is there any way to make electromagnetic waves reach a cell phone in Faraday cage although conductor surround cell phone everywhere? can we pass current through conductor to make charges move as a trick? then if electromagnetic field reach the conductor then no charge can prevent electric field as they are mobile not static charges.
[Physics] How to electromagnetic waves reach a cell phone in Faraday cage
conductorselectromagnetic-radiationelectromagnetismmicrowaves
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Faraday's cage is known to block static and non-static electric fields. The mechanism of blocking depends on whether the electric field is static or non-static (EM field). I suppose you question is about how the cage works in non-electrostatic case.
In EM case (time changing field), two scenarios could happen. The first is electric discharge where the current flows from a distant electrode to the cage. The second is an EM wave with high power propagating toward the cage generating its current locally within the conductor. I will explain how the cage works for both cases.
With respect to the first case, it can be mathematically described by charge continuity equation (equation 3 in this link). This equation basically relates the current flowing through a conductor to the charge accumulating in it.
What happens in the first scenario is that the external current (being moving charges) coming from the electrode accumulates at the point where it (the spark or the streamer) hit the cage. Because the cage is a conductor the charge continuity equation tells us that the local accumulation of charge where the spark hit the cage will cause current to flow within the conductor to remove that accumulation. The characteristic time required to remove the accumulation is called the relaxation time. It can be derived from charge continuity equation. For the derivation have a look at pages 57-59 of this document. I think that is taken from a book called Elements of Electromagnetics chapter 5.
If the conductor is made from a material with infinite conductivity, the relaxation time is zero. That means the current will keep flowing though the cage without any problem and that the electric field in the conductor is ALWAYS zero. In other words, the electrostatic point of view holds even for non-electrostatic case if the conductivity is infinite. That is a direct consequence from charge continuity equation. For non infinite conductivity cases, the electric field within the conductor will survive within the conductor with a time scale related directly to relaxation time of that conductor. I hope it is clear now with respect to first case.
The second case is related to EM waves where they generate their currents locally within the conductor, that is where the skin effect comes into play. An EM wave penetrates into a conductor the Skin effect occurs. In general, EM waves when they penetrate a conductor they are attenuated until their fields become almost zero. A characteristic depth of penetration is called Skin depth. The skin depth is the distance it takes an EM wave to be attenuated to certain value. This skin depth depends on many factors such as conductivity and frequency, the following figure taken from Wikipedia shows the skin depth of different materials for different frequencies:
For the cage to protect from EM waves, it is thickness has to be larger than multiples of skin depth at the particular frequency of interest.
So briefly with respect to the second scenario, the skin depth becomes relevant when we speak about shielding from electromagnetic waves rather than discharge current.
The first and the second scenarios can be put together in frequency spectrum, the first scenario describes why the cage protects current in low frequencies while the second scenario describes why it protects from both current and radiation at high frequencies.
I think the cage in the picture shows scenario 1. You can clearly see the distant electrodes and the point at which the spark hits the cage
I hope that answered your question
The most important concept relating Faraday cage hole size to cell phone signal attenuation is the idea of a cutoff frequency. For round holes, you would model them as cylindrical waveguides. For simplicity, we'll consider rectangular waveguides instead.
Matching the boundary conditions at the metal wall, you get so called transverse electric (TE) and transverse magnetic (TM) modes. These look like partially standing waves, with a traveling wave component for the third. For TE modes, they're of the form (for polarization in the y direction):
$$E = E_0 \sin(k_x x) \cos(k_y y) e^{i(k_z z-\omega t)}$$
There's a multitude of standing wave modes. These are described by different values of $k_x$ and $k_y$, which are solved by setting the above expression to zero at the walls of the waveguide (for the sine portion), or derivative zero (for the cosine portion). The solutions:
$$k_x = \frac{m\pi}{w}$$ $$k_y = \frac{n\pi}{h}$$
Where $w$ and $h$ are the width and height of the waveguide, and $m$ and $n$ are integers. Substituting the above expression into the wave equation, we get the relation between the different $k$ components and frequency.
$$\left( \frac{\omega}{c} \right)^2 = {k_x}^2+{k_y}^2+{k_z}^2 $$
The lowest possible such frequency is when $k_z = 0$ and $m=1,n=0$
$$f_c= \frac{c}{2w}$$
This is the cutoff frequency. Below this frequency, the signal exponentially decays as it propagates through the structure. To show this, solve for $k_z$, and write it in terms of cutoff frequency.
$$k_z = \frac{2 \pi}{c} \sqrt{f^2-{f_c}^2}$$
Evidently, at frequencies below cutoff, $k_z$ becomes imaginary. Substituting this into our travelling wave expression, it becomes exponential decay.
$$\alpha := \frac{2 \pi}{c} \sqrt{{f_c}^2-f^2}, f < f_c$$
$$E = E_0 \sin(k_x x) \cos(k_y y) e^{-\alpha z-i\omega t}$$
Notice that for our rectangular waveguide, the cutoff frequency depended only on the width. In general,
$$\lambda_c \approx \textrm{largest feature size}*2 $$
(This is exactly true for a rectangular waveguide, and should hold approximately for other shapes.)
For Faraday cages with openings in the size of centimeters, the cutoff frequency is around 20GHz, which is quite large compared cell phone signals in the range of 2GHz. We can approximate the decay constant $\alpha$
$$\alpha \simeq \frac{2 \pi f_c}{c} = \frac{2\pi}{\lambda _c} $$
To figure out the amount of decay, we need to assume some length $l$ to the opening (equivalently, thickness of the cage material), and then substitute $l$ for $z$ in the wave expression. Converting this to a decibel scale, we get the following power loss:
$$\frac{l}{\textrm{largest feature size}} (48.6 \textrm{dB})$$
Another important point is that the signal will mostly negatively interfere with itself in the cage, except at a few spots within the cage where it is effectively amplified (probably the center). If the cage features are fairly large, you might be able to notice this signal hotspot.
Edit: There are also complex effects where the fields in one hole can induce fields in another hole. The above analysis is a simplified description of a complex field problem, but I expect the general principles to hold.
Best Answer
Actually, cell phones do work in Faraday cages these days. What happens is that the conductor in the cage is not ideal, and there is some amount of leakage of electromagnetic radiation to and from the inside of the cage, specially at high frequencies. In order for the cage to be perfectly blocking it would need to have no holes at all (hence it is no longer a cage, but a box) and made of a perfectly conducting material, such as a superconductor, with a thickness larger than about 3 times the penetration depth of the radiation for that material.
The sizes of the holes need to be smaller than the wavelength of the EM radiation, so that you can neglect the holes. The wavelength is inversely proportional to the frequency of the radiation, thus, Faraday cages are more efficient for low frequencies, such as a DC field or lightning. This is why light, which is also a form of electromagnetic radiation, can pass through the centimeters-sized holes of an ordinary cage.