# 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

I had the same problem a few weeks ago. It is clear that an electromagnet does produce work, to solve the paradox you need to take into account the generator that runs the current and balance the energy ( Griffiths p211, introduction to electrodynamics). In other words, if the magnetic moment is created by a current, the system does produce a net work, but deeper analysis shows that it is not created by the magnetic force (qvxB) but by the generator.

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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
I have never had a clear answer, but the most probable explanation for a permanent magnet lifting a weight is a magnetic domain reconfiguration. In other word, the magnet looses some of its force each time you use it. – Shaktyai Aug 13 '12 at 6:56
I am not sure I understand why there is an electric force if V and B are parallel (VxB=0). However, the magnetic field created by a magnet is non uniform so indeed electrons, in the stell ball lifted up, experience an electric field. There is an other source of energy: the ball is occupying a volume of space which is filled with magnetic energy. If the ball changes of location, the energy stored in the magnetic field changes. However, if one takes all these possible source of energy for a permanent magnet one has also to do it for an electric coil and Griffith's solution is wrong. – Shaktyai Aug 13 '12 at 8:22
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