Can Mu-Metal Reflect Magnetic Fields?

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Mu-metal effectively shields magnetic fields due to its high permeability, absorbing magnetic field lines rather than reflecting them. When a current loop is placed inside a mu-metal cylindrical shield, the shield becomes polarized, creating a magnetic field that opposes the loop's field. This polarization leads to a cancellation of the external magnetic field, although some field lines may still escape at the ends of the cylinder. The magnetization within the mu-metal is not uniform, being strongest at the center and weaker towards the ends. Accurate modeling of this phenomenon would require numerical simulations to account for varying currents along the cylinder's length.
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Hi

I have a question regarding mu-metal, which I haven't beem able to find the answer to online. Say I have a current loop placed enclosed in a cylindrical magnetic shield made of mu-metal as shown in the attachment (red loop is the current loop).

The shielding will prevent any B-field to penetrate it, but won't it also act as a "B-field mirror", i.e. it will basically appear to a particle inside the shield that there are three coils: 2 with oppositely directed currents to the actual, physical one?


Niles.
 

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Mu metal has a very high permeability, so it basically sucks up magnetic field lines.

In your example, the mu metal cylinder would get polarized by the magnetic field from the current loop in its center. The polarization of the cylinder would be opposite to the current loop and such that on the outside, the sum of the magnetic fields would cancel (or nearly cancel). I think that some field lines would still leak out at the ends of the cylinder.

I guess the magnetization would not be constant along the length of the cylinder, but highest in the mid plane, and lower towards the ends. To simulate this exactly, you would need a whole lot of current loops along the length of the cylinder, each with a different current. The direction of the current is opposite to the current in the central loop.

To get the details right, one would have to do a numerical simulation.
 
Thanks!
 
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