# How can a slowly varying magnetic field exist in Faraday cage when there is no electric field in it?

When I search Faraday cage in Wiki, I noticed a sentence that 'Faraday cages cannot block stable or slowly varying magnetic fields'. Since Electric field and magnetic field are coupled due to Maxwell equation, how can a slowly varying magnetic field exist in a Faraday cage when there is no electric field in it.

At low frequencies, such as 50 or 60 Hz grid frequencies, the electric and magnetic field are very weakly coupled. For DC fields, there is zero coupling.

The Faraday-Maxwell law $$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$ states that a magnetic field that changes in time gives rise to a current density in conducting materials, called eddy currents. These eddy currents in turn give rise to a magnetic field that opposes the external field. Note that the amplitude of these eddy currents depend on the first order time derivative of the magnetic field, i.e. these currents are stronger at higher frequencies. Additionally, at much higher frequencies second order time derivatives that give rise to electromagnetic waves will start to play a role. Faraday cages are typically effective when used to mitigate these waves.

If parts of the Faraday cage have good galvanic contact, reasonable shielding can be obtained at frequencies as low as 50 Hz. This usually requires an aluminum or copper shield with good gaskets or soldered/welded connections.

DC magnetic fields do not produce eddy currents and can only be shielded by materials with high magnetic permeability.

• Thank you for your explanation. It's really helpful! I am a little bit confused about the "weakly coupled" concept. Does that mean the change of magnetic field doesn't produce a strong electrical field? Oct 18, 2021 at 12:17
• Yes, that is what it means. A higher frequency of the B-field will create a stronger electric field. The first order effect of this can be readily seen from the Maxwell-Faraday equation in my answer. If B is harmonic with amplitude $B_0$, the amplitude of E will be proportional to $fB_0$ (omitting some other constants) , where $f$ is the frequency. Oct 18, 2021 at 12:33
• Yes, but I think the curl of E will make a factor of wavenumber k and the k/f create a constant 1/c which connects E and B. Therefore, I am confused why the E and B are weakly coupled in low frequency. Oct 18, 2021 at 15:12
• That line of reasoning assumes far-field radiation and propagating waves. This is generally not the case for lower frequencies. At low enough frequencies the displacement currents can usually be neglected and then $E$ will be directly proportional to $fB$. As an example, take $B$ as, say, the field inside solenoid with a sinusoidal input current, something like $B_0\sin(2\pi f t) \hat{z}$. Oct 19, 2021 at 7:32

how can a slowly varying magnetic field exist in a Faraday cage when there is no electric field in it.

Ideal Faraday cage is made of a closed shell made of ideal conductor which has zero resistance, and this enables perfect shielding effect - it does not allow external sources to cause any electric field inside it. So there can't be varying magnetic field inside such ideal Faraday cage; there can be only static magnetic field there.

In reality, Faraday cage is made of real conductor, which has non-zero resistance. Because of this, the shielding effect is not perfect and external sources do cause non-zero electric field inside, and this electric field can change in time. Because of this, inside real Faraday cage, both electric and magnetic field can change in time due to external sources. The better the Faraday cage, the lower this time-varying EM field is inside.