# I A diamagnet inside a magnetized hollow cylinder

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1. Mar 25, 2016

### Nono713

Hi!

Suppose I have a cylinder of pyrolytic carbon graphite (strong diamagnet) that's vertically placed inside a hollow cylinder made out of approximately radially magnetized neodymium. Assume the graphite is frictionlessly held into place vertically somehow (so gravity can be ignored) but can still move in the horizontal plane.

1) Will the graphite find a stable equilibrium in the center of the neodymium ring, or is it unstable?
2) Is this diamagnetic restoring force proportional to the strength of the surrounding magnets? (I believe so but I'd like confirmation)
3) Do the answers to the previous questions change if the graphite is spinning about its axis (assume negligible vibration)?
4) Can this restoring force be quantified, for instance in Newtons per distance from equilibrium?
4b) If it can't be easily quantified, would it at least be possible to realistically get ~ 0.1N of force on the graphite or are passive diamagnets just that weak, assuming reasonable cylinder sizes (say a few cm diameter/height)?

This isn't a homework problem, it's an engineering issue and I have found it quite difficult to find adequate information about how to measure the strength of diamagnets in various situations.

Thanks for any insight!

2. Mar 25, 2016

### jwinter

An interesting proposal that I hope I can throw some light on to. Firstly it is almost impossible to make a radially magnetised tube magnet! But to your questions:

1) Diamagnetic (and paramagnetic) effects only appear where there is a change or gradient in field strength and with diamagnets the force is always away from volumes with stronger field (close field lines) towards volumes with weaker fields (spaced out lines). On the inside of radially magnetised tube one can imagine the field lines getting closer as they approach the centre - which means a diamagnetic material will tend to move towards the magnet walls and thus be unstable! But I think (from a surface monopole argument) this intuition is incorrect and there there will actually be no field variation in the radial direction and thus no radial force. There will however be a significant axial force which will tend to thrust the diamagnet out of whichever end of the tube it is closest to.
2) The force is proportional to the gradient of the field - which I expect to be zero in the radial direction and significant in the axial direction.
3) No, I think spinning a non-conductor in a magnetic field has almost no effect.
4) Yes, but I think it is zero. 4b) No!

But it looks as though you would like to make some sort of semi-passive magnetic bearing and you are prepared to spin one side of it? In this case there is a very simple design that could really serve you well. If you put a strong (axially magnetised) cylindrical magnet (or a stack of them N-to-N and S-to-S) as a horizontal stub axle (with its axis along the rotation axis), and then spin a horizontal copper tube concentric with it, the tube will be strongly levitated so that its axis approaches the axis of the magnet. (What happens is that as the spacing between the magnet and the copper sheet varies as the sheet travels from top to bottom and back to top, eddy currents are induced in the copper which act against the fixed magnet, thereby levitating the copper tube). Such bearings are available commercially I believe but it is very easy to make your own. It is definitely best to have the copper tube outside the fixed magnet, but it may work to some lesser extent to have a copper shaft spinning within a fixed (axially magnetised) stack of ring magnets (again stacked N-to-N and S-to-S to get the field variations).

Something that is very interesting about this bearing is that it is no good spinning the magnet - that has no effect at all. If the copper sleeve is not spun then there is no centring force. You can even spin the magnet at the same speed as the copper sleeve so there is no relative rotation - it doesn't alter the magnetic field and so makes no difference to the centring force!

Last edited: Mar 25, 2016
3. Mar 25, 2016

### Nono713

Thanks a lot for the thorough answer! I hadn't realized that the field would be zero inside the cylinder; naive me had expected it to kind of work like a normal magnet lining the inner cylinder walls :D good thing you mentioned the axial force as that could seriously hurt someone.

The design you suggest seems neat; so the copper "knows" it's rotating and you can't fake it by rotating the magnets instead, I find that really cool. I have a couple questions however:

1) how can I calculate how fast it needs to go before the copper starts aligning with the magnets? Do I just need to integrate the eddy forces over the entire copper tube as a function of magnet strength and distance from the magnets and solve for velocity? I know this won't give me an exact answer but it should work as a starting point right, or is it more complicated than that?

2) it seems to me that if you placed two of these bearings at either end of a shaft, they would passively stabilize all three axes of rotation (assuming one axis is supposed to spin) and two translational axes; the last translational axis (position along the copper tube) still needs to be stabilized; I think Earnshaw's theorem means you can't just use repelling magnets to repel the shaft at both ends, but can it be done passively or do I really need an electromagnet for that bit?

This is what I came up with using your bearing system (side cut, the entire thing is radially symmetric, everything connected to the shaft is spinning, everything else is fixed):

So as per (2) I am guessing this makes the shaft unstable in the up/down direction as it would oscillate up and down with no damping until it hits one of the magnets. Or wouldn't the eddy currents inside the copper tube resist any up/down movement of the tube along the magnets, naturally providing damping, as in the popular "drop magnet in copper tube" demonstration, or does that damping just go into the fixed magnets in the form of heat?

