Talk:Faraday paradox

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.

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)

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)[reply]

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)[reply]

OK, thanks for getting back to me. I have asked the question at WikiProject Physics. --Heron 18:50, 19 December 2005 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

That's fine with me. --Heron 19:38, 20 December 2005 (UTC)[reply]

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)[reply]

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)[reply]

What is a "translating solenoid"? Occultations 20:43, 22 December 2005 (UTC)[reply]

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)[reply]

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)[reply]

The text written by Art Carlson (talk · contribs) has been removed with the comment "Evidence abounds that spinning a magnet over a copper turntable can cause it to drag in the same direction in attempt to minimize magnetic flux changes. Lenz' law in action!". However the new text states that "charge separation will be proportional to the relative rotational velocity between the disk and the magnet." That makes no sense to me because the experiment shows that rotating the magnet alone induces no current. Are there any sources for this claim? I believe that Art Carlsons original text was correct.

Edit: This seems to support that spinning a magnet alone does have some effect, but the explanation given actually strengthens Art Carlsons explanation, that the absolute rotation in space is crucial and not the relative rotation of the disc and the magnet. Nescio224 (talk) 20:00, 29 March 2019 (UTC)[reply]

You are responding to a discussion that went dormant 14 years ago. The article has changed a lot since then. It would be more useful for you to start a new topic regarding your views on the article as it is now.Constant314 (talk) 20:39, 29 March 2019 (UTC)[reply]

The section I'm talking about is "Configuration without a return path" and it still exists. The first half of that section has still the same exact sentences as in 2005. I see a discussion here from 2005 where the consensus was the text written by Art Carlson. Then in 2015 the user Kmarinas86 (talk · contribs) has rewritten the last half of that section and made it wrong in my opinion. I don't see any discussion about that change here at all. And this is the only major edit to that section since 2005. So why do you think the discussion not relevant anymore?

I didn't say it wasn't relevant. I said it would be useful to start a new topic. Perhaps I should have said section. Constant314 (talk) 03:47, 30 March 2019 (UTC)[reply]

Are you sure about the section? It only has three sentences and it doesn't appear to be incorrect.Constant314 (talk) 04:01, 30 March 2019 (UTC)[reply]

This sentence is incorrect: "This charge separation will be proportional to the relative rotational velocity between the disk and the magnet." Previously it was this: "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." It might not be entirely correct, because according to this the rotation of the magnet might also play a role because it induces a polarization inside the rotating magnet, which might also induce a polarization in the disk above the magnet. However in that case both effects would add. Each effect depends on the absolute rotations of the disk and the magnet, not on the relative rotation between disk and magnet. Nescio224 (talk) 06:13, 30 March 2019 (UTC)[reply]

That sounds good. Put it back if you want. However, I think that unless someone has a reliable reference that Faraday carried out such an experiment, then the entire section should be deleted. Constant314 (talk) 17:35, 30 March 2019 (UTC)[reply]

I have rewritten that section:

Configuration without a return path

A Faraday disk can also be operated with neither a galvanometer nor a 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, for example, through the electromotive force generated between the rim and the axle (though not necessarily easy). In this configuration, the polarisation is determined by the absolute rotation of both the disk and the magnet, that is, the rotation relative to an inertial frame. The relative rotation of the disk and the magnet plays no role, as a rotating system is not an inertial frame. There is no "relativity of rotation".^[1] When the magnet spins, the particles inside the magnet move through the magnets own magnetic field creating moving current loops. This results in the magnet becoming polarized radially toward or away from the magnet’s spin axis.^[2] Thus there is a net radial electric field component above and below the magnet when it spins. If disk and magnet are spinning together, the polarization of the disk can explained by the sum of these electric and magnetic forces.

