Homework Help: Help,this problem cost me much time

1. Aug 15, 2006

goodboy

How to get this conclusion

if we know the magnetic moment m,and the magnetic induction B.then the force is
who can tell me how to get this.thanks very much

Last edited: Aug 16, 2006
2. Aug 20, 2006

quasar987

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

$$\nabla (m\cdot B) = m\times (\nabla \times B) + B \times (\nabla \times m)+(m\cdot \nabla)B+(B\cdot \nabla)m$$

The first term vanishes because $\nabla \times B=0$ 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