4. Mar 26, 2016

### jwinter

Another thought which didn't occur to me earlier is that if you changed the magnets in your original idea to a stack of N-to-N and S-to-S ring magnets with the pyrolitic graphite rod down the middle, then you would get some good falling off of the field strength towards the centre of the rod and thus a restoring centring force in that case. Better still would be to made a "bearing" tube out of pyrolytic graphite and put a stack of N-to-N and S-to-S cylindrical magnets down its centre (as you have shown in the diagram above but with a copper tube). The restoring forces would be very weak (as the diamagnetic effect is very weak indeed) but it would work statically when it is not being spun - which might be an advantage? The force calculation should be reasonably straightforward (apart from the horrible shaped fields). You use the same formula as for the force on a permanent magnet in a non-uniform magnetic field (m x dB/dx; where m is the magnetic moment and dB/dx is the field gradient). In the diamagnetic case you obtain the strength of the "permanent magnet" by multiplying the (average) field B in the diamagnet by its diamagnetic susceptibility (a number of order 10^-4 for pyrolytic graphite) times its volume.

With regard to your questions above:

1) Eddy current force calculations are notoriously difficult. You could probably get an order-of-magnitude value from very simplified assumptions, but to do a proper job you really need a multi-physics finite element package (expensive). Since it is so cheap and easy to buy some magnets and give it a spin and get a certain and correct answer, I would take that approach. If you do a search for "electrodynamic bearing" then you should come across the PhD thesis of Torbjörn Lembke which should give you plenty of information and probably even measured restoring forces with various magnets and at different rotation speeds.

2) The diagram you have shown above should work fine. Of course it will need some mechanical bearing surfaces to slide on while it is being spun-up before it self-centres. Earnshaw doesn't apply to either diamagnetism or eddy current repulsion (= strong AC diamagnetism!) and so your longitudinal stabilisation should also work just fine. (From Earnshaw the radial directions would be unstable until you got it spinning) If you have money to spend you could even hire Torbjörn's company magnetal.com to do the design for you! (It is possible that he has some patents on the ideas but I have not seen any mention of such).

I am sure you are correct that axial motion will also be strongly damped by the "drop magnet in copper tube" effect!

Good Luck!

5. Mar 26, 2016

### Nono713

Thanks again for your help! One last question: I assume the N to N and S to S magnet layout is required in order to get the correct field configuration for the desired eddy currents and/or diamagnetism effects, however is it possible to space the magnets a little bit with something nonmagnetic like aluminum so I don't need to have the two poles touching each other (which would be physically challenging if I want to put strong magnets in there)?

The graphite tube idea is very good, I hadn't thought of it. But pyrolytic graphite isn't that conductive (30x to 60x less than copper if I found the right numbers) so maybe two concentric tubes might be better: the thick graphite one inside, with the copper tube on the outside. I need to start measuring the force to see if the diamagnetic force would be strong enough to justify the added build complexity though.

I'll check out Torbjörn Lembke's thesis; any data I can get my hands on before settling on a bearing design is more than welcome :)

6. Mar 27, 2016

### jwinter

A magnetic field can only generate a force in volumes which have a field gradient. A uniform field can generate a torque, but it cannot generate a linear force. This is the reason that there would be no centring force from your radially magnetised tube magnet - it had no gradient for the diamagnet to want to slide down. The field lines were evenly spaced in the radial direction.

In the case of a row of magnets axially magnetised, you have two choices. You can assemble them N-to-S to make one very long magnet with a single pole at each end, or you can force them together N-to-N and S-to-S to make multiple north and south poles along the length of the assembly. Consider how the field falls-off (ie its gradient) in each case:

The single poles widely spaced has a long range effect, its field being detectable far from the rod because it falls-off so slowly. But the field from the rapidly alternating poles falls off much quicker - it falls off with a scale length proportional to the spacing between the poles. In your case you want a good fraction of the fall-off to occur within the diamagnetic material itself, because it does no good anywhere else. If you make the N-S alternation too close, then the majority of the field gradient will occur within the air-gap around the bearing shaft - which also does no good. So you need to scale the spacing to match the gap and thickness of the diamagnetic material.

Diamagnetic (and eddy current) forces have an additional consideration that permanent magnets applied in the same situation do not have. And that is that the strength of the diamagnetic magnetisation is proportional to the absolute field strength (rather than its gradient). So if you like the force obtained is proportional to the field strength (which creates the magnetisation) times the field gradient (which the magnetisation then wants to slide down). So you want to err on the side of having a stronger field with maybe less gradient, than having maximum gradient with only average field strength.

It is certainly reasonable to put small gaps between the N-to-N poles of the magnets. This would relieve the demagnetising stress with very little loss of useful force.

I am not sure why you think conductivity is important (except for eddy current effects)? The diamagnetic effect is like an induced opposing permanent magnet and permanent magnets work just fine with zero conductivity - eg ferrite ones.

With permanent magnets on the inside and outside of a diamagnetic tube, you will need to line up the inner and outer poles optimally. I think what you want for best field gradient effect again is that everything should be repelling as strongly as possible - ie north on the outside corresponding to north on the inside, etc. You also want the diamagnetic tube wall thickness to be relatively large.

You must know that the diamagnetic effect in pyrolytic graphite is direction dependent? - I think the strongest effect occurs when the graphite planes are perpendicular to the field gradient. I think this is the way it would lay down if you had it manufactured to shape, but I think it would be wrong if you machined a cylinder from a block.