a cage, it is an enclosure. The whole point of a Faraday cage is to offer as much as possible of the shielding provided by an enclosure, while also allowing you to access the interior.

A characteristic of conductive enclosures is that any incident EM wave sets up eddy currents in the surface. These eddy currents in turn set up magnetic (B) fields which tend to cancel out (or reflect) the incoming B-field. A resistive material will attenuate the eddy currents, reducing the level of cancellation and allowing some of the B-field to leak through. A magnetic material will bind in this B-field better than a non-magnetic one.

Mild steel is slightly resistive and is magnetic, so is quite a good magnetic shield. Aluminium is more conductive, so the eddy currents are stronger and and the resultant B-field weaker. But it is also non-magnetic, so the attenuation of the B-field is less and more may well end up getting through.

I can vouch from my professional experience that interference still gets through both materials, having designed and EMI tested RF power enclosures made from both. (Silver-plated copper has high surface conductivity, but it is expensive and hence impractical for most applications. And it is non-magnetic, so the residual field still gets through). The trick is to ensure the residual interference is harmless, if necessary by hardening your circuit as required.

In the case of the modern ICT (Information and Communications Technology) you ask about, the signal levels penetrating a half-decent enclosure are so pathetic that there is not going to be a problem unless your circuit is a SQUID (superconducting quantum interference device) or similar super-sensitive toy.

Now, a cheaply-made Faraday cage full of unintended gaps and badly-bonded pieces, that is a different threat level altogether!

As long as the cage is electrically conductive, it will act as a shield to EM waves. How conductive? Any common metal will do.

Aluminium has a conductivity $$\sigma \simeq 4\times 10^7$$ S/m. It will act as an excellent conductor whenever $$\omega \ll \sigma/\epsilon_0 = 4.5\times 10^{18}$$ rad/s. In such circumstances, the impedance is given by $$\eta \simeq \sqrt{\frac{\mu_0 \omega}{\sigma}}\exp(-i\pi/4)\, ,$$ which is much lower than the impedance of vacuum. If we consider a wave incident from medium 1 normally upon an interface with medium 2, then the amplitude reflection coefficient is $$r = \frac{\eta_2 - \eta_1}{\eta_1+\eta_2}\, .$$ If $$\eta_2 \ll \eta_1$$, then $$r \simeq -1$$ and the wave is totally reflected.
Second, even if we admit that a small fraction is transmitted, then that fraction is given by $$t \simeq 2\frac{\eta_2}{\eta_1}\, .$$ But this weak transmitted wave is then attenuated whilst travelling through the finite thickness of the aluminium sheets. The attenuation factor is approximately $$\exp(-x/\delta)$$, where the skin depth $$\delta$$ is given by $$\delta = \sqrt{\frac{2}{\mu_0 \sigma \omega}} \simeq 2.5 \left(\frac{f}{{\rm GHz}}\right)^{-1/2}\ \mu{\rm m}\ .$$ Thus at GHz frequencies you have a thousand skin depths between your source of interference and it getting through your 2mm aluminium sheet. If you went down to kHz frequencies then some penetration is possible, but then the reflectivity would be that much higher.
To incorporate the relative gain in using a magnetic material, just substitute $$\mu_r \mu_0$$ for $$\mu_0$$ in the equations above. The impedance would increase, making it less reflective, but the skin depth would increase leading to less transmission.