# How do magnets work in classical electromagnetism?

A charged particle moving in a magnetic field experiences a Lorentz force $F=qv\times B$. A charged particle in a electric field experiences a force given by Coulomb's inverse square law.

But for a magnet we have what seems to be adhoc ways to calculate how the magnet will rotate. In the above case we speak concretely of a charged particle being influenced by a magnetic field or an electric field. But what is the analogous "thing" or "particle" causing the magnet to rotate?

It feels like classical electromagnetism is Maxwell's equations plus laws for how a charged particle behaves in a magnetic field and electric field, but notably missing corresponding Coulomb and Lorentz force laws for "magnets".

• "A charged particle in a electric field experiences a force given by Coulomb's inverse square law." Actually, the force in this case is $F=qE$. The inverse square law is only for when you have two pointlike charged particles. – David Z Nov 8 '13 at 0:42

A magnet is in essence made up of atoms with orbital and spin angular momentum, (mainly due to the electrons) and the forces that act on these electrons can be derived from Dirac's equation, but give you what is essentially the Lorentz force law. You can think of the magnetic field acting on every single electron individually, and all these forces added up will apply a net force and a net torque to the magnet as a rigid body.

If you want to do some calculations, you can imagine each atom as having a microscopic current proportional to the dipole moment of that atom (basically, the magnetization). This "bound current" is

$\vec{J} = \nabla \times \vec M$.

Again, imagine an electron orbiting the nucleus as providing a "current" flowing around the nucleus -- the Bohr magneton. It's just an idealization but you can make it rigorous if you want. Now, currents of adjacent atoms cancel out where they intersect, because they are going in opposite directions, but on the boundary of the magnet they do not cancel out, because there is no neighboring atom there. Thus it is perfectly mathematically and physically sensible to model a permanent magnet as a sheet of current moving along the magnet's surface, roughly like a solenoid does, and given by

$\vec{K} = \vec{n}\times \vec{M}$,

where $\vec{n}$ is the normal to the surface. Now this surface current is a real current, and therefore it will experience a force when subjected to an external magnetic field (and it's own magnetic field, btw. This is why transformers hum).

It is not that the corresponding law for magnets is missing. All magnetic fields in classical electrodynamics are produced by charges in motion, so to compute the force on a magnet is just computing the force acting on these charges in motion, there is nothing else or new about $F= q(E + v \times B)$.

• But you can place two magnets next to each other and they will repel even though there is no motion at least at the macroscopic scale. – user782220 Nov 8 '13 at 0:56
• The motion is macroscopic in the sense that all the microscopic currents in the material add up or cancel out in just the right way to give a macroscopic current, i.e., a loop of current that is as large as the magnet itself. – lionelbrits Nov 8 '13 at 1:02