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Help,this problem cost me much time

  1. Aug 15, 2006 #1
    How to get this conclusion

    if we know the magnetic moment m,and the magnetic induction B.then the force is 未命名.jpg
    who can tell me how to get this.thanks very much
    Last edited: Aug 16, 2006
  2. jcsd
  3. Aug 20, 2006 #2


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    Hopefully you know that for a current loop of steady current in a uniform magnetic field, the mechanical energy of interaction is [itex]U_{mech}= -m\cdot B[/itex], making the force on the loop [itex]\vec{F}=\nabla(m\cdot B)[/itex]. Now use the vector identity that says

    [tex]\nabla (m\cdot B) = m\times (\nabla \times B) + B \times (\nabla \times m)+(m\cdot \nabla)B+(B\cdot \nabla)m[/tex]

    The first term vanishes because [itex]\nabla \times B=0[/itex] in the absence of current or changing electric fields. m taken as a vector field is constant wrt change in position, so all its derivative wrt position are null, making the 2nd and 4th term vanish. Remains the 3rd term
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