Creating a Poleless Magnet Ring

[mrspeedybob]

All magnetic field lines of a permanent magnet must pass through the magnetic material, since the iron atoms are the source of the field lines, and therefore the lines must pass through the iron atoms. Any external field lines must therefore pass through the surface of the torus. There is no way for this to happen without violating symmetry. Thus, there can be no external field.

Mike

I think there is a serious flaw in your reasoning here. It is true that all field lines must pass through the material from which they originate but magnetic field lines add like vectors so two fields of opposite direction can pass through the boundary of the material at the same point and cancel each other out resulting in the illusion of a field that remains entirely outside the material. I made a picture to illustrate.

In this picture, if you were to measure magnetic flux at the surface of the bar you would read zero because all flux lines are canceled out by opposing flux lines. They have no effect, that doesn't mean they don't exist. There existence is demonstrated by the existence of the external field which doesn't appear to enter the material, but actually does.

 

[MS La Moreaux]

stevenb,

Regarding the bar magnet version of the disk generator, the output would have to be constant by symmetry.

In general, we seem to be coming from different directions and talking past each other. I realize that this frustrates you. It does me, too, and I am sorry. You have more education in the subject and bring in references to things that I have not even heard of. I, of course, cannot respond in kind, much as that would be desirable. I am just going by what is in my freshman physics and undergraduate electromagnetics textbooks. I have not done any experiments, to speak of. Basically, I only know what I read and have been taught. I rely upon what is in those textbooks for the most part, although I find inconsistencies in them. I take it that you disagree with my acquaintance, Professor Riles, who said that Faraday’s Law is an oversimplification for undergraduate students. If you do not disagree with him, then I cannot hope to argue with you at the higher level because of your superior knowledge and would have to defer to your way of thinking. If you do disagree with him, then I have to assume that my textbooks are correct, for the most part. My physics textbook states that Faraday’s Law only applies to circuits. The general closed path, whether conductive or not, is reserved for Maxwell’s Law for transformer emf. This would seem to obviate your elaborate arbitrary contours. In the case of the disk generator with a uniform magnetic field across the disk (the field lines orthogonal to the disk), consider making the disk nonconductive. You seem to be saying that there would still be the same emf according to Faraday’s Law. Is this so?

Mike

 

[MS La Moreaux]

mrspeedybob,

Oh, dear, I thought that this had been settled. Your reasoning is very clever and insightful, and I certainly did not consider the resultant vectors. The field lines of the iron atoms seem analogous to those around the turns of a toroidal electromagnet. In that case, the external magnetic field is severely attenuated by symmetry. I cannot find a flaw in your analysis, probably because I cannot visualize the field sufficiently, but since I know that the effect in the toroidal electromagnet is real, I must conclude that somehow your analysis is incorrect. Perhaps someone else can shed more light on this matter.

Mike

 

[stevenb]

Regarding the bar magnet version of the disk generator, the output would have to be constant by symmetry.

I agree with you. I would expect an experiment to give constant output. An experiment is always a good idea, especially in a case like this where it is so simple, but I would be surprised to see a different result.

Again, I would have to put more thought into this example to be sure of how to analyze this using Faraday's Law. One thing is clear. The answer you get with Faraday's Law (or by any other method, such as using Lorentz force directly) is path dependent. So there is no clearly defined EMF through Faraday's law, unless you clearly identify the path a charge would take. So one needs to think carefully about what it means to measure a voltage with meter, and define a "measured EMF" that would consider all paths via some formalism. I'm confident that the proper analysis over all paths will yield the constant value you are seeking. This is not a trivial problem to solve, but it is an interesting one. I plan to spend some time thinking about this. But, I must stress that this example does not invalidate Faraday's law, even if it does cause a person to pause and think, or even if it stumps a person completely. This issue is discussed by both Munley and Scanlon et al.

In general, we seem to be coming from different directions and talking past each other. I realize that this frustrates you. It does me, too, and I am sorry. You have more education in the subject and bring in references to things that I have not even heard of. I, of course, cannot respond in kind, much as that would be desirable. I am just going by what is in my freshman physics and undergraduate electromagnetics textbooks. I have not done any experiments, to speak of. Basically, I only know what I read and have been taught. I rely upon what is in those textbooks for the most part, although I find inconsistencies in them.

