When you release a magnetic dipole in a nonuniform magnetic field, it will accelerate.
I understand that for current loops (and other such macroscopic objects) the magnetic moment comes from moving charges, and since magnetic fields do no work on charges ($F$?perpendicular to $v$) it follows that the work done on the dipole (that caused its gain in kinetic energy) must have come from somewhere other than the magnetic forces (like electric forces in the material).
However, what about a pure magnetic moment? I‘m thinking of a particle with intrinsic spin. Of course, such a thing should be treated with quantum mechanics, but shouldn't classical electrodynamics be able to accommodate a pure magnetic dipole? If so, when I release the pure dipole in a nonuniform B-field and it speeds up, what force did the work? Is it correct to say that magnetic fields DO do work, but only on pure dipoles (not on charges)? Or should we stick with "magnetic forces never do work", and the work in this case is done by some other force (what?)?
Thanks to anyone who can alleviate my confusion!
Best Answer
Yes, of course that if a field - magnetic field - is able to make a bar magnet rotate or move, it is doing work. The statement that magnetic fields don't do any work only applies to point-like pure electric charges.
Magnetic moments may be visualized as objects with a forced motion of charges (solenoids have the same magnetic field as bar magnets), and if something is moving, the magnetic force is becoming a force that does work.
In terms of formulae, the magnetic force on a charge is $q\vec v\times \vec B$ which is identically perpendicular to $\vec v$ and that's why it does no work. However, forces on magnetic dipoles and more general objects don't have the form $\vec v\times$ - they're not perpendicular to $\vec v$, so they do work in general.