Magnetic moments and angular momentum
The first theorem we want to prove from classical mechanics is the following: If an electron is moving in a circular orbit (for example, revolving around a nucleus under the influence of a central force), there is a definite ratio between the magnetic moment and the angular momentum. Let’s call $\FLPJ$ the angular momentum and $\FLPmu$ the magnetic moment of the electron in the orbit. The magnitude of the angular momentum is the mass of the electron times the velocity times the radius. (See Fig. 34-2.) It is directed perpendicular to the plane of the orbit. \begin{equation} \label{Eq:II:34:1} J=mvr. \end{equation} (This is, of course, a nonrelativistic formula, but it is a good approximation for atoms, because for the electrons involved $v/c$ is generally of the order of $e^2/\hbar c\approx1/137$, or about $1$ percent.)  The magnetic moment of the same orbit is the current times the area. (See Section 14-5.) The current is the charge per unit time which passes any point on the orbit, namely, the charge $q$ times the frequency of rotation. The frequency is the velocity divided by the circumference of the orbit; so \begin{equation*} I=q\,\frac{v}{2\pi r}. \end{equation*} The area is $\pi r^2$, so the magnetic moment is \begin{equation} \label{Eq:II:34:2} \mu=\frac{qvr}{2}. \end{equation} It is also directed perpendicular to the plane of the orbit. So $\FLPJ$ and $\FLPmu$ are in the same direction: \begin{equation} \label{Eq:II:34:3} \FLPmu=\frac{q}{2m}\,\FLPJ\:(\text{orbit}). \end{equation} Their ratio depends neither on the velocity nor on the radius. For any particle moving in a circular orbit the magnetic moment is equal to $q/2m$ times the angular momentum. For an electron, the charge is negative—we can call it $-q_e$; so for an electron \begin{equation} \label{Eq:II:34:4} \FLPmu=-\frac{q_e}{2m}\,\FLPJ\:(\text{electron orbit}). \end{equation}  That’s what we would expect classically and, miraculously enough, it is also true quantum-mechanically. It’s one of those things. However, if you keep going with the classical physics, you find other places where it gives the wrong answers, and it is a great game to try to remember which things are right and which things are wrong. We might as well give you immediately what is true in general in quantum mechanics. First, Eq. (34.4) is true for orbital motion, but that’s not the only magnetism that exists. The electron also has a spin rotation about its own axis (something like the earth rotating on its axis), and as a result of that spin it has both an angular momentum and a magnetic moment. But for reasons that are purely quantum-mechanical—there is no classical explanation—the ratio of $\FLPmu$ to $\FLPJ$ for the electron spin is twice as large as it is for orbital motion of the spinning electron: \begin{equation} \label{Eq:II:34:5} \FLPmu=-\frac{q_e}{m}\,\FLPJ\:(\text{electron spin}). \end{equation}  In any atom there are, generally speaking, several electrons and some combination of spin and orbit rotations which builds up a total angular momentum and a total magnetic moment. Although there is no classical reason why it should be so, it is always true in quantum mechanics that (for an isolated atom) the direction of the magnetic moment is exactly opposite to the direction of the angular momentum. The ratio of the two is not necessarily either $-q_e/m$ or $-q_e/2m$, but somewhere in between, because there is a mixture of the contributions from the orbits and the spins. We can write \begin{equation} \label{Eq:II:34:6} \FLPmu=-g\biggl(\frac{q_e}{2m}\biggr)\FLPJ, \end{equation} where $g$ is a factor which is characteristic of the state of the atom. It would be $1$ for a pure orbital moment, or $2$ for a pure spin moment, or some other number in between for a complicated system like an atom. This formula does not, of course, tell us very much. It says that the magnetic moment is parallel to the angular momentum, but can have any magnitude. The form of Eq. (34.6) is convenient, however, because $g$—called the “Landé $g$-factor”—is a dimensionless constant whose magnitude is of the order of one. It is one of the jobs of quantum mechanics to predict the $g$-factor for any particular atomic state.  You might also be interested in what happens in nuclei. In nuclei there are protons and neutrons which may move around in some kind of orbit and at the same time, like an electron, have an intrinsic spin. Again the magnetic moment is parallel to the angular momentum. Only now the order of magnitude of the ratio of the two is what you would expect for a proton going around in a circle, with $m$ in Eq. (34.3) equal to the proton mass. Therefore it is usual to write for nuclei \begin{equation} \label{Eq:II:34:7} \FLPmu=g\biggl(\frac{q_e}{2m_p}\biggr)\FLPJ, \end{equation} where $m_p$ is the mass of the proton, and $g$—called the nuclear $g$-factor—is a number near one, to be determined for each nucleus.  Another important difference for a nucleus is that the spin magnetic moment of the proton does not have a $g$-factor of $2$, as the electron does. For a proton, $g=2\cdot(2.79)$. Surprisingly enough, the neutron also has a spin magnetic moment, and its magnetic moment relative to its angular momentum is $2\cdot(-1.91)$. The neutron, in other words, is not exactly “neutral” in the magnetic sense. It is like a little magnet, and it has the kind of magnetic moment that a rotating negative charge would have. 