I find inconsistencies in textbooks also, and Scanlon et al, discuss this at great length. This just tells you that the subject is not simple. I can't stress enough how unintuitive Faraday's Law is, particularly when you deal with nonconservative fields. If you want to say that teachers need to do a better job, or that textbooks need to be improved, then I agree. The references I provided are well respected in the scientific community, so they would be a good place to base your understanding off of. That doesn't mean that they are necessarily unquestionable, and nothing in science is unquestionable anyway. However, I do disagree with prematurely forming the conclusion that accepted scientific laws (such as Faraday's) have exceptions, without adequate scientific proof using accepted scientific methods.

I take it that you disagree with my acquaintance, Professor Riles, who said that Faraday’s Law is an oversimplification for undergraduate students. If you do not disagree with him, then I cannot hope to argue with you at the higher level because of your superior knowledge and would have to defer to your way of thinking. If you do disagree with him, then I have to assume that my textbooks are correct, for the most part.

I don't disagree with anything that you quoted from Prof. Riles. I don't see anywhere in his quote where he say "Faraday's Law is an oversimplification for undergraduate students". I doubt very much that he would say this. I also did a search on him and read some of his comments at his website. I don't recall reading anything I would necessarily disagree with (at least not strongly). A couple of his comments are ambiguous to me (which may be my failing, not his), so I can't say whether I agree or disagree on those, but there is nothing in his statements that can be compared with your claims about Faraday's Law being false or having exceptions. Also, I noted at his website that he mention's Jackson as a respected graduate level text.

So, I don't think it's fair to lump Prof. Riles in with your ideas, based on hearsay. You can contact him and have him give a direct opinion here, but without that, I will assume his views are not as extreme as yours.

My physics textbook states that Faraday’s Law only applies to circuits. The general closed path, whether conductive or not, is reserved for Maxwell’s Law for transformer emf. This would seem to obviate your elaborate arbitrary contours. In the case of the disk generator with a uniform magnetic field across the disk (the field lines orthogonal to the disk), consider making the disk nonconductive. You seem to be saying that there would still be the same emf according to Faraday’s Law. Is this so?

To me, that group of sentences is just misconception layered on top of misinformation.

1. Faraday's law applied to circuits is called Kirchoff's Voltage Law (which is also mistreated in many texts, but let's not go there). Please provide the reference to your physics book, so we can identify a book that should be avoided.

2. Your statement about what general closed paths are "reserved" for is completely unfounded. What is this based on? You seem to just pull these rules out of the air.

3. What does it mean to say the disk is non-conductive? Where in your analysis does the conductivity of the disk come into play? Did you assume it is a perfect conductor? Did you assume that the conductivity was that of copper at room temperature? Or, did you assume it was steel, lead, tin, aluminum? Does non-conductive mean just a good insulator, or a perfect insulator? How do you then measure the voltage? Do you use a real volt meter, or a hypothetical perfect voltmeter with infinite internal resistance? How do you calculate the voltage from a voltage source with infinite source resistance using a meter with infinite load resistance? What if I take a real copper disk and short it out with a superconductor? What then is the output voltage?

Ultimately, what do you care what I think anyway? I must stress that all this sidesteps the big issue here. What I think or say, or what a textbook author thinks or says, or what you come up with in thought experiments is not the sum total of what the current status is of the scientific validity of Faraday's Law. There is a mountain of scientific evidence out there that takes priority over all of this. So, ask questions, raise objections, devise thought experiments, do real experiments and talk about this subject all you want, but hold back on stating "Faraday's Law is false", until you have a scientifically credible basis to state it. Neither one of us is in a position to make such a statement. However, I am in a position to state I am not aware of any known scientific evidence that contradicts Faraday's law. Even in studying it in the context of General Relativity, it is found to be correct. The one area where we might be able to question it is in the context of quantum field theory. This is an area that I've been trying to learn on my own, but it is most challenging and I can't say how Faraday's law holds up in this domain. This is a question I'm curious about myself and hope to answer (or have answered for me), someday.

