# Faraday cage - reducing EMI?

I am curious about the working of a Faraday cage as the shield for EMI in many systems. As far as I studied, it depends on Faraday's and Lenz's laws.

When the electromagnetic waves are incident on the Faraday cage, which is highly conductive metal, it produces eddy current which opposes the incident magnetic field (according to Lenz's law), thus stopping the EM waves entering into the environment.

I think the above is how a Faraday cage works. Any other ideas? I am asking because when I surf through the internet, everyone is giving different justifications.

Also I need to know why we are connecting the shield metal to the ground in the PCB? I thought that connecting to the ground layer forms a bottom layer of the Faraday cage (as the ground layer is made of copper) and forms a closed metal. Is there any other reason for connecting the shield to ground?

Also I need to know that if the above is the reason for connecting the shield to ground, will it be efficient to connect the shield as below? If your PCB has 8 layers, normally 2nd and 7th layers are ground, so connecting the shield to the 7th layer instead of 2nd layer surrounds the signals in all above 6 layers and it will acts are cage of all the above 6 layers. Will it be useful?

Please shed some light on this. This confuses me a lot.

• A faraday cage provides a ground path to absorb loop currents, reflect RF back thru the dielectric to create a capacitance and also attenuate the stray E-field by connecting it to a low impedance Earth ground. SO the overall model is complex, but inside parallel grounds offers microstrip to have a controlled impedance with track W/gap ratios and and also attenuates digital noise outside the 2 planes. Commented Jul 1, 2019 at 14:08
• en.wikipedia.org/wiki/Faraday_cage Commented Jul 1, 2019 at 21:47
• Electric field boundary conditions require that the tangential electric field at the interface between a dielectric and a conductor is zero. If a plane wave is incident on the surface of a Faraday cage at a right angle, the electric field must go to zero. Since no power is dissipated and energy must be conserved, the entire wave is reflected. If a perfect magnetic conductor existed we would be able to use the same argument for the magnetic field, but it doesn't. Commented Jul 2, 2019 at 18:53