Are you ok with that? Nescio224 (talk) 22:04, 1 April 2019 (UTC)[reply]

I think that you should leave the conductivity of the magnet out of the discussion. It is not important for understanding the paradox. But don't put a lot of work into it. I'm going to put a needs citation template on it, soon. Constant314 (talk) 02:14, 2 April 2019 (UTC)[reply]

I think this section is actually important fully understand the paradox. That is because it is only in the case without return path that you realize that the phenomenon can't be fully explained by considering only the relative rotations of the parts to each other. For example if you rotate the whole apparatus (including the return path), you don't measure a current, so you might come to the wrong conclusion that rotating the whole setup makes no difference, as nothing is roating relative to anything. However this is only because in that case the induced voltage in the disk is equal but opposite to that in the return path. The net effect of that is not zero, because it causes the whole apparatus to become polarized and a electric dipole field will develop. You can see how confusing this can be right here in the beginning of this thread. I was confused about this paradox too, so I read this article and it didn't help until I looked up more info about this case specifically.

Regarding the citations: I have given two sources in the text and here is a third one, which might be even better than the previous two. https://arxiv.org/pdf/0803.3616.pdf They all agree on this subject.

About the conductivity, I can cite the the third source: "An electromotive force arises even when the disk is made of a dielectric not supporting electric current."

To be honest, I think the whole article is not very well ordered and there are also more instances where the article is misleading. For example the section "Taking the return path into account" has this sentence "an EMF is generated whenever the disc moves relative to the return path, regardless of the rotation of the magnet", which is misleading/wrong, because an EMF is still generated when the whole apparatus is rotated, without relative movement of the disk relative to the return path. It's just such an EMF that doesn't generate a current. Also the rotating magnet will produce an EMF itself.

Anyways, I'm not going to fix the entire article. I feel like this discussion is contraproductive, because you don't give any factual arguments. Nescio224 (talk) 17:25, 2 April 2019 (UTC)[reply]

References

^ R.P.Feynman, "Feynman Lectures Vol. II 14-4", http://www.feynmanlectures.caltech.edu/II_14.html
^ George Russell Dixon, "Electromagnetism and Rotational Relativity", http://www.maxwellsociety.net/PhysicsCorner/CurrentLoopPolarization/ElectromagnetismAndRotationalRelativity.html

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)[reply]

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)[reply]

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)[reply]

In the Faraday reference the disk contacts are apparently not sliding but fixed. So in the experiments the disks are effectively fat wires and the induction law applies. The figures in Plate I drawn by Faraday are easily misinterpreted to describe full rotation when only short excursions were used. Some figures in Plate II actually indicate alternating rotation directions. In all experiments the magnetic field was highly localized to a small region within the disc. I believe there is no experimental evidence that a "Faraday disc dynamo" has ever worked.HCPotter (talk) 08:35, 26 October 2010 (UTC)[reply]

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

Removed Unsourced Original Research

The following text, which appears to be incorrect as well as unsourced was removed. If it turns out that the Mr. Cully was some historic author or critic in the early 1900's or something, I cannot find any referenece as such anywhere - so I have deleted the section as orignal research.

These three observations are explained by the fact that electrons have mass and are subject to inertial forces under rotation - behaving in a fluid-like manner, they are flung to the outer edge of the disc under rotation - in either direction of rotation, the centripetal acceleration acts outwards towards the edge of the disc, leading to a pressure (potential) difference between the axis of rotation and the edge of the disc. Therefore, when edge and axis are connected, a direct current will flow, always from edge to axis regardless of magnetic field orientation or direction of disc rotation. The magnet is not required for current production, but may add or subtract somewhat to the effect, though step 2 suggests otherwise. Possibly the main effect is eddy currents and consequent disc heating. Step 3 shows that relative motion between magnetic lines of flux and conductor are not necessary to induce current in the case of the Faraday disc. Test observation shows current registers solely by rotation of the disc. It is interesting that this so-called paradox is never represented with the observations of current and voltage of the rotating disc alone in the abscence of the magnetic field. Testing shows that both voltage and current increase with disc rotation speed in abscence of magnets, as expected from effect caused by centripetal acceleration acting on a mass. The electrons flung to the outer edge of the disc follow a circular path, and moving electrons create a magnetic field. The polarity of the magnetic field being dependant on the direction of rotation. Magnetic lines of flux are not 'imaginary', and they move with the magnet, and this can be observed by simple testing and observation, as in moving a magnet under a paper sprinkled with iron filings. The special theory of relativity has no value in explaining this so called paradox, which is not a mathematical and conceptual abstraction, but a real physical system - Christopher Culley

The question of field rotating with magnet

I'd just like to point out an inconsistency in the article. Under section "Using Lorentz Forces" a reference is given to the literature and the question (apparently unresolved) of whether or not the magnetic field rotates with a magnet. It notes that several methods have been proposed but none have been established experimentally (yet).

