# Relation between magnetic moment and angular momentum -- classic theory

The easiest way to see the equality is to use a more general formula for the magnetic dipole moment of a particle. For a flat planar loop of current, it's true that $\mu = IA$, with the direction of the dipole normal to the loop. However, the more general case is that of a a *volume* current $\vec{J}$ in some finite region of space. In this case, the the general formula for the magnetic dipole moment of the configuration is
$$
\vec{\mu} = \frac{1}{2} \int \vec{r} \times \vec{J} \,d^3r.
$$
(Showing that this reduces to the above formula for a flat planar loop is left as an exercise to the reader.) If we further assume that the current density is due to a number of particles with number density $n$, charge $q$, velocity $\vec{v}$, and mass $m$, then we have current density $\vec{J} = nq\vec{v}$; thus,
$$
\vec{\mu} = \frac{1}{2} \int \vec{r} \times (n q \vec{v}) \, d^3 r = \frac{q}{2 m} \int \vec{r} \times (n m \vec{v}) \, d^3 r.
$$
But $n m \vec{v} = \rho \vec{v}$, where $\rho$ is the mass density of the cloud; thus, the above integral can be rewritten as
$$
\vec{\mu} = \frac{q}{2 m} \int \rho \vec{r} \times \vec{v} \, d^3 r = \frac{q}{2m} \vec{L}.
$$
QED.

The above relationship will hold so long as we can model the object as made out of particles with a definite charge-to-mass ratio, or (which is equivalent) so long as the ratio of charge density to mass density is constant throughout the body. The easiest way for this to happen is, of course, for both densities to be constant.