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.

HarisCould you please check your curl of the volume current. I believe it should have a minus sign.

Also what happened to the little r when you take the curl?

gwrowePost authorIf you mean equation 10 (it’s really helpful if you quote equation numbers!), which is the curl of the magnetization, not the volume current, it’s correct. My variable is what Griffiths calls and if you look up the formula for the curl in cylindrical coordinates (inside the front cover of Griffiths, for example), the curl reduces to (since the only non-zero component of is ):

HarisMy apologies I meant curl for the volume current not of.

Thank you very much for that, I made a silly mistake while calculating the curl.

JoshHow did you arrive at the RHS of eq. 4?

gwrowePost authorThe current enclosed by a circular path of radius inside the wire is the fraction of the total current equal to the fraction of the cross-sectional area enclosed by the loop, which is .