Talk:Faraday paradox

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I just created a new temporary page, based on a survey of many web sites and so not a copyvio. I have tried to explain both the origin of the paradox and its solution. With a bit more material it could become a Faraday disc article. --Heron 21:01, 26 Jun 2005 (UTC)

Thanks for your efforts. The step by step explanation is rather clear, but looks somewhat out of line with the typical encyclopedical style. Perhaps someone can mold into in form which combines both virtues. --Pjacobi June 29, 2005 20:47 (UTC)
And I'm not totally sure, whether Faraday Paradox and Homopolar generator should be separate pages. --Pjacobi June 29, 2005 20:58 (UTC)

I don't mind if you want to merge the two articles. The paradox article just acted as a seed for the disc article, so it has served its purpose. --Heron 29 June 2005 21:27 (UTC)

"In the third step, the disc and magnet are spun together"--spun together in the same direction or in opposite directions? Please clarify.


Contents

[edit] Rewrite of 2005 June 30

I have rewritten the modern explanation. If you don't like it, revert it and we can discuss it here. I believe that the observations can be explained using Faraday's law and that electrons (discreteness of charge) are not necessary. We could put back the explanation using the Lorentz force if you like (There are umpteen correct ways to analyze the problem.), but I don't think it gets to the heart of the paradox. Shouldn't the electrons feel the Lorentz force regardless of whether they are moving relative to the field of the field relative to them? The Lorentz contraction business is pretty heavy stuff (and I haven't even told the whole story). Maybe we should leave it out. Art Carlson 2005 June 30 08:56 (UTC)

I like your new 'Modern explanation' section, as it's much easier to understand than my version but seems to be equally valid. I am afraid that my explanation gave way to hand-waving towards the end, when I just assumed that the words "special relativity" would explain everything. Would it be possible to derive an equation for the current output based on your explanation? I have seen it done for the Lorentz force version using E = qv x B.
I was initially unhappy with what you said above, and in the article under 'Configuration without a return path', about relative motion, because I had convinced myself that rotating a symmetrical magnetic field about its axis could have no externally detectable effect. However, I then read "Electromagnetism and Rotational Relativity" by G.R.Dixon, 1/5/04, which seems to give experimental evidence that it can. It seems that you can't generate current in a closed circuit by rotating the magnet, but you can generate electric polarisation. Doesn't this mean that the polarisation ought to generate a small (and perhaps only transient) current in Faraday's disc when the disc is stationary and the magnet is rotated? Perhaps it does, but nobody has noticed. --Heron 30 June 2005 19:41 (UTC)
I was initially happy with what I said above, and in the article under 'Configuration without a return path', about relative motion, until I read your reference. The considerations there seem to imply that a uniform and constant magnetic field, when viewed in a rotating frame of reference, would look like a uniform charge density in space. Is that right? What happens to general relativity, that says that all frames of reference - even non-inertial frames - are equivalent? What does E&M look like when it is formulated in a rotating frame? Taking it back to my edits, since I have no way to charge up the disk, there can be no electric field between the disk and the solenoid. Unless, like I fear, empty space gets charged. Does anyone understand this any better than I do? Art Carlson 2005 July 1 11:37 (UTC)

I'm glad you see the point of my objection, Art. Unfortunately, I am now out of my depth and will have to wait to be rescued by someone with a better grasp of physics than mine. I look forward to seeing how this turns out. --Heron 2 July 2005 11:53 (UTC)

[edit] Is this right?

In "Configuration without a return path" you say "rotating the magnet must really produce the same charge distribution as rotating the disk". Surely this can't be right. Rotating the cylindrical magnet around its axis has no effect on the magnetic field that it produces. I guess this could be tested by connecting the axle to an electroscope.

I'm not happy with that section, either, although I'm not that hot on relativity. I have pointed out our concerns to the author, Art Carlson (talk · contribs), so we'll see what happens. --Heron 22:50, 18 December 2005 (UTC)
I already expressed some misgivings about what I had written in the previous section of this talk page. On the other hand, simply asserting that "Rotating the cylindrical magnet around its axis has no effect on the magnetic field that it produces." is also not adequate. I haven't thought about the problem since then and don't know if I can find the time to get to the bottom of it now. It would be great if we could call in some real expertise, someone who gets paid for thinking about things like this. Somebody said the real physics jocks hang out at Wikipedia talk:WikiProject Physics. You could try there or in some newsgroup. Sorry I can't give a more definitive answer at this time. --Art Carlson 08:35, 19 December 2005 (UTC)
OK, thanks for getting back to me. I have asked the question at WikiProject Physics. --Heron 18:50, 19 December 2005 (UTC)

Rotating the magnet alone, without rotating the disk, will not produce a charge distribution on the edge of the disk. Only a rotating disk will get a charge distribtion, irrespective of the motion of the magnet.

