# Spinning metallic disk without magnetic field

1. Aug 7, 2012

### esorolla

Hello everybody.

I have to admit that I feel quite troubled since long time, actually since I read the solution of a problem that I don't understand and whose wording I immediately pass to briefly relate you. I guess that many of you have heard about the Faraday's disk, that is, a spinning metallic disk subject to an external magnetic field where a current is produced by the Lorent's force (or at least this is a way to explain the presence of the current). The rim of the disk is connected to the center of the disk by a wire of negligible resistance, closing the loop and establishing the circuit.

However, my problem is quite troubling because I was enjoying of the books that I love to read where the authors, J. M Lévy-Leblond and André Butoli, propose the same problem assuming no magnetic field at all. The puzzle that they propose asks whether there will be any current, and in my naive ignorance I was convinced that the answer was negative. My surprise was huge when I saw that the answer provided by the authors was that there would be some current, though it would be negligible because the centrifugal force would be balanced with the small potential difference produced between the rime of the disk (where the electrons are supposed to be gathered by the centrifugal force) and the center of the disk (where a net positive charge remained).

I have to admit that I proposed the puzzle to some people in my lab and only one girl dared to think about this problem, but surprisingly for me, she replied almost immediately that, indeed, there would be a current, though she didn't say anything about how big or small it would be. My problem in the beginning was that I expected that the electric field produced by the small potential difference between the rim and the center of the disk would not produce any current since the attraction to the center of the disk is produced through the disk as well as through the wire that is supposed to close the circuit connecting the rim to the axis of rotation of the bar which makes the disk spin. I expected that there would be a position of equilibrium where the centrifugal force (the inertia of the electrons to move away from the center) and the electrical attraction towards the center of the disk through the wire would compensate the attraction towards the disk through the disk itself. I imagined that it would be as if I was the electron pulled my left arm on one side and my right arm from the other. If the centrifugal force plus the attraction trhough the wire are bigger than the attraction to the center of the disk through the disk itself, there would be a net current, however, I expect that the attraction to the center would depend on the distance between the electrons and the distance. Thus, I expected an equilibrium position.

Nevertheless, I guess that this reasoning has an error, because by the same reasonings I would expect that in the case of the Faraday's disk, there couldn't be any current, and I know that it exists. So I'm lost. On the one hand I know that Faraday's disk works and I don't understand why, and on the other hand I respect a lot the word of Lévy-Leblond, but I don't see how there might be current if no magnetic field exists. Actually many people in many forums talking about Feynmann's and Faraday's paradoxes comment that without magnetic field there is no current. And the worst thing is that my reasoning to understand why there can't be current is non-valid, since it would invalidate the possibility that in case of the existence of magnetic field the current existed!

I hope that I didn't make it too much complicated. Do you think that Lévy-Leblond is wrong or the fact that my explanation is non-valid doesn't invalidate the assertion that in the Faraday's disk, if no magnetic field exists, there is no current?

Thank you very much if you arrived to the end and good summer.

Regards

Last edited: Aug 7, 2012
2. Aug 7, 2012

### Simon Bridge

I figure the approach would have been to treat the valence electrons in the metal as a Fermi gas in a box - if you spin the disk you'd expect the gas to shift towards the rim, giving you a potential difference. (This would suggest that you do slightly more work spinning-up a conducting disk than an insulating one...)

Like you I would not expect to be able to set up a circuit from the rim to the hub of the disk since the wire is also turning for the same effect.

However, I can imagine running a stationary wire along the bench (say) from rim to hub - with the electrical connection provided by brushes?

Your fun authors would be these guys?

Last edited: Aug 7, 2012
3. Aug 8, 2012

### esorolla

Indeed, those are the authors.

Concerning the wire, yes, it should be fixed to make the problem simpler.

Thank you, at least, for the support to the same idea that I defend, but I'm still confused.

Regards

4. Aug 8, 2012

### Emilyjoint

Are you certain that the Earths magnetic field was not having an effect.
Or maybe some other stray magnetic field.
How did you check that there was definitely no magnetic field?

5. Aug 8, 2012

### esorolla

That's a good point! I didn't think about that. Perhaps the authors assumed that the reader was aware of the importance of the earth's magnetic field. I have to admit that I have not made the experience. I don't have at my disposal the tools to recreate the experience for the moment, though I'd really like to try it.

Thanks

6. Aug 8, 2012

### Simon Bridge

If you did the experiment - you could change the orientation of the apparatus to see if the earth's field is having an effect ... though a back-of envelope calculation should be able to show you the scale of this effect.

7. Aug 8, 2012

### Staff: Mentor

As a model, I think it is fine to treat the rotation as centrifugal force.

To get an estimate for the magnitude: Assume a disk of 1m diameter with 100/s angular velocity. For electrons, this leads to a maximal centrifugal force of 10^(-26)N = 6*10^(-8) eV/m. As you can see, this leads to voltages of the order of nanovolts. I would expect that the mechanical contact between disk and cable produces more noise and other effects.

8. Aug 8, 2012

### Simon Bridge

Not to ention static charging from friction with the brushes...
More useful would be to figure out how fast the spin has to be to get a measureable effect ... also compare with the same disk spinning in a magnetic field 30-50μT :)

9. Aug 8, 2012

### Staff: Mentor

Solid, homogeneous, rotating disks are limited to ~4km/s velocity at the rim if I remember correctly, this limit just depends on the ratio of tensile strength to density and the geometry, but not the radius. As an interesting coincidence, the voltage just depends on the velocity, too, where 4km/s correspond to 45µeV (maybe with a factor of 2 somewhere).

2km/s * electron charge * 50µT * 1m = 0.2 eV. Here, the voltage depends on the radius, and it is larger by 4 orders of magnitude.