Later at least one of these (electrostatic) methods is described but no reference is provided to any actual experiments in the title "Configuration without a return path". It is asserted there that the field does not rotate with the magnet. No reference is given.

And we see after the title "Some Observations" the assertion that the field does not rotate.

And further down under the title "An Additional Rule" the assertion is made that the field does indeed rotate with a magnet.

Thus, the article includes three opinions: 1. The question hasn't been resolved. 2. The field does rotate. 3. The field does not rotate.

I would suggest that somehow (hopefully with references to actual experiments) this inconsistency within the article needs to be resolved. 69.213.184.155 (talk) 17:34, 20 May 2009 (UTC)[reply]

This is indeed an inconsistency in the article.

The formalism of the vector field does not capture the notion of rotating or non-rotating. If B(r) is independent of time at every point and axisymmetric then that is all there is to say. So as far as the vector field is concerned the question has been resolved, but the answer is neither 2 nor 3. If you want to make the question meaningful you have to add something else, such as "lines of force" that have a reality beyond their derivation as field lines. (For example, in ideal MHD, in a certain sense the fluid and the field are locked together, so-called "flux-freezing", and the distinction between rotating and non-rotating is meaningful because it is described by the fluid velocity. In this case each field line is associated with particular fluid elements and therefore retains its identity from one time to the next.)

It is the same if you ask whether a uniform magnetic field is "moving". If B = B₀ (a constant) everywhere then that fully specifies the field.

In short: there are no "field velocities" in Maxwell's equations. If you want to talk about motion of the field, you have to mean more by "field" than the vector B.

The issue is non-obvious because a moving bar magnet, for example, has a "moving field" in a different sense, namely that its time-variation at any point is entirely determined by the time-variation of the position of the magnet, because in the rest frame of the magnet the field is time-independent. It is the "shape" of the field that moves, like the shape of a wave on the sea. There is no "shape" to move in the case of a uniform field, or of an axisymmetric field under rotation. 82.10.111.112 (talk) 21:31, 28 August 2009 (UTC)[reply]

If the magnet is rotated, the magnetic field rotates with it. This is the conclusion of Kelly's paper ("Faraday's final riddle", listed under "Further reading" at the end of the article). To say that the formalism of vector field does not capture the notion of rotating or non-rotating is beside the point. The vector field description of the electromagnetic field requires a reference frame. If one takes a reference frame in which the magnet is stationary (that is, either both the magnet and the reference frame are stationary, or both the magnet and the reference frame are rotating together) then the electromagnetic field is a pure magnetic field and zero electric field. If one takes a reference frame which rotates with respect to the magnet (for example, the reference frame is stationary and the magnet is rotating), then the electromagnetic field also has an electric field component, in addition to a (slightly changed) magnetic field. This follows from the transformation laws of the electromagnetic field between different reference frames (see electromagnetic tensor). So the explanation given by Faraday in the section "Faraday's explanation" in the article is wrong in case the magnet is rotating.

Faraday's paradox can be explained in three different ways: 1) by the laws of electromagnetic induction, 2) by the Lorentz forces acting on electrons in the moving conductors, or 3) by the electric forces present in a reference frame that rotates with respect to the magnet. From a theoretical viewpoint, these three explanations are the same. As Kelly in the quoted article points out, for this the whole circuit needs to be taken into account, since the electromagnetic forces act on the whole circuit, not just the part of the circuit in the disk. For example, if the rest of the circuit is stationary with respect to the disk (that is, if both the disk and the circuit are rotating and the magnet is stationary, or if only the magnet is rotating), there is no current, since the electromagnetic forces acting on the disk are canceled by the electromagnetic forces acting on the rest of the circuit. Mateat (talk) 10:27, 26 October 2009 (UTC)[reply]

