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Magnetic dipole derivations

  1. Mar 9, 2007 #1

    Kolahal Bhattacharya

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    this is not homework help.I want to know.

    Hello,can any one suggest why for a dipole m placed in a magnetic field B
    F=grad(m.B)
    and N=mxB?
     
    Kolahal Bhattacharya, Mar 9, 2007
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  3. Mar 9, 2007 #2

    siddharth

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    Take an infinitesimal current loop of arbitrary shape and use the Lorentz force law.
     
    siddharth, Mar 9, 2007
  4. Mar 9, 2007 #3

    Meir Achuz

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    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.
     
    Meir Achuz, Mar 9, 2007
  5. Mar 9, 2007 #4

    Kolahal Bhattacharya

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    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.
     
    Kolahal Bhattacharya, Mar 9, 2007
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