# Electromagnetic Shielding

• I
cuallito
I've been thinking about electromagnetic shielding lately.

But after staring at Maxwell's equations awhile, I'm thinking:

We're after a static E field, right?

So if we set ## \frac{\partial \textbf{E}}{\partial t}=0## in Ampere's Law, we get

$$\nabla\times \textbf{B}=\mu_0 \textbf{J}$$

Implying that if we can hold the current density ##\textbf{J}## static around a region of space, say with a constant current source, we'd effectively 'fix' the field in that region of space, thus blocking out any incident EM radiation (theoretically)?

For example, if we had a cylinder wrapped around in conducting wire, hooked up to a constant current source, would the field in the interior of the cylinder stay static regardless of what the field outside it was doing?

Staff Emeritus
Homework Helper
Gold Member
We're after a static E field, right?
No. We are after shielding against electromagnetic radiation.

Implying that if we can hold the current density J static around a region of space, say with a constant current source, we'd effectively 'fix' the field in that region of space, thus blocking out any incident EM radiation (theoretically)?
No. It is not implying that and I am not sure how you arrived at such a conclusion.

For example, if we had a cylinder wrapped around in conducting wire, hooked up to a constant current source, would the field in the interior of the cylinder stay static regardless of what the field outside it was doing?
No.

Perhaps it would be beneficial to first understand how a Faraday cage works.

vanhees71
Mentor
Implying that if we can hold the current J density static around a region of space, say with a constant current source, we'd effectively 'fix' the field in that region of space, thus blocking out any incident EM radiation (theoretically)?
Fields add vectorally, so adding a static field does nothing to "shield" other fields that are present. It just adds a DC field term to the vector sum.

artis
I've been thinking about electromagnetic shielding lately.

But after staring at Maxwell's equations awhile, I'm thinking:

We're after a static E field, right?

So if we set ## \frac{\partial \textbf{E}}{\partial t}=0## in Ampere's Law, we get

$$\nabla\times \textbf{B}=\mu_0 \textbf{J}$$

Implying that if we can hold the current density ##\textbf{J}## static around a region of space, say with a constant current source, we'd effectively 'fix' the field in that region of space, thus blocking out any incident EM radiation (theoretically)?

For example, if we had a cylinder wrapped around in conducting wire, hooked up to a constant current source, would the field in the interior of the cylinder stay static regardless of what the field outside it was doing?
A Faraday cage is essentially the same as a mirror for visible light frequencies. The way it shields is the charges present on the surface of the conducting shield reorient in response to the external field, this causes an equal and opposite field to appear at the surface of the shield which blocks the primary EM that tries to propagate through.

Somewhat the same way a waveguide for RF frequencies work. The RF wave doesn't get outside the enclosed guide because the inner walls are conducting and they reflect the wave everytime it hits the walls , which it does so after even distances which are related to the wavelength of the wave