Should a Faraday cage block a radio's signal?

Today, I tried creating a very basic Faraday cage by surrounding a radio with two baking trays made out of iron. It didn't seem to affect the radio's signal (AM was being used, not FM).

In theory, should what I used block the radio's signal?

I can't remember how thick the baking trays were. There was some gaps totalling a few square centimetres, because the two baking trays were not identical. I didn't attempt to "Ground" the baking trays.

• See Walter Lewin's lecture on Faraday cages at ocw.mit.edu. – JamalS Jun 29 '14 at 7:46
• How did you "surround" it? Was it completely enclosed? How thick were the baking trays? Did you ground them? – Rob Jeffries Jun 29 '14 at 8:06
• What type of radio , what type of antenna? Di the trays touch the antenna is what I am aiming at. – anna v Jun 29 '14 at 9:24
• Was this a battery-op radio or did an A/C cord sneak in between the baking trays? Despite RobJeffries' answer, I would suspect that a tiny gap between the trays is more than sufficient for an AM signal to get thru. – Carl Witthoft Jun 29 '14 at 11:49
• It would be nice to get some feedback from the OP. FYI I just put some Al kitchen foil around a radio and it completely cuts the signal. You don't even have to wrap it up, just flop the sheet loosely over the top is basically enough. On the other hand I put the radio in a biscuit tin of unknown construction and could still hear the radio inside. So, are the baking trays iron? – Rob Jeffries Jun 29 '14 at 20:58

I'll assume that you were using a radio tuned to a 1Mhz frequency ($\omega = 6.3\times 10^6$ s$^{-1}$) and that the radio was completely enclosed inside $t=3\,$mm of pure iron.

There are two important effects to consider. (i) How much power is reflected from the iron surface. (ii) How much of the transmitted power makes it through the iron.

To figure this out we need the properties of iron; a conductivity $\sigma = 10^7$ S/m, a relative permittivity $\epsilon_r \simeq 1$ and a relative permeability $\mu_r \simeq 10^4$ (for 99.9% pure iron).

First we check whether iron works as a good conductor at these frequencies by noting that $\sigma/{\epsilon_r \epsilon_0 \omega} = 1.8\times 10^{11}$; i.e. $\gg 1$ and therefore a good conductor.

The modulus of the impedance of a conductor is given by $\eta_{\rm Fe} = (\mu_r \mu_0 \sigma / \omega)^{1/2} = 0.14$ $\Omega$.

So, now the relevant equations are: Electric field transmission at the air/iron interface (assuming normal incidence)

$$\frac{E_t}{E_i} = \frac{2 \eta_{\rm Fe}}{\eta_0 + \eta_{\rm Fe}} \simeq 2\frac{\eta_{\rm Fe}}{\eta_0}\, ,$$. where $\eta_0 = 377$ $\Omega$.

The EM waves then propogate into the metal but are exponentially attenuated on a scale defined by the "skin depth" $\delta = (2/\mu_r \mu_0 \sigma \omega)^{1/2} = 1.59 \times 10^{-6}\,$m.

Thus, after traversing a thickness $t$, the E-field is attenuated by $\exp(-t/\delta)$.

Finally the wave emerges through the iron/air interface on the other side and we use the transmission formula again but with the labels swapped on the impedance values.

Hence the ratio of the net transmitted electric field to the incident electric field is given approximately by $$R = 2 \frac{\eta_{\rm Fe}}{\eta_0}\, \exp(-t/\delta)\, 2 \frac{\eta_0}{\eta_0 + \eta_{\rm Fe}} = 4 \frac{\eta_{\rm Fe}}{\eta_0}\, \exp(-t/\delta)\, .$$

For the numbers I've assumed $R \simeq 0$ because the wave traverse $>1000$ skin depths to get through the iron! The transmitted power is $\propto R^2$.

So my conclusion is that enclosing within 3mm of pure iron would certainly block AM radio.

How might this not work? Perhaps the iron you used is very impure and the permeability is orders of magnitudes lower? If $\mu_r =1$ then $\eta_{\rm Fe} = 0.0014\,\Omega$, $\delta = 1.59\times 10^{-4}\,$m. Thus 3mm is still 18 skin depths. The conductivity I assumed is unlikely to be much lower, so I'm a bit confused as to why it wouldn't work.

The demo I use in my lectures is wrapping a mobile phone in aluminium foil. In principle, this is much more marginal because though the frequencies are higher, the foil thickness is much lower, but it certainly works.

• Flat surface Faraday cages (like the simple iron box you describe or a foil surface that is laid down smoothly) suffer a little from the signal inducing motion in the conduction electrons that tracks the signal and re-radiate in-phase (but about half in each direction). Using a wire cage with spacing much smaller than the wavelength (or foil that has been previously crinkled up and is no longer smooth on the wavelength scale) strongly damps this and blocks the signal better. It's the usual coherent versus incoherent bit. – dmckee Jun 29 '14 at 15:16
• I'll have to experiment. The calc. above assumes the surface is flat and normal incidence. I believe the treatment accounts for induced currents. The ohmic dissipation integrated over say a skin depth is equal to the loss of flux implied by the diminishing Poynting vector. – Rob Jeffries Jun 29 '14 at 20:39
• I think you are right abut getting the correct answer for your calculation, but the point is that you can do better per unit mass with less simple geometries. The difference between covering a flash lamp with a neat and smooth layer of Al-foil and a crinkled up messy layer was about an order of magnitude. – dmckee Jun 29 '14 at 21:14

AM radio is in a band from about 500kHz to 1500kHz which corresponds to wavelengths from about 200m to 600m, vastly longer than your baking trays. This affects the manner of the interaction between the waves and the trays, and how much the trays will attenuate the signal.

For the sake of comparison, your microwave oven is a Faraday cage; it effectively confines some 800W to 1000W of microwave power. A microwave oven emitter typically operates around 2GHz or so, corresponding to a wavelength on the order of a few centimeters. That's much smaller than the cavity, The holes in the window screening are very much smaller than a wavelength - practically invisible to the wave. This relationship in sizes allows the cage to effectively reflect the microwave energy, preventing the escape of all but a tiny amount of energy.

AM receivers are designed to work on miniscule signals. Any local broadcasts would have to be attenuated considerably before the receiver is no longer able to work with it. An improvised cage might prevent the radio from receiving distant stations, but it would take something more carefully constructed to effectively block a stronger, more local signal.

• As I mentioned above. Just loosely folding a bit of aluminium foil over my AM radio effectively cut out a radio signal at 909kHz. I did not wrap it up and I could still see the radio through a small gap between the foil. – Rob Jeffries Jun 30 '14 at 6:35

"There was some gaps totalling a few square cm"

A rule of thumb is that if the screen isn't airtight those radio waves will get in - or out

• My understanding is that some Faraday cages aren't airtight, but still work. Why is it different in this circumstance? Is it because of the wavelength of the signal used by AM radio? – Andrew Grimm Jun 30 '14 at 2:14
• Small gaps should not matter greatly. As I've said above, I tried this with Al foil and it blocked the signal without having to wrap the radio up. The wavelengths here are large ~hundreds of meteres. – Rob Jeffries Jun 30 '14 at 11:08
• @AndrewGrimm as long as the gaps are significantly smaller than a wavelength then they don't matter. That's why your microwave (15cm) has a mesh with 1mm holes in the door. But for high frequencies, and ultra low noise, it is difficult to make an effective screen – Martin Beckett Jun 30 '14 at 22:35
• -1 Because that just isn't true. – Rob Jeffries Nov 30 '14 at 12:32