References: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 6.17.

Another example of using the auxiliary field to find the magnetic field. This time, we have a long wire of radius carrying a uniform current . The wire is made of a linear metal so that the magnetization is directly proportional to .

where is the *magnetic susceptibility*. Given this proportionality, the field is also directly proportional to :

where is the magnetic permeability.

From the symmetry of the setup, is circumferential, so we can take a circular integration path to get

From this we get the field for

Outside the wire, and the enclosed free current is just , so for

The magnetization is

so the bound currents are

The total bound current is

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Anonymousi think you meant to say bound current where you said “so the free currents are”

growescienceQuite right; fixed now.

OllieI believe the B field outside the wire (r>a) is actually (Muo*I)/(2pi*r)…

growescienceFixed now. Thanks.