Magneti Shielding

[Wannabeagenius]

Hi All,

A conductor cannot shield a circuit against a magnetic field as it does for an electric field. When used in this fashion, the conductor is called a faraday shield.

In an electronics circuit, do the induced magnetic fields pose a problem and, if so, how are they handled?

Thank you,
Bob

[Bob S]

Some solid state devices, like Hall Effect sensors, are affected by magnetic fields. I have not had a problem with other solid state devices in low magnetic fields (a few Gauss). Some electrical devices, like photomultipliers, are verry sensitive to magnetic fields, and need magnetic shielding. Also, an ac magnetic field (dB/dt) can induce ac voltages in circuits via the Faraday Law. Be careful to minimize potential ground loops from ac transformer stray field, among other things, by planning your grounding strategy.
Bob S

[Wannabeagenius]

Some electrical devices, like photomultipliers, are verry sensitive to magnetic fields, and need magnetic shielding.

How is this accomplished?

Bob

[f95toli]

Usually by using materials like mumetall, permalloy or similar to make a shielding enclosure.

[Adjuster]

Among the components most often affected by magnetic fields are those containing coils of wire, such as inductors (eg radio tuning coils), tape heads, dynamic microphones etc.

The effect can be quite pronounced with an unscreened winding - this is utilised to advantage in such things as ferrite bar antennas for radio receivers, metal detector search coils, tape heads and many other devices.

Screening can help to reduce unwanted pickup, as can giving the inductor a closed magnetic circuit (eg in a pot core coil). Perfectly toroidal windings theoretically don't emit or pick up magnetic fields, and finally carefully positioning coils within equipment can minimise interactions.

[Bob S]

How is this accomplished?
Here is a brief tutorial on magnetic shielding without illustration.

Suppose the magnetic field outside a soft iron shield were 10 Gauss, and we need to have 0.1 Gauss inside for a photomultiplier; i.e., a 100 times reduction

Consider external field B1 perpendicular to the axis of a cylindrical soft iron shield of diameter D. The flux 2·B1·D is concentrated in the iron shield itself at midpoint.
B is continuous because div·B = 0
If the shield has a thickness t then
B2 = (D/t)·B1 in the iron
This should not exceed 10,000 Gauss for soft iron, 2,000 Gauss for mumetal. So in this example using soft iron with B1= 10 Gauss, t>= 0.001·D (one of two limits)
Now, If we want B3= 0.1 Gauss (inside shield)
B3=B2/μr because H-parallel is continuous; Curl H = 0
where μr is relative permeability, assume 2500 for soft iron
So B2 in iron must not exceed 250 Gauss
So t>=D/25

Try same calculation with mumetal.

Bob S