 

[MagnetDave]

Mike,

I'll resist the urge to comment on your quixotic mission to knock over Faraday's Law - which is how I originally found the forum. I found this post because, well, it touches on my expertise, so I'll confine this to the original question:

Yes, you can make a toroidal magnet where the field is very nearly completely internal. If you envision slicing the ring into a succession of wedges, and magnetizing each wedge in the tangential direction, and then assembling them, this will produce your desired ring.

However - when you say "a magnet with no poles" you are revealing a fair amount of misunderstanding of magnets. This is more analogous to a snake eating its tail than a magnet with no poles. (In fact, we do this routinely to ship magnets safely, and the practice is called snaking.)

I strongly recommend you try your concept out experimentally. However, on my limited understanding of it, your zero result seems to be exactly what Faraday's Law would predict, so I'm not sure how it disproves it.

 

[MS La Moreaux]

stevenb,

My physics textbook is University Physics, Third Edition, by Francis Weston Sears and Mark W. Zemansky, Addison-Wesley Publishing Company, Inc., Copyright 1964, Second Printing - March 1965. It states, “Hence we can say: The induced electromotive force in a circuit equals the negative of the time rate of change of magnetic flux through the area bounded by the circuit. The preceding statement is known as the Faraday law.”

When Faraday’s Law is restricted to circuits it is bad enough, but when that restriction is removed, as you seem to imply that it is, it is even worse. It would seem that you would say that if the disk of the disk generator were removed, you could still use the same contour of Scanlon’s analysis and rotate it at the same rotational speed and get the same emf. If this is the case, then in contrast to what you said previously, Faraday’s Law does seem to state that the flux change produces the emf. (I am not quibbling over cause and effect but merely using language like “responsible for” and “produces” as shorthand for convenience.) I have to go back to the basics, again. Motional emf requires a conductor. Maxwell’s Law for transformer emf does not. Your version of Faraday’s Law seems to be a generalization of that Maxwell’s Law extended to closed paths with motion. But Maxwell’s Law requires an intrinsic flux change. The extended version does not. If there is no transformer emf and no motional emf, what is responsible for the emf? Transformer emf is a function of an electric field associated with the changing magnetic field. Motional emf and Scanlon’s contour are not associated with an electric field. The emf in motional emf is the result of magnetic force on the electrons in the conductor. What is the supposed emf of Scanlon’s contour due to? The question of measurement is valid, but only from a practical standpoint. Conceptually, an emf is real whether or not it can be measured practically. A stationary electron situated somewhere along the contour at some instant should experience a force. What kind of force would it be, electric or magnetic? If either, what would be the mechanism? Scientists do prefer mechanisms, where possible, rather than just results. Quantum mechanics is a notable exception. Often results are questioned when there is no apparent mechanism to produce them.

Mike

 

[MS La Moreaux]

MagnetDave,

I have already gone over how to make the magnet. That is not the problem.

I do not see a problem with the magnet having no poles. Poles involve a concentration of field lines emerging from the surface, and symmetry prevents this.

As I have already posted, I now realize that Faraday’s Law does not apply to my example because of the slip ring.

Mike

 

[stevenb]

My physics textbook states that Faraday’s Law only applies to circuits. The general closed path, whether conductive or not, is reserved for Maxwell’s Law for transformer emf.

It states, “Hence we can say: The induced electromotive force in a circuit equals the negative of the time rate of change of magnetic flux through the area bounded by the circuit. The preceding statement is known as the Faraday law.”

The above are your two statements about your physics book. The second statement indicates that your first statement is a misinterpretation (yet another among the many others). The book is not stating that Faraday's Law ONLY applies to (conductive) circuits. It is saying that Faraday's law CAN be applied to conductive circuits.

So, not only do you spread misinformation about Faraday's law, but you also misquote your teachers and your textbooks. Very sad indeed.

As I have already posted, I now realize that Faraday’s Law does not apply to my example because of the slip ring.

Yet again, complete nonsense is stated. First, your example proves Faraday's Law is false, then Faraday's law does not apply to your example at all. What sense can anyone make of your message? Hey, here's a thought (one that you are sure to ignore), maybe Faraday's law is just fine and does apply to your example, hence providing one more drop of water to the ocean that is the scientific evidence supporting Faraday's law.