On the other hand, do people really talk about charge distribution in this thing? Its kind of like talking about the charge distribution in a battery: its there, but would be hard to measure for a low-voltage batery. linas 19:10, 19 December 2005 (UTC)

The section's pretty muddled, but it looks to me like it's predicated on the assumption that you can't tell non-inertial from inertial reference frames. — Laura Scudder ☎ 19:50, 19 December 2005 (UTC)

Actually I think my explanation was wrong. Even if there is a change in charge density, which I no longer believe is possible, it would make no electric field inside a cylinder. That suggests that there is charge separation if and only if there is absolute rotation of the disk, i.e. compared to an inertial frame. If that is true, then the right answer is that an emf can be generated even if there is no relative motion of the disk and the magnet! --Art Carlson 20:54, 19 December 2005 (UTC)

I propose the following:

Configuration without a return path
A Faraday disk can also be operated without a galvanometer and its return path. When the disk spins, the electrons collect along the rim and leave a deficit near the axis (or the other way around). It is possible in principle to measure the distribution of charge (though not necessarily easy). This charge separation will be proportional to the magnetic field and the rotational velocity of the disk. The magnetic field will be independent of any rotation of the magnet. In this configuration, the polarisation is determined by the absolute rotation of the disk, that is, the rotation relative to an inertial frame. The relative rotation of the disk and the magnet plays no role.

--Art Carlson 15:39, 20 December 2005 (UTC)

That's fine with me. --Heron 19:38, 20 December 2005 (UTC)
I agree as well. I was inventing a thought experiment, modelling a magnet as a current flowing in superconducting loop of wire, and then rotating it about its rotational axis. I convinced myself that it doesn't affect the resulting magnetic field and that there's no charge separation in the magnet. You seem to accept that as well now. Occultations 21:16, 20 December 2005 (UTC)
More or less. I think. My problem is this. A translating solenoid will have charge separation and an electric field in the vXB direction. (Details on request.) Now, I should be able to split my big, rotating solenoid up into a bunch of little, azimuthally translating solenoids, which would imply a radial electric field proportional to r, and a constant charge density in the interior. I might even believe that this story is really true for a permanent magnet, but where should the charge density come from in a solenoid with a vacuum inside? --Art Carlson 11:23, 21 December 2005 (UTC)
What is a "translating solenoid"? Occultations 20:43, 22 December 2005 (UTC)
A solenoid is a current-carrying coil of wire, an electromagnet. Translating means moving, in this case perpendicular to the axis of the solenoid. The physics must be the same whether I am sitting on the solenoid and watching what happens, or driving past, or the solenoid is moving past me. The only way for that to be true is, assuming there is no electric field when the solenoid is stationary, is if there the an electric field in the moving solenoid equal to vXB. --Art Carlson 22:06, 22 December 2005 (UTC)
I agree with the statement as written, with a minor quibble that "the charge separation will be proportional to the magnetic field" assumes a constant magnetic field, otherwise the charge separation is more complicated.
  • With regard to Art Carlson's problem, I think a small translating solenoid gives an electric dipole field, and the electric field of two dipoles aligned end to end is not double the field of one dipole. Also, there is the fact that the dipole strength decreases to zero at the center, etc. so I question the idea that a radial electric field proportional to r would be implied necessarily.
  • The idea that the disc and magnet rotating at the same rate would give still give charge separation bothers me, but not in a fundamental way. I'm just trying to visualize it from the point of view of a co-rotating (therefore non-inertial) frame. I'm not good at general relativity but I always had a notion that if I am in a non-inertial frame, I can concoct some gravitation-like force field and pretend that I am in an inertial frame. But if we do that in the above case, there will be like a negative mass at the center of the disc repelling all masses (=the coriolis force) which will account for the small charge separation to be expected if there is no magnetic field. But if there is a magnetic field, what accounts for the stronger charge separation? The electrons in the disc are motionless so it cannot be qVxB forces. It must be that in the non-inertial frame an electric-like force field dependent on the magnetic field must be introduced as well. PAR 14:12, 22 December 2005 (UTC)

[edit] Is this still controversial?

Looking at the references I added to the article, it appears that this subject is still somewhat controversial. Are there situations where only the Lorentz law can be used? Note that the Scanlon reference appears to be the source of the explanation based on Faraday's Law given in this article.--J S Lundeen 02:59, 23 January 2006 (UTC)

As far as I can tell there _is_ no classical explanation. The one given in the article reads like so much hand-waving to me. There _is_ no change in the linked flux. Nothing is "tilting". You need the Lorentz force. See the Feynman reference. -- D.keenan 11:03, 4 December 2006 (UTC)

Please see the "Sources of Confusion" section of homopolar motor and the "Common Misconceptions" section of line of force. -- D.keenan 11:09, 4 December 2006 (UTC)

[edit] Merge

I've removed the tag to merge to Homopolar motor as it's been sitting dormant for a year and no one seems to be interested in it. —Quarl (talk) 2007-02-26 00:48Z