It may be useful to review magnetic dipoles which I wasn’t able to include in my chapter 5 lectures, but which I included at the end of the crash course for Chapter 5.

Review of torque on a dipole:

\[\vec{N} = \vec{m}\times\vec{B}\]

Force on a dipole:

\[\vec{F} = \nabla(\vec{m}\cdot \vec{B})\]

Compare to it’s electrical ``twin”: \(\vec{F} = \nabla(\vec{p}\cdot \vec{E})\)

The “Gilbert” model:

You can do a lot of magnetostatic problems by using electrostatic results and replacing:

  • \(\vec{p}\) with \(\vec{m}\)
  • \(1/\epsilon_0\) with \(\mu_0\)
  • \(E\) with \(B\)

Once you get really close-up to the dipole (where it looks like a ring of current instead of a perfect magnetic dipole) then this breaks down. Griffiths’ advice is to use the Gilbert model to get intuition, but don’t rely on it for quantitative results.

Paramagnetism

We saw in chapter 4 that an electric field can induce polarization in individual atoms by sending all the electrons to one side and all the nuclei to the other. How would you guess a magnetic field induces magnetism in an atom?

There are actually (at least) two answers. The one you’ve probably heard of more is that that a magnetic field can make the spins of an electron align with the magnetic field (see the torque equation above, it’s effect is to align the dipole moment with the field).

Why does an electron spinning have a magnetic dipole moment? It’s like a spinning sphere of charge.

The spin alignment with the magnetic field creates an additional magnetic field in the same direction. This is called paramagnetism. Para in latin means “defense, protection agains”. I’m not sure I would’ve called it that.

What about the “current” of the electron going around the nucleus? (Spoiler alert/punchline. This gives you diamagnetism, where the magnetization is opposite the applied field. Why??? That doesn’t make any sense!! It does actually. It turns out that it’s hard to actually change the orientation of an orbit, so the torque that happens in paramagnetism isn’t the most important thing. The most important thing is the increase in velocity of the electron.)

See if you can come up with what the magnetic dipole of an electron in an orbit would be.

Ans: You can think of the electron’s orbit as a current. Let’s look at that!! Current is charge per time, or linear current density times velocity.

\(I = \lambda v = ev/2\pi R\).

And magnetic moment is

\[\vec{m} = I\vec{a}\]

where \(\vec{a}\) is the vector area (direction chosen by right-hand-rule from current, I). The area is \(\pi R^2\). So the magnetic moment of an electron orbit is….

\[\vec{m} = -\frac{evR}{2}\hat{z}\]

So now we just have to figure out how much the velocity changes when we add the magnetic field.

The argument goes like this…

Let’s start without a magnetic field, just an electric field. As you know, a central \(1/r^2\) force can give you an orbit. To find its radius you equate the force to \(mv^2/r\). I’m using \(v_e\) in this equation to specify the velocity in the presence of the Electric field only.

\[\frac{1}{4\pi\epsilon_0} \frac{e^2}{R^2} = m_e\frac{v_e^2}{R}\]

The \(v_e\) means “The velocity when only the electric field was involved.”

The thing on the left is the electric field of a point charge multiplied by charge \(e\). (We’re assuming there’s an electron orbiting a proton, and the way we’ve put the sign of the evB term, we’ve assumed that the magnetic field is increasing the magnetic force toward the center, which means that the magnetic moment of the atom is pointing opposite the magnetic field. This is a super important point - you have to get the signs right for this argument to be interesting).

To this expression we can add a magnetic field contribution:

Let’s say the magnetic field is perpendicular to the plane of the orbit, so v and B are always perpendicular to each other.

\[\frac{1}{4\pi\epsilon_0} \frac{e^2}{R^2} + ev_BB= m_e\frac{v_B^2}{R}\]

The \(v_B\) means “The velocity when the magnetic field is added.”

