Does an electronic device housed within a sealed Faraday cage leak radiation outside the cage?
Link.
Yes, it works both ways. BTW, folks around here appreciate it if you make your thread title informative as to what the topic is. You might want to acquaint yourself with the rules before posting more threads. Just FYI.
As John Mace said, Faraday cages work both ways. They prevent radio frequency radiation outside the cage from getting inside of it, and they prevent radio frequency radiation inside the cage from getting outside of it. In the real world, Faraday cages usually aren’t perfect. You end up with a small bit of leakage around seams and holes (most Faraday cages have holes for things like power cords and such, otherwise they aren’t of much use).
Also, the frequency matters. You generally want the holes in the cage to be significantly smaller than the smallest wavelength you want to block. A factor of 10 usually works pretty well in the real world.
A good common example is a microwave oven. The box that the food goes into and the screen on the door all combine to make a Faraday cage. It works well enough to reduce the microwave radiation inside the box (which are strong enough to cook food) down to a level where it is not harmful at all to humans outside the box. Often, though, if you stick a cell phone inside the microwave the phone can still receive incoming calls. The signal strength will be greatly reduced, but often there’s enough signal leaking through that the cell phone’s sensitive receivers can pick it up. Note - If you are in an area with weak cell phone reception the microwave oven’s shielding may attenuate the radio signals to the point where this won’t work for you. Sometimes it works, sometimes it doesn’t. Also, DO NOT TURN THE MICROWAVE OVEN ON WHEN YOU ARE TRYING THIS OR YOU WILL DESTROY THE PHONE VERY QUICKLY.
Cell phones and microwave ovens both use signals around 2.4 GHz, by the way. It’s the same stuff, just different power levels.
For some definition of “around”. The typical cell phone signal is usually more in the 700 MHz - 2.1 GHz range, but of course this is still in the microwave region.
WiFi, however is (usually) 2.4 GHz on the dot, and so microwave oven interference can be quite strong.
I think a more interesting question is: Does any radiation leak from a completely sealed (100% welded) box? I know that the H-field will not be completely attenuated, but what about the E-field?
By Gauss’s law–yes, it must leak. Nothing can completely shield an electric charge (not even a black hole). Though you could make it completely unchanging (I suspect that a superconductor would make for a perfectly static E-field).
This doesn’t stop people from doing it.
A DC H-field won’t be attenuated just by an electrically conducting box. But for radiation there will be currents set up that will attenuate the H-field as well. In a sealed box, whether anything will leak will depend on how thick the walls are, and what the skin depth is at the frequency you’re interested in. But with enough thickness, you could completely attenuate any radiation generated in the box by a cell-phone or whatever.
Not by Gauss’s law. Radiation has no charge, and you can generate radiation by moving charges around within the box. Gauss’s law would only come into play if you were adding or removing charge from the box. You’re right that you can’t shield any charge, but that’s not the same as radiation.
True enough, but **beowulff **first said “radiation” and then “E-field”, and as you note they are not the same thing, since radiation is only emitted from moving charges. My comment assumed we were talking about the E-field in general. I probably should have made this distinction more clear.
<expanding on (usually)>
For some definitions of WiFi. And I don’t have my channel map handy, but it could only be for one specific channel. (The “on the dot” part).
</expanding on (usually)>
That being said…the microwave isn’t tuned to that exact frequency, and should provide lots of shielding for WiFi (b/g) and Bluetooth.
-D/a
Even a dipole antenna sealed inside a metal box will leak a tiny bit, to the extent that the metal isn’t perfectly conducted. The amplitude will fall off exponentially as a plane wave travels through a metal, with length scale of the skin depth. For most metals and reasonable frequencies, this length is fairly small, and if you make your box walls more than a few skin depths thick, you’ll get negligible leakage. But the radiation outside the box won’t be exactly zero unless your box material has zero resistance.
As i have posted on this Board before, I have repaired Faraday cages. I am still amazed at how easily they can be compromised. You’d think that something as simple as a Faraday cage would be easy to construct and have working properly. And if it were simply a metal box with no breaks, that’s be easy. But people want light and air to flow easily in and out, and to put electrical devices in and have computers as part of a network inside.
One year I built and repaired several cages. I was floored to find that I could bring my cell phone and a radio into what was apparently a sealed metal-mesh encased roo m and have noth continue to work. The electrically-conducting seals on the dors were OK, so what was the problem? Several things:
a.) Any access holes that are too large
b.) and access holdes that have insufficiently long isolating pipes extend from them
c.) computer network connections not running over fiberoptic links
d.) electrical power not brought in through isolating transformers
and, most insidious of all –
e.) any rips or tears in the wire mesh.
Even if the riop leaves the rest of the mesh intact, a slice through the mesh actually acts as an antenna, ‘broadcasting signal into (or out of) the cage. I had to meticulously go over the mesh, because it might appear to be intact to the naked eye, but have a cut running through a portion. I had to "sew’ or "weave the ends back together, or use a cwuire mesh patch to restore integrity. When I could turn on the radio or cell phone and cet no signal, I knew i’d sealed all the gaps.
And when the cage wasn’t secure, I could call out of the cage, so signal definitely went both ways
I have a tangential question. I remember a physics lecture where the prof showed how the charges redistribute to cancel an EM field. I think I understood it then but remember little of it now, and never took any EE courses that might have solidified it.
The thing is, some experience seems to contradict the theory. I’m confident the theory is correct, so wondering why my experience seems to contradict it.
Background
I have a Fender Jazzmaster guitar, which has single-coil pickups. A single-coil pickup is particularly subject to EM buzz, for the obvious reason that it’s a coil intended to pick up EM fields (caused by the string vibrating nearby). Double-coil pickups (aka “humbuckers”) use two coils in proximity, wired in parallel but with one inverted, so that they cancel out “common mode” EM but pick up the local EM field of the string. In addition to the Jazzmaster pickups being single-coil, they’re also “fat” meaning the coils are wide ovals wrapped around a short spool – the wires wind around themselves deeply rather than being distributed shallowly along a long spool. That is, the diameter of the coil is high, so it has great common-mode sensitivity, which makes it sound cool but noise-sensitive.
To make matters worse, I moved to a house where the noise was unusually bad. I tried eliminating everything in the house that might cause noise, but to no avail (including killing all circuits but the one. I guess I must have been close to some source outside the house, but don’t know for sure what the cause was.)
Anyway, to reduce the problem, I experimented with shielding the pickups. First I removed them from the guitar and wrapped them completely in aluminum foil, and grounded that foil to the guitar cable shield (two-conductor cable). The result was no reduction in noise, as far as I could tell.
How can that be? How can completely wrapping a coil in foil not reduce the EM transmission? (The coils were separated from the foil by several inches, if that matters.)
Clearly, I’m missing something, and would appreciate any illumination. Sorry if this is too much of a hijack.
Maybe Cal answered my question. There was one hole, where the cable came in (say, 1/4" diameter), and the cable was not transformer-isolated. It was plugged into a hi-Z guitar amp input (say, >100K Ohms)
The coil is sensitive to magnetic fields, which are not going to be attenuated by aluminum foil. I suspect that the reason your house has a lot of hum is because it has large current loops - I’ve seen this in a commercial building, where the hot and neutral wires were not run next to each other, but rather in a big loop.
Learjeff, the calculation you saw in physics class was probably for a static charge distribution. Electromagnetic waves (or, really, anything dynamic) are more complicated.