Magnetism

In electrostatics, we defined polarization,
$\myv P \equiv$ electric dipole moment per unit volume.

Now, Magnetization:
$\myv M \equiv$ magnetic dipole moment per unit volume.

There are at least three different kinds of magnetic behavior of materials:

  • $\frac {d \myv M}{d \myv B} \gt 0$: para-magnetism,
  • $\frac{d \myv M}{d \myv B} \lt 0$: dia-magnetism, and
  • $\myv M \neq 0$ even when $\myv B=0$: ferro-magnetism.

Torques on a constant-current loop

So, let's calculate the torque in a magnetic field of one such rectangular loop:

The magnetic force on a line of current is: $$\myv F_m = \int dq\,\myv v\times\myv B=I \int d\myv l \times \myv B.$$

The loop shown is tilted by an angle $\theta$ from the $x$-$y$-plane. There is a magnetic field in the $\uv{z}$ direction, so $\myv B = B\uv{z}$. The forces are all towards the outside.

Only the forces on the sides of length $b$ will exert a torque on the loop. The force on one is: $$F=IbB.$$

The torque resulting from these two forces is: $$\myv \Gamma = Fa \sin \theta\uv{x}.$$

Putting these together... $$\myv \Gamma = Iab B \sin \theta\uv{x} = \myv m \times \myv B .$$

The net effect of this torque is a tendency for magnetic dipoles to line up with the magnetic field. This accounts for paramagnetism in a general way. (Why doesn't the current change instead of the tilt... huh?)

In real materials, because of the Pauli exclusion principle, every filled orbital will have just as many spin up as spin down electrons. This combination acts like something with no net magnetic dipole moment. So paramagnetism should only occur in materials that have an unpaired electron (materials with odd $Z$).

Mutual spin orientation

If the magnetic field is due to a current loop at the origin with a dipole-ish field, and we place a second small loop at ${\cal A}$ which is free to rotate, what orientation will the small dipole take on? What orientation would a small dipole at ${\cal B}$ take on?

"Orbiting" charge

QM does not support the idea of an electron orbiting the nucleus like a planet about the sun.

Still this classical picture can account qualitatively for diamagnetism:

  • The electric force of the positive nucleus: $$\frac{e^2}{4\pi\epsilon_0 R^2} = m_e\frac{v^2}{R}.$$
  • The hurtling electron looks like a current: $$I=\lambda v=\frac{ev}{2\pi R},$$
  • with a dipole moment $m=I\pi R^2$: $$\myv m=-\frac{1}{2}evR \uv{z}.$$

Now, imagine that a magnetic field is turned on:

  • $\myv v \times \myv B$ points outward,
  • But $F_m = -e\myv v \times \myv B$ points inward.
  • With a greater inward force, the electron's velocity increases. [Why can't it change radius? Ultimately QM...]
  • This means that the orbital magnetic dipole moment increases in the negative $\uv{z}$ direction--opposite the magnetic field.
  • The magnetic field due to this increased speed electron opposes the applied $\myv B$ field.

This is apparently diamagnetism: Magnetization increases opposite to the magnetic field.

This affects both electrons (even if opposite spin orientation) in the same orbital angular momentum state. It is most prominent in materials with closed electron shells (that is, no spare electrons that are subject to paramagnetism).

Most materials (including frogs) are weakly diamagnetic.

Diamagnetism is a weak effect. If a material responds both diamagnetically and paramagnectically to some extent, its net behaviour is usual paramagnetic.

There is no diamagnetic effect for an isolated electron. Why?

We can't justify classically this business of assuming that the electron has to stay in an orbit at the same distance. Diamagnetism is ultimately a quantum phenomenon. But the behavior of this classical picture does agree with the empirical behavior of real (quantum) diamagnets.

Problem 6.6

Which would you expect to be diamagnetic or paramagnetic?: Al, Cu, CuCl2, C, Pb, N2, NaCl, Na (metal), sulfer, H2O ?