Magnetic dipole moment of a ferromagnetic cylinder

[amicus_tobias]

I have a Nickel cylinder, 5 micron in diameter and 10 micron in length. I know the external field \vec{B}is 80 gauss, what is the formula for the magnetic dipole moment of this cylinder in the field?

Now I only have the formula for magnetic dipole of spheres, which is \mu = \frac{4}{3}\pi a^3 \chi B, so I sort of make estimates by modeling the cylinder as a dimer of two spheres. But I would like to get a formula particular to cylinders.

Please help. Thank you!

 

[amicus_tobias]

Can someone help here? thanks!

 

[cragar]

when you say magnetic dipole moment
do you mean M= \int I da

 

[amicus_tobias]

when you say magnetic dipole moment
do you mean M= \int I da

I think that is just another way to calculate the induced magnetic dipole moment if you know the current flowing. But now I have a magnetic field, I know exactly what it is, and I want to know what is the magnetic moment in a cylinder which happens to be ferromagnetic.

 

[cragar]

The formula i gave and the one you gave have different units.
there off by a \mu_0
So because you have a ferromagnetic material when we place this in an external B field it will cause the magnetic domains to line up and cause the cylinder to have its on B field.
could you just multiply it by the volume of a cylinder or will that not work.
I flipped through Griffiths electrodynamics and couldn't really find anything on it.