The average speed of electrons in metals is very slow. For example, if a copper wire of one millimeter radius carries a one Ampere current, then the electrons move 84 millimeters in an hour, on average (see here). So, for all practical purposes, when considering the electromotive force caused by the Lorentz force on the electrons, one may assume that the electrons move as fast as the copper disk. Mateat (talk) 17:32, 26 October 2009 (UTC)[reply]

Let me give a few new observations. Let me assume a perfectly conducting disk for simplicity, so we don't have to worry about electron motion (an infinite conductivity implies a zero electron velocity, for a finite current density). When the disk is rotating and the magnet is stationary, then the electrons inside the disk feel a radial force vector due to the magnetic field that is given by q (v x B). However, there will also be a radial electric field inside the disk, given by E = - v x B. (This is a general expression for the electric field that is inside any perfect conductor that is moving with a velocity v with respect to a magnetic field.) The total force on the electrons (due to the electric and the magnetic fields) will be zero, as it must be since the disk is perfectly conducting. The Thevenin (open-circuit) voltage of the generator can be easily calculated by applying Faraday's law to a closed path that goes through the output terminals, the leads, and a radial path inside the disk. This shows that the Thevenin output voltage is that obtained by integrating the radial electric field along a radial path between the inner and outer radii of the disk, and this leads to a simple formula for the Thevenin voltage (result omitted). Next, let us assume that the disk is stationary and the magnet is rotating. Again, there is no force on the electrons inside the disk since it is perfectly conducting. Since the velocity v of the electrons is now zero (with respect to the laboratory frame of reference), there is no force on the electrons due to the magnetic field that is in the laboratory frame of reference (which is approximately the same as the magnetic field that is in the magnet's frame of reference). Since the total force on the electrons is zero, it now follows then the electric field inside the disk is exactly zero. Therefore, an application of Faraday's law will show that the Thevenin voltage of the generator is exactly zero. The Thevenin resistance of the generator can be found in terms of the disk conductivity, radii, and thickness (details omitted). Elee1l5 (talk) 19:25, 13 November 2009 (UTC)[reply]

paradox

I didn't really studied the paradox, so I ;ight be wrong, but a situation that looks close to this paradox is easily solved with special relativity, it is about the magnetic lines moving with the magnet. For example a wire perpendicular to an (almost) infinite wire with current, that moves in a parallel motion in respect of the wire, experiences a lorentz force. But if the infinite wire moves with the same velocity, there exists a restframe were nothing moves. The problem is solved by applying lorentz tranformation to the quadri vector potential, or the maxwell tensor, hence there appears an electric field that exactly cancel the induction. Electromagnetism is stupid, you have the fields, you apply the lawsKlinfran (talk) 11:26, 31 May 2010 (UTC)[reply]

There is no induction at all

In my opinion, Faraday's disc has nothing to do with electromagnetic induction at all. From the view of the laboratory system, the magnetic field is constant in time $(\partial {\vec {B}}/\partial t=0)$ . Thus the induction law ${\text{curl}}~{\vec {E}}=-\partial {\vec {B}}/\partial t$ equals zeros at all places and all times, so that the electrical field is a conservative field with some charges being separated in space by the Lorentz force. The flux rule

\oint _{C}{{\vec {E}}'}{\text{d}}{\vec {s}}\approx -\int _{A}{\frac {\partial {\vec {B}}}{\partial t}}{\text{d}}{\vec {A}}+\oint _{C}{\vec {v}}\times {\vec {B}}\ {\text{d}}{\vec {s}}=-{\frac {\text{d}}{{\text{d}}t}}\int _{A}{\vec {B}}\cdot {\text{d}}{\vec {A}}

that is commonly used to explain why there is a rotational electric field, has no explanatory power concerning whether there is induction or not, because E' is measured by an observer in the (moving) system of the path ${\text{d}}{\vec {s}}$ (that means: not in one common reference system). Furthermore, it is only an approximation for non-relativistic (that means: small) velocities.

It is better to describe all variables from a common reference system (e. g. the laboratory system) and use the "true" induction law

\oint _{C}{\vec {E}}\ {\text{d}}{\vec {s}}=-\int _{A}{\frac {\partial {\vec {B}}}{\partial t}}\ {\text{d}}{\vec {A}}

observed in an arbitrary but common inertial system. Note that the "true" induction law does not consider any change of the borders of contour C.

