# I A magnetic dipole in a magnetic field

1. Mar 30, 2016

### barnflakes

Let's say we have a north south bar magnet and we place it in a uniform magnetic field such that the magnetic moment is perpendicular to the magnetic field.

What happens to the orientation of the magnet?

In my view, the magnet receives a torque pointing out of/in to the page that causes the magnet to rotate so that its magnetic dipole moment aligns with the external field.

In this scenario, there is absolutely no precession.

In my understanding, there should only be precession if that magnetic dipole already has some angular momentum.

So what is the correct answer?

I ask this after watching the following video by Leonard Susskind:

In which at 5:44 he claims that a bar magnet will precess when placed in a magnetic field. He seems to be conflating electrons with bar magnets in this example, though, and I agree that in the case of an electron it will indeed precess. But for a simple bar magnet it should not.

2. Mar 30, 2016

I think there is agreement that there will be a torque $\tau$ proportional to $M \times B$. This is equal to the time rate of change in angular momentum of the system. The angular momentum can change in two ways in this system-one is a mechanical rotation of the mass (of the magnet) in the direction mentioned in the OP. The other is a precession where the direction of $M$ changes. The magnetization $M$ contains an intrinsic angular momentum $L$ that is proportional to $M$ and points along the magnetization $M$ that isn't completely obvious because it doesn't involve any macroscopic mechanical motion. What I think may happen is a combination of precession with a damping effect that eventually makes the pole magnet align with the applied magnetic field. (An extra detail, important for quantitative calculations is that the total magnetic moment is a product of the magnetization $M$ per unit volume multiplied by the volume of the magnet.)... A follow-on: I must admit, this is the first time I have tried this particular calculation. Putting in some typical numbers for the angular momentum for a healthy magnetization $M$(c.g.s. units $M$=1000 is a good number, and multiplying by 2mc/e where m is the electron mass, c =speed of light and e=electron charge, and using a typical magnet volume of 50 cm^3) seems to suggest that the angular momentum from the magnetization is rather weak. Thereby, the torque would most likely result in a mechanical motion as the OP suggests, and would tend to rotate to align the magnet with the magnetic field. The change in angular momentum that results will be almost entirely due to the rotation of the mass of the magnet.

Last edited: Mar 30, 2016
3. Mar 30, 2016

Additional comment is the deHaas-Einstein experiment confirmed that magnetization does indeed have an angular momentum associated with it. When the magnetization is reversed in the deHaas-Einstein experiment by reversing the magnetic field, the above calculations suggest the rotation that occurs in the deHaas-Einstein experiment (of the magnetized cylinder hanging on a thread) would not be a rapid one, but rather a somewhat weak response. The professor in the OP does have an interesting idea, but it would appear the angular momentum that is present from the magnetization is not of sufficient strength to cause appreciable precession of a bar magnet. I'd be interested in hearing if other physics people agree with this assessment.

4. Mar 31, 2016

### barnflakes

Let's just consider an idealised dipole and forget about the cause of the magnetic moment. In this case, the bar magnet should not precess. Do we agree on that?

5. Mar 31, 2016

### Staff: Mentor

In order to have precession, it seems to me there needs to be an angular momentum for the torque to act on. Does your idealized dipole also have angular momentum, in addition to a dipole moment?

If you have an object with a charge distribution and a mass distribution, and let it rotate, then it has both a magnetic dipole moment and angular momentum. Put the rotating object in a magnetic field and it precesses, in general. A Google search turns up some examples including an exercise from the intro physics text by Tipler & Mosca (problem 69 in the linked Google Books page).

6. Mar 31, 2016

If you have a magnetic dipole, i.e. plus and minus poles of a bar magnet, it necessarily arises from magnetization that always comes with angular momentum. The question here is really whether there is sufficient angular momentum $L$in the magnetization, so that if the bar precesses (and thereby $L$ changes its direction and has a $dL/dt$), that the $dL/dt$ is of the same order of magnitude as the torque. It appears that it is not, so instead, the mechanical motion needs to be such as for the rotating mass to have a $dL/dt$ that will equal the torque from the magnetic dipole in the magnetic field. If the bar precesses, it will also have a mechanical $dL/dt$ that is larger than the magnetization's $dL/dt$ that is in the wrong direction and won't match the torque.

7. Mar 31, 2016

### barnflakes

No initial angular momentum.

I don't have access to the text you linked, could you perhaps summarise the results of the problem?

8. Mar 31, 2016

### Staff: Mentor

I haven't actually worked out the solution for that problem, but simply posted it as an example of a magnetic dipole (with associated angular momentum) that precesses in a magnetic field. Paraphrased slightly:

A uniform disk of mass m, radius R, and surface charge density σ rotates about its center with angular velocity ω. A uniform magnetic field B makes an angle θ with respect to the rotational axis of the disk. Calculate (a) the net torque acting on the disk and (b) the precession frequency of the disk.

I suppose one could simplify this by considering a thin rigid charged ring instead of a disk. Then it wouldn't be necessary to integrate to get the dipole moment.

9. Mar 31, 2016

### barnflakes

The disk will have angular momentum, causing the precession. The bar magnet won't. That is the heat of my question.

10. Mar 31, 2016