Does a Faraday cage block all magnetic field frequencies lower than x? 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?
 A: 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). Since frequency is involved, your question is about EM case.
If you read the third paragraph of Wikipedia page I refered to, you will see that there are two conditions for Faraday's cage to shield EM fields, which are:


*

*The conductor is thick enough.

*The holes in the cage are much smaller than the wavelength.


Let me start by the second, if your cage is designed such that it blocks 100 kHz, that means the holes are small enough to make this condition satisfied, that means it will also be satisfied automatically for 50 Hz because the wavelength at 50 Hz is much longer that wavelength of 100 kHz.
So for condition 2, yes it will shield 50 Hz if it shields 100 kHz.
Getting back to first condition, the thickness of the conductor is related to a phenomena called Skin effect. 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:
 
Assuming your cage is made of copper and designed to shield 100 kHz, the skin depth at that frequency is 0.11 mm. the cage needs to be a multiple of skin depth so say 1 mm thick cage will be enough.
For 50 Hz however (the red vertical line), the skin depth is almost 10 mm. If your cage is 1 mm thick to shield 100 kHz it WON'T automatically shield 50 Hz. 
So for condition 1, no it won't shield 50 Hz if it shields 100 kHz. In general, lower frequencies require thicker metal for shielding. So the design should be based on minimum frequency.
The brief answer is No. Hopefully that helped
A: 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.