By using the Lorentz transformation for the electromagnetical field (shown here), it is possible to change from one system to another. Doing this, the integration path can easily be completed by a direct line between the two electric contacts (that means by a straight unmoved line through the disc). The electric field ${\vec {E}}'=0$ of the ideally conducting disc (measured in he moved system) will transform to an electric field ${\vec {E}}\approx {\vec {E}}'-{\vec {v}}\times {\vec {B}}=-{\vec {v}}\times {\vec {B}}$ measured in the laboratory system. By doing this, it can be shown that the voltage displayed by the voltmeter is the same voltage that is comprised in the rotating disc. This shows, again, that the field is not rotational, and there is no induction. Michael Lenz (talk) 00:04, 28 May 2011 (UTC)[reply]

Moved from article, needs rewrite for tone and references

--Wtshymanski (talk) 18:37, 8 July 2011 (UTC) Correction Plus Simple Explanation for the non-technical person. [My focus is on an accurate statement of the facts, please, feel free to improve the form of this presentatin.][reply]

Faraday's Paradox

Faraday's law of induction states: "The induced electromotive force (EMF) in any closed circuit is equal to the time rate of change of the magnetic flux through the circuit." A homopolar generator is given the name of Faraday's Paradox because it violates Faraday's law of induction, in that there is no change of flux within the enclosed loop of the electrical circuit.

The reason why the homopolar generator can violate Faraday's law of induction is that as stated, the law is not a precisely accurate description of physical reality. To be precise, Faraday's law of induction should state: The net induced electromotive force (EMF) in any closed circuit is equal to the time rate of change of the magnetic flux through the circuit.

The subtle difference is this: If a closed loop, with its plane perpendicular to a uniform magnetic field, is moved parallel to its plane, then equal and opposing voltages are induced in opposite legs of that loop. Since these voltages are connected in series opposition, there is no net voltage to be measured at the ends of the loop. But voltages are actually induced in the various sections of the loop!

A more important practical issue is that certain measured output voltages of a homopolar generator are counter intuitive when the original experimental set-up is used. The entire problem exists because the original experimental set-up reinforces the experimenter's error of focusing attention on only the conducting disk and the magnet to the exclusion of recognizing what occurs between the magnetic field and the return wire. Replacing this asymmetric configuration of the experiment with a set-up that focuses the experimenter's attention on the symmetry of the circuit immediately shows that the observed measurements are easily and fully explained by the known laws of electronic circuits.

Replace the return wire by a second conducting disk parallel to the first disk with a bar magnet in between the disks as in the original lossy experiment. Or use a concentric cylindrical magnet to eliminate the loop back currents. Using brushes, connect a voltmeter between the rims of the two disks.

The previously confusing measured results become obvious and are easily and simply explained by the fundamental laws of electronic circuits.

Case 1: Rotate one disk with the second disk and the magnet stationary. The relative motion between the stationary magnet [and magnetic field] and the rotating disk induces a voltage in the disk, which is measured by the voltmeter.

Case 2: Rotate one disk and the magnet together with the second disk stationary. There is no relative motion between the rotating magnet and the rotating disk and hence no voltage is induced in that disk. There is a relative motion between the rotating magnet [and the magnetic field] and the stationary disk, and hence a voltage is induced in the stationary disk. If the direction of rotation is the same as in Case 1, the relative motion of the stationary disk is in the opposite direction, and the induced voltage is of the opposite polarity. This is measured by the voltmeter as the same value and polarity of voltage as in Case 1 above.

Case 3: Both disks are stationary and the magnet is rotated. The relative motion between the stationary disks and the rotating magnet [and magnetic field] is the same for both disks so that the same magnitude and polarity of voltage is induced in both disks. Because these voltages are connected in series opposition, the net voltage measured by the voltmeter is 0 V.

Contrary to many claims on the internet, there is no violation of the laws of electrical circuits and physics.

