How can magnetic dipole derivations be solved without using the force equation?

[Kolahal Bhattacharya]

this is not homework help.I want to know.

Hello,can anyone suggest why for a dipole m placed in a magnetic field B
F=grad(m.B)
and N=mxB?

 

[siddharth]

Hello,can anyone suggest why for a dipole m placed in a magnetic field B
F=grad(m.B)
and N=mxB?

Take an infinitesimal current loop of arbitrary shape and use the Lorentz force law.

 

[Meir Achuz]

Take an infinitesimal current loop of arbitrary shape and use the Lorentz force law.

That will work, but the derivation is a bit involved for an arbitrary shape.
If you are satisfied with doing it for a rectangular loop, that is easier for the torque. Once you know the torque, you can show U=-mu.B, and then
use F=-grad U.
If you model the magnetic dipole as two magnetic poles, then the derivations are just the same as for electric dipoles, which are easier.

 

[Kolahal Bhattacharya]

To Siddharth:I found your approach in jackson.However,Griffiths also asks in his exercise to do in the same way.Only Jackson used J while Griffiths prefers I (in his hint).
I got upto:
F=I* closed int{dl x [(r. grad_0)B](r_0)} where dl is infinitesimal loop element; and the right portion is obtained after Taylor series expansion.
Now griffiths and Jackson says to use Levi-Civita symbols...
I was trying to insert the result: vector area a= (1/2)int{ r x I) inside the integral extracting (dl x r) so that I can get 'a' inside the integral and then have dm inside.But got stuck.It appears I am near the way but cannot get it.
Please help.

The other approach ultimately assumes F=grad(m.B).So, I prefer not to use it.