- #1

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Hello,can any one suggest why for a dipole m placed in a magnetic field B

F=grad(m.B)

and N=mxB?

- Thread starter Kolahal Bhattacharya
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- #1

- 135

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Hello,can any one suggest why for a dipole m placed in a magnetic field B

F=grad(m.B)

and N=mxB?

- #2

siddharth

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Take an infinitesimal current loop of arbitrary shape and use the Lorentz force law.Hello,can any one suggest why for a dipole m placed in a magnetic field B

F=grad(m.B)

and N=mxB?

- #3

Meir Achuz

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That will work, but the derivation is a bit involved for an arbitrary shape.Take an infinitesimal current loop of arbitrary shape and use the Lorentz force law.

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.

- #4

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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.

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