Faraday paradox . If a large amount of mass is under the sea a hill can develop on the water's horizon . A water surfer on top of this hill can not surf down as from a gravity point of view the hill is flat . If a magnetic field were to develop a voltage difference in a copper plate under static conditions , nothing moving , then the voltage difference can not lead to current flow in the same way the water surfer can not surf down the gravity hill . A hall effect transistor takes advantage of this to detect the strength of a magnetic field . The reason I bring this up is to liberate those who would have concerns with the conservation of energy if a static magnetic field were to produce a voltage difference in a copper disk . Energy is conserved as current can not flow from a voltage difference created in a copper dist under static conditions in the same way a surfer can not surf down a gravity hill .

John H — Preceding unsigned comment added by 65.95.228.34 (talk) 19:41, 28 December 2013 (UTC)[reply]

Vanja Janežič

This appears to be a private individual's web site. While I have no reason to doubt its veracity, it does not constitute a reliable source.Constant314 (talk) 02:01, 4 June 2014 (UTC)[reply]

Inapplicability of Faraday's law

This describes an elegant experiment. But it is not one of Feynman's examples. If it is based on Feynman's example then it is WP:SYN. A reliable source is needed.Constant314 (talk) 02:04, 4 June 2014 (UTC)[reply]

An additional rule

Good picture, but you do not need the math. When the switch is flipped, one circuit is destroyed and another created. During the time that either circuit exists, the flux enclosed by the circuit does not change. Hence Faraday's law would predict 0 EMF. Constant314 (talk) 04:15, 4 June 2014 (UTC)[reply]

Why is this paradoxical?

This illustrates nicely why "should" should be avoided in encyclopedic articles.

Does "rotating the disc relative to the magnet, whether by rotating the magnet or the disc, should produce an EMF"

mean that the experiment should produce EMF or that Faraday's law should predict an EMF?

Anyway, this paradox is easily explained by the magnetic field having two ways to produce an EMF:

1, Faraday's Law

2. v x B

Constant314 (talk) 04:34, 4 June 2014 (UTC)[reply]

“

Although Faraday's law does not apply to all situations, the Maxwell–Faraday equation and Lorentz force law are always correct and can always be used directly.<ref name=Feynman>"The flux rule" is the terminology that Feynman uses to refer to the law relating magnetic flux to EMF.{{cite book|author=Richard Phillips Feynman, Leighton R B & Sands M L|title=The Feynman Lectures on Physics|year=2006|page=Vol. II, pp. 17-2|publisher=Pearson/Addison-Wesley|location=San Francisco|url=http://books.google.com/?id=zUt7AAAACAAJ&dq=intitle:Feynman+intitle:Lectures+intitle:on+intitle:Physics|isbn=0-8053-9049-9|nopp=true}}</ref>

”

From the article: Faraday's law of induction. F.Y.I.^talk2siNkarma86—Expert Sectioneer of Wikipedia 19:45, 6 December 2015 (UTC)[reply]

Using the relativistically-correct electric field

This new section seems to probably violate Wikipedia rules on own work WP:OR and synthesis WP:SYN. It also reaches the non-sensible conclusion that "the induced electric field cannot be described by the Maxwell–Faraday equation". Constant314 (talk) 03:48, 6 December 2015 (UTC)[reply]

The Maxwell–Faraday equation is zero-valued when the electric field E is curl-free and the magnetic flux density B is constant.^talk2siNkarma86—Expert Sectioneer of Wikipedia 18:00, 6 December 2015 (UTC)[reply]

Perhaps you can elaborate. I am presuming you are discussing the case where there is zero actual EMF which would agree completely with a zero-value E field.Constant314 (talk) 18:28, 7 December 2015 (UTC)[reply]

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Indention of subsections needs correction.