So how much faster is the electron going with the addition of the magnetic field, i.e. how much larger is \(v_B\) than \(v_e\)? Subtract the the first equation from the second…

\[ev_BB = m_e\frac{v_B^2 - v_e^2}{R}\] \[ev_BB = m_e\frac{(v_B - v_e)(v_e + v_B)}{R}\]

As long as the difference is small (which it surely is. Recall our earlier discussion on how much small magnetic forces are than electrical forces) we can approximate…

\[v \approx v_B \approx v_e\]

And let

\[\Delta v = v_B - v_e\] \[ev_BB = m_e\frac{(\Delta v)(2v)}{R}\] \[\Delta v = \frac{eRB}{2m_e}\]

If the velocity changes by that much then the dipole moment changes by

\[\Delta \vec{m} = -\frac{1}{2} e(\Delta v)R\hat{z} = -\frac{e^2R^2}{4m_e}\vec{B}\]

Notice the change in \(\vec{m}\) is opposite the direction of \(\vec{B}\). This is called diamagnetism.

How does this produce an overall magnetic field in the materials?

Excellent question. It’s not obvious. The atoms are all aligned in different ways, so the bulk magnetization is zero. When you apply the magnetic field, this effect (above) adds a small magnetic moment to most of them, all in the same direction so the material becomes polarized. (Does it really add it in the same direction? Yes, consider an atom where \(m\) and \(B\) are aligned. In this case, the magnetic force will decrease the central force and the velocity will decrease. So aligned \(\vec{m}\)’s will decrease and anti-aligned \(\vec{m}\)’s will increase, thereby increasing the overall magnetization opposite the direction of the applied magnetic field.)

Andrea thinking about this later….

The equation we wrote up there…

\[\frac{1}{4\pi\epsilon_0} \frac{e^2}{R^2} + ev_BB= m_e\frac{v_B^2}{R}\]

is the equation for a classical stable orbit. It’s actually the requirement for a stable orbit. If the electron starts out with a particular velocity, \(v_0\) and then you increase the central force, the electron will no longer be going fast enough to escape falling down onto the nucleus. At least that’s what would happen in the classical case. Obviously that’s not what happens. It somehow increases the energy of the orbital.

But this also addresses the problem of magnetic fields not being able to do work (and not being able to speed up electrons). It’s not really speeding it up. Rather, it’s increasing the central force and therefore increasing the energy of the orbit.

Magnetization

\(\vec{M}\) is the magnetic dipole moment per unit volume (just like \(\vec{P}\) is the electric dipole moment per unit volume.)

The vector potential due to some magnetized matter is:

\[\vec{A}(\vec{r}) = \frac{\mu_0}{4\pi} \int \frac{\vec{M}(\vec{r}^\prime) \times \hat{\mathscr{r}}} {\mathscr{r}^2} d\tau^\prime\]

In a manner very similar to what we did for electric dipoles, we can define a bound volume current and a bound surface current.

\[\vec{J}_b = \nabla \times \vec{M}\] \[\vec{K}_b = \vec{M} \times \hat{n}\]

where \(\hat{n}\) is the normal unit vector (normal to the surface.)

I drew some sketches here that were surprisingly helpful. First I drew side by side loops and showed that everything internal cancels out and you’re left with surface current. Then..I drew a series of current loops with \(\vec{m}\) coming out of the board that increase in area going to the left. You can look at how the current side by side is cancelling out and see that the volume bound current will be up, perpendicular to both the derivative and the direction of \(\vec{m}\).

With those definitions:

\[\vec{A}(\vec{r}) = \frac{\mu_0}{4\pi} \int_V \frac{\vec{J}_b(\vec{r}^\prime)}{\mathscr{r}} d\tau^\prime + \frac{\mu_0}{4\pi} \int_S \frac{\vec{K}_b(\vec{r}^\prime)}{\mathscr{r}} da^\prime\]

Should you be excited about using that equation to calculate the vector potential?

No.

Why? Because it’s really laborious. First you have to calculate J and K, and then you have to integrate over each of them, and then you have to take the curl of the result. Oi.

Mostly I would avoid dealing with it if you can. If you can use the auxiliary field \(H\) to calculate \(B\) in a problem then you should. \(\vec{H}\) was invented to make your life easier so that you don’t have to calculate the bound current.

The idea of bound current and bound charge give you a way to think about why the field is changed by the material, but it’s not a good way to calculate anything.

After break Dave is going to tell you about the auxilliary field \(H\) which is the better way to calculate the field in magnetized matter.