Diamagnetism
Next we want to look at diamagnetism from the classical point of view. It can be worked out in several ways, but one of the nice ways is the following. Suppose that we slowly turn on a magnetic field in the vicinity of an atom. As the magnetic field changes an electric field is generated by magnetic induction. From Faraday’s law, the line integral of $\FLPE$ around any closed path is the rate of change of the magnetic flux through the path. Suppose we pick a path $\Gamma$ which is a circle of radius $r$ concentric with the center of the atom, as shown in Fig. 34-4. The average tangential electric field $E$ around this path is given by \begin{equation*} E2\pi r=-\ddt{}{t}\,(B\pi r^2), \end{equation*} and there is a circulating electric field whose strength is \begin{equation*} E=-\frac{r}{2}\,\ddt{B}{t}. \end{equation*}  The induced electric field acting on an electron in the atom produces a torque equal to $-q_eEr$, which must equal the rate of change of the angular momentum $dJ/dt$: \begin{equation} \label{Eq:II:34:14} \ddt{J}{t}=\frac{q_er^2}{2}\,\ddt{B}{t}. \end{equation} Integrating with respect to time from zero field, we find that the change in angular momentum due to turning on the field is \begin{equation} \label{Eq:II:34:15} \Delta J=\frac{q_er^2}{2}\,B. \end{equation} This is the extra angular momentum from the twist given to the electrons as the field is turned on.  This added angular momentum makes an extra magnetic moment which, because it is an orbital motion, is just $-q_e/2m$ times the angular momentum. The induced diamagnetic moment is \begin{equation} \label{Eq:II:34:16} \Delta\mu=-\frac{q_e}{2m}\,\Delta J=-\frac{q_e^2r^2}{4m}\,B. \end{equation} The minus sign (as you can see is right by using Lenz’s law) means that the added moment is opposite to the magnetic field.  We would like to write Eq. (34.16) a little differently. The $r^2$ which appears is the radius from an axis through the atom parallel to $\FLPB$, so if $\FLPB$ is along the $z$-direction, it is $x^2+y^2$. If we consider spherically symmetric atoms (or average over atoms with their natural axes in all directions) the average of $x^2+y^2$ is $2/3$ of the average of the square of the true radial distance from the center point of the atom. It is therefore usually more convenient to write Eq. (34.16) as \begin{equation} \label{Eq:II:34:17} \Delta\mu=-\frac{q_e^2}{6m}\av{r^2}B. \end{equation}  In any case, we have found an induced atomic moment proportional to the magnetic field $B$ and opposing it. This is diamagnetism of matter. It is this magnetic effect that is responsible for the small force on a piece of bismuth in a nonuniform magnetic field. (You could compute the force by working out the energy of the induced moments in the field and seeing how the energy changes as the material is moved into or out of the high-field region.)  We are still left with the problem: What is the mean square radius, $\av{r^2}$? Classical mechanics cannot supply an answer. We must go back and start over with quantum mechanics. In an atom we cannot really say where an electron is, but only know the probability that it will be at some place. If we interpret $\av{r^2}$ to mean the average of the square of the distance from the center for the probability distribution, the diamagnetic moment given by quantum mechanics is just the same as formula (34.17). This equation, of course, is the moment for one electron. The total moment is given by the sum over all the electrons in the atom. The surprising thing is that the classical argument and quantum mechanics give the same answer, although, as we shall see, the classical argument that gives Eq. (34.17) is not really valid in classical mechanics.  The same diamagnetic effect occurs even when an atom already has a permanent moment. Then the system will precess in the magnetic field. As the whole atom precesses, it takes up an additional small angular velocity, and that slow turning gives a small current which represents a correction to the magnetic moment. This is just the diamagnetic effect represented in another way. But we don’t really have to worry about that when we talk about paramagnetism. If the diamagnetic effect is first computed, as we have done here, we don’t have to worry about the fact that there is an extra little current from the precession. That has already been included in the diamagnetic term. 