After reading through the article, it is clear that sections 3, 4, and 5 are actually subsections of section 2. I'm going to fix that. It will be ugly, but it will actually reflect the structure of the article. Constant314 (talk) 01:59, 7 February 2019 (UTC)[reply]

It has been done. Constant314 (talk) 02:04, 7 February 2019 (UTC)[reply]

"When the magnet is rotating, but flux lines are stationary, and the conductor is stationary" needs citation

If you read this section carefully, it states that the actual effect of a rotating magnet is different from the classical prediction. That needs a citation. Everything after the first paragraph appears to be WP:OR. If no citation is forthcoming, I will remove everything but the first paragraph.Constant314 (talk) 17:57, 29 March 2019 (UTC)[reply]

I agree with that. These paragraphs seem dubious to me anyways. For example the part about flipping the magnet. It is clear that the flip of the field lines can't occur instantly at a distance, because the change in the magnetic field can only travel at light speed. It is unclear to my why the author has to consider the movement of the nuclei or the lorentz contraction to explain that. The author also claims that "the induced electric field cannot be described by the Maxwell–Faraday equation", however the maxwell equations are all lorentz invariant and should be coherent with any relativistic explanation. Nescio224 (talk) 21:27, 29 March 2019 (UTC)[reply]

In the special case of a disc magnet rotating on a fixed axis of symmetry, the magnet 1) generates an electric field but 2) no changing magnetic field. This is because of two things, respectively: 1) magnetic polarization acquires an electric polarization under Lorentz transformation, and 2) the rotation axis of the magnet is the same as its axis of cylindrical symmetry (which, by the way, is not an example of "flipping" or swapping the poles of a magnet).

The Maxwell-Faraday Equation is the solution for the value of the electric fields' curl in terms of a changing magnetic field. In this case, that equation returns a value of zero because the magnetic field generated is static. This equation, which returns a value of zero here, does not "describe" the electric field at all as the electric field is non-zero, and that was the basis for saying "the induced electric field cannot be described by the Maxwell–Faraday equation" which should have been written "the induced electric field cannot be derived by the Maxwell–Faraday equation".

Also, if the magnetic field lines literally rotated with the magnet, then charges would appear in vacuum, which is clearly not what happens, and that would not be relativistically-correct anyway.

So it is wrong to think of the magnetic field lines as rotating even though an electric field is produced. And even though an electric field is produced, that electric field is conservative and so cannot explain the EMF for our closed current path. Thus, as we are left with the remaining option that the EMF is due to the magnetic force, relative motion of conductors is necessary to explain the EMF.

In the case of a magnet rotating end-over-end, the magnetic field changes, so the electric field has a rotational component, and in addition to that it has an irrotational component which is due to an electric polarization arising due to Lorentz transformation of the magnetic polarization. Lorentz symmetry is valid in a classical model, so this result does not lie outside the domain of classical physics even when classical physics itself is not generally valid. The classical result of the Lorentz transform of each moving source dipole comprising the magnet is, at the experimental scale of the Faraday Paradox, no different than the one from quantum mechanics.

Sincerely, ^talk2siNkarma86—Expert Sectioneer of Wikipedia 12:34, 25 May 2019 (UTC)[reply]

"An additional rule", unnecessary math

This section starts with the equation

\mathbf {F} _{21}={\frac {\mu _{0}}{4\pi }}I_{1}I_{2}\oint _{C_{1}}\oint _{C_{2}}{\frac {d\mathbf {l_{1}} \ \mathbf {\times } \ (d\mathbf {l_{2}} \ \mathbf {\times } \ {\hat {\mathbf {r} }}_{21})}{r_{21}^{2}}}\

However, for the apparatus shown, both currents are zero so this equation reduces to zero. The entire derivation is unnecessary.Constant314 (talk) 22:00, 30 March 2019 (UTC)[reply]

Sliding contact

The sliding contact at the rim of the disc is a "third party". Actually there are 8 configurations.

nothing rotates
only the disc rotates
only the magnet rotates
only the contact rotates
only the disc rests
only the magnet rests
only the contact rests
all three rotate

What will happen? At first sight I guess:

1 and 8 are equivalent, no emk

as are

2 and 5, induced emk

3 and 6, no emk

4 and 7, induced emk

Madyno (talk) 21:55, 10 July 2022 (UTC)[reply]

[Feynman-1] R.P.Feynman, "Feynman Lectures Vol. II 14-4", http://www.feynmanlectures.caltech.edu/II_14.html

[Dixon-2] George Russell Dixon, "Electromagnetism and Rotational Relativity", http://www.maxwellsociety.net/PhysicsCorner/CurrentLoopPolarization/ElectromagnetismAndRotationalRelativity.html

[1]

[2]