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Can someone explain why my am radio doesn't work when covered by a layer of foil that is less than one 'skin depth' at the appropriate frequency?

According to wikipedia and other websites on the subject of 'skin depth' kitchen foil which is 0.01mm thick is less than half of the skin depth of a 200khz em wave, so according to the graph and article should not completely attenuate said wave.

why then, does it attenuate it in my radio(s)?

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  • $\begingroup$ do you know what a faraday cage is? $\endgroup$ – niels nielsen Sep 17 '19 at 3:43
  • $\begingroup$ yes. it's continuous metal enclosure which redistributes incident electrical charge. However I am trying to integrate the concept of 'skin depth' into my understanding of attenuation and am under the impression that low frequency em waves aren't sufficiently attenuated when the thickness of a material is less than 1 skin depth of the said wave. $\endgroup$ – SIRT Sep 17 '19 at 3:56
  • $\begingroup$ I will think about this. $\endgroup$ – niels nielsen Sep 17 '19 at 4:42
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The skin depth of a good conductor is given by the expression $$ d = \sqrt{\frac{2}{\mu_r \mu_0 \sigma \omega}},$$ where $\omega$ is the angular frequency of the EM wave and $\sigma$ is the conductivity. For an angular frequency of $6\times 10^5$ rad/s and aluminium foil with $\mu_r = 1$ and $\sigma \sim 4\times 10^6$ S/m, then $d \sim 0.8$ mm and much bigger than the thickness of aluminium foil.

However, don't forget that there is an important reflection effect from the surfaces of conducting materials too. For a good conductor, the modulus of the transmission coefficient (from air/vacuum into the conductor) is approximately $$ |T| = 5.3\times 10^{-3} \sqrt{\frac{\mu_r \mu_0 \omega}{\sigma}}$$ Thus a more conductive material will transmit less field at the interface between the conductor and air/vacuum. This effect is more important for attenuating the E-field at low frequencies or cases where the conductor is not thick compared with the skin depth.

For the situation of Al foil and am radio, I get $|T| \sim 2\times 10^{-6}$. This transmission coefficient is for the field strength, so almost all the power is reflected.

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The concept of "skin depth" only applies for conductors whose thickness is significantly greater than the skin depth at the AC frequency of operation. For the case where the conductor is thinner than what the skin depth would be for a thick conductor, this does not mean the current is somehow protruding out of the conductor. It instead means the current is bunched within the available thickness and does not follow the exponential density curve for the skin effect.

This means that stopping the transmission of an EM wave through a conductive shield does not require the shield to be thicker than the skin depth. This is why a single layer of aluminum foil can stop radio waves.

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  • $\begingroup$ Another user, 'User2147' said "Faraday cage works as a Faraday cages if it's considerably thicker than the skin depth of the conductor in question at the frequency in question." user 'gotaquestion' also said "there are two conditions for Faraday's cage to shield EM fields, which are: 1. The conductor is thick enough. 2.The holes in the cage are much smaller than the wavelength. are they wrong? $\endgroup$ – SIRT Sep 17 '19 at 5:29
  • $\begingroup$ A shield is a faraday cage with no holes in it. have a look at the wikipedia page for skin effect and skin depth. $\endgroup$ – niels nielsen Sep 17 '19 at 5:55
  • $\begingroup$ cage and shield are used interchangeably. I will experiment with some loop antenna to investigate the effect skin depth has on propagation as there doesn't seem to be any consensus on google or otherwise. $\endgroup$ – SIRT Sep 17 '19 at 6:29

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