# How does one prove:$\nabla(\vec{\mu_m}\cdot\vec{B})\cdot\vec{dr}=0$?

Work done by a magnetic force(even over an infinitesimally short displacement)=0

Net Force on a current loop in an external magnetic field is given by: $$\vec{F}=\nabla(\vec{\mu_m } \cdot \vec{B})$$

How does one prove: $$dW=\nabla(\vec{\mu_m} \cdot \vec{B})\cdot\vec{dr}=0$$

$\vec{\mu_m}$: Magnetic Moment of the current loop.

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Magnetic force cannot perform work but electric field can do work.Such a field may result from the spatial variation of B from the curl B equation[comment to the answer may be considered] – Anamitra Palit Aug 14 '12 at 3:23

You can think of redistribution of charges and currents elsewhere(gadgets producing the field) to produce an electric field due to variations of magnetic field at some fixed point. But a pure magnetic field which does not vary with time cannot perform any work.What would be your mechanism for accounting for such variations of $\vec{B}$ in evaluating:$\nabla (\vec{\mu_m}.\vec{B}).\vec{dr}$. What would be the mechanism of including electric field in the said evaluation of elementary work? – Anamitra Palit Aug 13 '12 at 6:53
Let the magnetic field be in the x direction.If the observer moves along the y direction with a uniform speed v,we have $\vec{E}=\gamma(\vec{v}\times\vec{B})$. The electric field accelerates the charged changing its speed and doing work. Initially there was a acceleration normal to the direction of motion. But now we have acceleration due to change of speed also . If the net acceleration changes the radiating power should be different from the two frames.But energy is not an invariant wrt frame transformation though it is a conserved quantity. It might have different values in differentframes – Anamitra Palit Aug 13 '12 at 10:51
Spatial variation of $\vec{B}$ can produce an electric field:$\nabla \times \vec{B}=\mu_0 \vec{j}+\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$.If $\vec{E}\cdot\vec{dr}=\nabla(\vec{\mu_m}\cdot\vec{B})\cdot\vec{dr}$ to produce non-zero work then,$\vec{E}=\nabla(\vec{\mu_m}\cdot\vec{B})$. How do you prove that by direct calculation?[here you have a view of the problem through the differential equations which are supposed to represent the physical conditions accurately.] – Anamitra Palit Aug 14 '12 at 3:16