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Recently when re-reading Griffth electromagnetism, where he explained about how if the magnetic system is cylindrical, solenodial, planar or toroidally symmetric (4th edition, p.283; 1999 reprint, p.273), then

$$\nabla \cdot \vec{M}=0$$

always and thus the $\vec{H}$ field can be obtained via the usual Ampere's Law

However, on closer inspection, I am wondering whether said conditions by Griffth is a sufficient condition to ensure the divergence of magnetisation to vanish everywhere.

Consider a disk of material with electron spins being engineered to give the following radial magnetisation.

$$\vec{M}=a\hat{r}$$

where $a$ is a constant. Then such system clearly has nonvanishing divergence at the circular rim of the disk and also at the axis.

Now stack these discs vertically as shown, then a system obeying cylindrical symmetry can be formed with nonvanishing divergence. This is still kinda ok because at the flat ends, the south pole can still be found thus it still act like a bar magnet.

enter image description here

But things get very strange when the discs were arranged in a closed loop. The resulting system is not only toroidally symmetric, have nonvanishing divergence of $\vec{M}$. Due to the south pole end of each magnitsation vector being buried inside the core of the torus, to the external environment, the torus is effectively a magnetic north monopole.

  1. How to recoincile Griffth's statement that a magnetic system with some symmetry will ensure the vanishing of the divergence of $\vec{M}$ with the examples shown in this post?

  2. For the torus with radial $\vec{M}$ what is its magnetic field. Will it end up having no magnetic field (thus becoming some kind of anapole) due to how the south pole end is a ring buried inside the core of the torus?

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You can't use magnetization to produce electromagnetic fields that can't also be produced by an arrangement of charges and currents, because magnetization is made, microscopically, from charges and currents. So you haven't made a magnetic monopole. Let's try to figure out what, exactly, you have made; it's interesting.

I think that you can't have have $\vec M = a\hat r$ at the center of the disks, where the direction of $\hat r$ is not defined. So each of your disks is a torus of its own. You could produce such a field by rotating a disk-shaped, charged capacitor about its axis: adjacent, parallel sheets of charge, rotating to produce opposing currents. If it's important to have a uniform radial magnetic magnetization, $\vec M = a\hat r$ rather than $\vec M = \frac{ar}{R}\hat r$, you can arrange for your capacitor to have different charge distributions at different radii.

The principle of superposition tells us that if we were to place many capacitors in series with each other, we'd get your stack of disks with uniform radial magnetization. It's helpful to remember that an infinite, uniform current sheet produces a constant change in the magnetic field. Far from the ends of the stack of disks, we can break the symmetry between radial-in and radial-out magnetization by having different distances between the "capacitor gaps" that we used to define the disks in the previous paragraph and the spacing between the capacitors:

          ^           (spin axis: up)
o o o o o | x x x x x (positive plate)
  B-->    |   <--B
o o o o o | x x x x x (negative plate)
o o o o o | x x x x x (positive plate)
  B-->    |   <--B
o o o o o | x x x x x (negative plate)

compared to

          ^           (spin axis: up)
o o o o o | x x x x x (positive plate)
  B-->    |   <--B
o o o o o | x x x x x (negative plate)
  <--B    |   B-->
o o o o o | x x x x x (positive plate)
  B-->    |   <--B
o o o o o | x x x x x (negative plate)

compared to

          ^           (spin axis: up)
o o o o o | x x x x x (negative plate)
  <--B    |   B-->
o o o o o | x x x x x (positive plate)
o o o o o | x x x x x (negative plate)
  <--B    |   B-->
o o o o o | x x x x x (positive plate)

Now we take our very long stack of radial-out magnetized rotating capacitors, bend it gently into a loop so that the approximation that our current sheets are much larger than the gaps between them is still good, and we have your outward-magnetized torus.

We can re-analyze your magnetized torus as a superposition of opposing toroidal currents. Apparently the anapole moment cancels, but the external magnetic field depends on the "phase" of the superposition. If the "coils" of the two toroids overlap, then superposition says that must be the same as having no current at all. If the two toroids are equally spaced, as in the middle sketch, then the near field is complicated but the far field is zero. But if the two toroids are unequally spaced, as in the first and third sketches, the arguments above suggest that the near field, at least, ought to have more "out" than "in" component.

If you think of the monopole, dipole, and quadrupole moments of a charge distribution as the first, second, and third moments of the charge (or current), what you've produced here is something like the "second anapole moment." It's very interesting. A computation that shows how the "monopole" field goes away would be an illuminating project.

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