I want to block the magnetic field of a very strong magnet, can I put it inside a Faraday cage to block its magnetic field?
[Physics] Does a Faraday cage block magnetic field
electric-fieldselectromagnetismmagnetic fields
Related Solutions
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:
Actually ~50 Hz magnetic fields can be effectively shielded with a good conductor like copper due to eddy current shielding. High permeability (and expensive) mu-metal is not necessary
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.
The theory for a tinfoil screen is fairly straightforward.
There are two things going on: first, a lot (most) of the signal is reflected. Second, what penetrates into the foil is attenuated and dissipated by currents.
The impedance of the aluminium "tinfoil" is given by $\eta_{\rm Al} = (\mu_r \mu_0 \sigma / \omega)^{1/2}$, where $\omega$ is the angular signal frequency, and $\mu_r=1$ and conductivity $\sigma= 3.5 \times 10^{7}$ S/m are reasonable values for Al. Therefore $\eta_{\rm Al}= 44\omega^{-1/2}$ $\Omega$.
The transmitted E-field fraction is given by the following equation $$\frac{E_t}{E_i} = \frac{2 \eta_{\rm Al}}{\eta_0 + \eta_{\rm Al}} \simeq 2\frac{\eta_{\rm Al}}{\eta_0}\, ,$$ where the impedance of free space (or air), $\eta_0 \simeq 377$ $\Omega$
Once the field gets into the foil it is exponentially attenuated according to the skin depth $\delta = (2/\mu_r \mu_0 \sigma \omega)^{1/2} = 0.045 \omega^{-1/2}\,$m.
Let's now make the assumption that we ignore reflection from the foil/air interface on the way out. In that case the transmission factor at that interface is given by $2 \eta_{0}/(\eta_{\rm Al} + \eta_{0}) \simeq 2$$.
Putting this all together we get a final transmission fraction of $$\frac{E_t}{E_i} \simeq 4 \frac{\eta_{\rm Al}}{\eta_0} \exp(-t/\delta) = 0.47 \omega^{-1/2} \exp(-22 \omega^{1/2} t),$$ where $t$ is the foil thickness. The transmitted power fraction would be the square of this.
Take an example: Typical domestic Al foil has $t= 3\times10^{-5}$ m and let's use a low radio frequency of 1 MHz or $\omega = 6.3 \times 10^{6}$ rad/s. The foil is only just over a skin depth thick at this frequency, but most of the signal is reflected and $E_t/E_i \simeq 3.6\times 10^{-5}$. Even if the foil was much thinner, only $2\times 10^{-4}$ of the field would be transmitted.
At a frequency of 1 GHz the foil is many ($\sim 50$) skin depths thick and is more reflective, so the attenuation of the field is more than 20 orders of magnitude.
Best Answer
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