Electromagnetic Shielding

  • I
  • Thread starter cuallito
  • Start date
  • #1
cuallito
95
1
I've been thinking about electromagnetic shielding lately.

Of course, you have your good, old-fashioned Faraday cage.

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?
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
20,004
10,650
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.
 
  • #3
berkeman
Mentor
64,140
15,353
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.
 
  • Like
Likes Vanadium 50
  • #4
artis
1,326
878
I've been thinking about electromagnetic shielding lately.

Of course, you have your good, old-fashioned Faraday cage.

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
 

Suggested for: Electromagnetic Shielding

  • Last Post
Replies
8
Views
528
Replies
2
Views
263
Replies
1
Views
280
Replies
3
Views
127
Replies
3
Views
312
Replies
1
Views
124
Replies
2
Views
518
Replies
222
Views
6K
Replies
12
Views
499
Top