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
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
You ground these because Faraday cages are not perfect. One near field effect which matters in this case is capacitive coupling between the shield and the wires. If the shield is ungrounded, it's effectively an antenna, and thus acts as coupled antenna to your wires. Grounding the shielding fixes the antenna like effect.
In theory you could also resolve this issue with an ungrounded shield that is further away from the wires, but that takes up more space.
Of course, you have to be careful. If you ground it on both sides, you get a ground loop, which can cause all sorts of problems for some analog hardware.