1
$\begingroup$

As part of a project I'm trying to prevent WiFi transmission of frequency 2.4 GHz from reaching a Raspberry Pi via a Faraday cage.

Would a 20 micron aluminum foil do the job?

$\endgroup$

closed as off-topic by Chris, Emilio Pisanty, Kyle Kanos, Jon Custer, Cosmas Zachos Mar 1 '18 at 15:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question appears to be about engineering, which is the application of scientific knowledge to construct a solution to solve a specific problem. As such, it is off topic for this site, which deals with the science, whether theoretical or experimental, of how the natural world works. For more information, see this meta post." – Chris, Emilio Pisanty, Kyle Kanos, Jon Custer, Cosmas Zachos
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ i almost feel like this would be more suited to electronic engineering, anyone else? $\endgroup$ – Alex Robinson Mar 1 '18 at 9:37
  • 3
    $\begingroup$ I don't know the answer, but why not just try covering virtually any WiFi appliance with such foil and seeing if it works? $\endgroup$ – The Vee Mar 1 '18 at 9:59
  • $\begingroup$ I'd think covering the Raspberry Pi would be the simpler thing to do. Covering the transmitter may not be good for their output stage since the reflected signal is going to increase. $\endgroup$ – Jon Custer Mar 1 '18 at 14:34
2
$\begingroup$

Yes.
The quantity that measures how far an electromagnetic wave of frequency $f$ can pass through a conductor is called the skin depth:

$$ d = \sqrt{{\rho \over \pi f \mu}} $$ Where $\rho$ is the resistivity and $\mu$ the magnetic permeability of the conductor. This equation is true at frequencies lower than $1/\rho\epsilon$ where $\epsilon$ is the permitivity.

If you don't feel like looking up those values for aluminum, use an online calculator. I get a skin depth of 1.647 microns. The skin depth will dictate how quickly the amplitude of the EM wave decreases, and with amplitude related to intensity, we can relate how many dB drop 20 microns of aluminum would generate: $$\text{dB} = -\frac{10}{\ln{10}}\frac{20 \mu\text{m}}{1.647 \mu\text{m}} = -52 \text{dB}$$ Which is down in the noise.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.