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We're given a situation where the circuit can move under the influence of a magnetic force. Now I need to show that the force on the magnetic dipole is

$$\vec{F} = (\vec{m} \nabla) \vec{B} $$

Could I start from a given Force for a loop with a dipole moment in the B-field, where $$\vec{F} = \nabla (\vec{m} \cdot \vec{B})$$ and apply the identity $$\vec{F} = \vec{m} \times (\nabla \times \vec{B}) +\vec{B} \times(\nabla \times \vec{m}) + (\vec{m} \cdot \nabla)\vec{B} + (\vec{B}\cdot\nabla)\vec{m}$$ where terms 1,2,and 4 vanishes? 1 vanishes from assuming that $J=0$, and that the $\vec{m}$ are not dependent on positions?

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