Where x is a function of the dimension of its holes.
I mean, if I build a faraday cage able to shield magnetic field at 100kHz, will it we able to shield also all frequencies lower than 100kHz, such as 50Hz?
[Physics] Does a Faraday cage block all magnetic field frequencies lower than x
electromagnetism
Related Solutions
X-rays
In order to make metal radioactive one have to turn it into another element or isotope. This can be performed only with high-energy particles (including photons).
X-rays can be produces if an electron enters metal with very high speed in two ways:
- deceleration radiation (Bremsstrahlung)
- an atom absorbs part of the electron's kinetic energy, moves to one of the excited states and moves back to ground state emitting a high-energy photon
In any case the energy of the incident radiation (or particles) must be comparable to the energy of X-ray or gamma photons. This is far from cellular phone frequency range for sure.
Shields and cages
Solid metallic shield reflects electromagnetic waves back. The material of the shield is important since it should have good conductivity. Most of metals works well. As far as I know, copper and gold are the best especially for high frequencies (microwaves).
If the frequency is quite low there is no need for solid shield.
The effective area that reflects the wave is proportional to $\frac{\lambda^2}{4\pi}$, where $\lambda$ is the wavelength.
It can be much larger than the antenna size. So metallic lattice works well if the distance between the wires is lower than $\lambda$.
GSM phones uses frequencies about 1 GHz which corresponds to $\lambda\approx$ 30 cm.
For low frequencies the diffraction effects are important. If the size of the shield is comparable to $\lambda$ the radiation can just bypass the obstacle as it happens with sound. In this case metallic box is the best solution. It can be a cage with appropriate cell size. There should be no big holes like doors and windows.
Grounding
Grounding removes charges from the outside surface of Faraday cage, but if you are inside there is no way to determine whether it is grounded or not. This is more concerned with safety.
Shape of the antenna
This is important if you need something more interesting than just screening.
Using the effect of Bragg diffraction, it is possible to build a shield that reflects one frequency and does not affect others (this will work only for some directions).
The shape of the shield also allows to control polarization of the radiation.
Edit 1. Answering the questions in the comments
Is the 'energy' of incident radiation proportional to the frequency of transmitted EMF waves, or what people commonly also call 'radiated power' measured in Watts/meter-sq (as well) ?
There are two energy characteristics for EMWs:
- Intensity - the amount of energy incident on unit area per unit time (measured in W/m$^2$). It describes total power of the radiation.
- Energy of quantum - the energy of single photon (measured in Joules). It is equal to product of Planck's constant and frequency: $h\nu$ or $\hbar\omega$. This value is very important in quantum mechanics since quantum system can absorb only integer number of photons. If one photon is not enough to change system state then even 1000 photons will just pass through with no effect.
if EMF radiation that is several hundred/thousand times in excess of international norms, could cause the X-ray generation
This is possible if you have a system that can collect energy of low-frequency radiation and turns it to something else. For example, if EMW induces plasma discharge between some metallic details and electrons collect enough energy before collision there can be X-rays (I'm not sure such situation is possible).
Edit 2. Answering the questions in the comments
would the lattice structure/size computation be good enough if I base it on the highest frequency (thus get the smallest lattice size needed) ?
Also, does it matter if the material (s.a. common steel mesh) has the cross-over joints fused or insulated from each other ?
If lattice period is smaller than $\lambda$ then it should work as a good screen.
Since you need computations and optimization it's better to ask someone who specializes in electrodynamics and antenna theory.
May be it's better to ask this as a separate question.
Edit 3. Answering the questions in the comments
is it possible to make practical application of Bragg diffraction to cause destructive interference of the EMF wave, when the waves are for large no. of different carrier waves, and clustered around 4-5 group of central frequencies ?
This can be done with multiple Bragg mirrors one for each frequency. AFAIK it is done for infrared radiation.
Apart from Bragg diffraction are there other ways in which the EMF can be reduced / nullified in a small region (say within a radius of 5-6 meters), where the EMF energy is captured using an antenna, converted to electrical energy, and converted to heat/light
Single antenna affects EMF only within $\lambda$ distance. 5 meters is too much for 1 GHz which corresponds to $\lambda\approx$ 30 cm.
No. The point of a Faraday cage is that it's made of a conductor, which responds to electric fields. A strong magnetostatic field is different, and will barely be affected by the Faraday cage. (The cage may have some magnetic properties, but that's not what makes it a Faraday cage, and it's unlikely to have a significant impact.)
There's a little info on Wikipedia about magnetic shielding: http://en.wikipedia.org/wiki/Electromagnetic_shielding#Magnetic_shielding
Best Answer
No. Faraday cages cannot block low frequency magnetic fields. Faraday cages work through the redistribution of electrical charge throughout their electrically conductive structure, so they mainly shield against electric fields. Your relationship is true for electric but not magnetic fields. High frequency magnetic fields in the form of farfield electromagnetic radiation can be shielded against, for such radiation cannot propagate when its constituent plane waves have their electric field components "tethered" to a small value by reaction from moving charge in the conductive cage. Each such plane wave component must have $|\vec{E}| = c|\vec{B}|$, so the magnetic fields are quelled if the electric fields are.
If you need to shield against low frequency magnetic fields, as is done for an oscilloscope, you can use a continuous (as opposed to a cage-like) shield of mu-metal.
Response to Comment
User1247 makes the interesting comment:
This is true in some applications. But that's effectively the same as a Faraday cage, as the eddy currents are essentially the Faraday cage action; the redistribution of charge. And you're going to need a great deal of copper at 50Hz: I make the skin depth of copper to be about 9mm at 50Hz.
This comment points the way towards a quantitative version of my answer: a continuous surface 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. The electromagnetic field is shielded because it can't propagate away from its source: away from sources Maxwell's equations become the following complex wavenumber Helmholtz equation (we're writing the equations for a Cartesian component of the field "trying" to propagate through a metal):
$$\left(\nabla^2 + i\,\omega\,\mu\,(\sigma + i\,\omega\,\epsilon)\right)\,H = 0$$
where $\omega,\,\mu,\,\epsilon,\sigma$ are angular frequency, magnetic constant, electric constant and conductivity of the medium, respectively. Thus, plane waves have wavenumbers (often called propagation constants in this context):
$$k = \pm\sqrt{i\,\omega\,\mu\,(\sigma + i\,\omega\,\epsilon)}\approx\pm\,e^{i\,\frac{\pi}{4}}\,\sqrt{\omega\,\mu\,\sigma}$$
where the approximation holds for a conductor like copper and $k$ thus has a fiercely attenuating real part. Incoming electromagnetic fields will decay to $1/e$ of their amplitude after having propagated the skin depth:
$$\delta = \sqrt{\frac{2}{\omega\,\mu\,\sigma}}$$
which is about 9mm for copper at 50Hz.
So depending on how big the incoming field is, and how much you need to attenuate it by, the copper may have to be several centimeters thick.