Your other comments are so full of additional misinterpretations and misrepresentations and misinformation, that I can't even comment anymore, as we seem to be moving backwards.

I think MagnetDave has found the perfect adjective when he describes your quest as quixotic. I'm now realizing that my attempt to help you is equally quixotic. Hence, I will now bow out of this thread and give poor Rocinante a much needed rest.

 

[MS La Moreaux]

stevenb,

I believe it is you that is misinterpreting the textbooks, as well as my posts. I agree that we are not getting anywhere, so farewell.

Mike

 

[stevenb]

stevenb,

I believe it is you that is misinterpreting the textbooks, as well as my posts.

No, the situation is crystal clear to me.

But, thank you for grouping me on the side of Faraday, Jackson, Munley, Scanlon et al. Those are good terms to leave on. I'm also happy to be not too far from the viewpoints of Riles, Sears and Zamansky as clear from your quotes of them, as well as your obvious misinterpretations of those quotes.

Any statement pronouncing to the casual reader that I'm not your Sancho Panza is most welcome.

 

[mrspeedybob]

mrspeedybob,

Oh, dear, I thought that this had been settled. Your reasoning is very clever and insightful, and I certainly did not consider the resultant vectors. The field lines of the iron atoms seem analogous to those around the turns of a toroidal electromagnet. In that case, the external magnetic field is severely attenuated by symmetry. I cannot find a flaw in your analysis, probably because I cannot visualize the field sufficiently, but since I know that the effect in the toroidal electromagnet is real, I must conclude that somehow your analysis is incorrect. Perhaps someone else can shed more light on this matter.

Mike

The reason the field is significantly stronger within the material is that the fields produced from opposite sides of the material (or from opposite sides of the coil) add together at points in between them and partially cancel for points not in between them. They do not however completely cancel because one side of the coil or material is always closer then the other side. If you actually construct this magnet you will find that it does indeed have an external field,

 

[MS La Moreaux]

In analyzing Faraday's disk dynamo (the case with a uniform magnetic field through the disk), I have come up with the following. The force on a free electron in the disk is radial. Its path in the disk is therefore along a radius, relative to the disk. The drift speed of an electron is on the order of microns per second. Therefore, it will be dwarfed by the rotation speed of the disk and will be dragged along with the disk as the disk rotates. If we consider the drift path of a single electron as the circuit, it will be a very tight spiral of millions of turns between the axle and the rim, as viewed in space rather than relative to the disk. This spiral will be stationary in space, not rotating with the disk. Thus, there will be no change with time of the flux linking the circuit.

Mike

 

[MS La Moreaux]

mrspeedybob,

It has occurred to me that the external magnetic field of a ring magnet would only be due to the atoms at the surface and would thus be negligible. In the interior, the atoms would be like little bar magnets lined up head to tail. Thus, the flux would flow from one to another all the way around the ring, keeping their flux entirely confined to the material of the ring.

My physics textbook states that an infinitely long electromagnet would have no external magnetic field. If a finite electromagnet is bent around so that its ends are joined seamlessly, it would also have no external field.

Mike

 

[thehacker3]

mrspeedybob,

It has occurred to me that the external magnetic field of a ring magnet would only be due to the atoms at the surface and would thus be negligible. In the interior, the atoms would be like little bar magnets lined up head to tail. Thus, the flux would flow from one to another all the way around the ring, keeping their flux entirely confined to the material of the ring.

My physics textbook states that an infinitely long electromagnet would have no external magnetic field. If a finite electromagnet is bent around so that its ends are joined seamlessly, it would also have no external field.

Mike

But ring magnets DO have a decently strong magnetic field - so what gives?

 

[MS La Moreaux]

thehacker3,

Are you referring to the internal or external field?

Mike

 

[thehacker3]

thehacker3,

Are you referring to the internal or external field?

Mike

External

 

[MS La Moreaux]

thehacker3,

Where do you get the idea that ring magnets have a decently strong external field? Do you understand that the ring magnet is magnetized circularly so that there are no poles? Toroidal transformers are used because they have severely reduced external fields.

Mike