Can a magnet's magnetic field perform work on another magnet?

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The discussion centers on whether a magnet can perform work on another magnet, with participants debating the implications of classical electromagnetism and the Lorentz force. It is established that while a magnetic field does not do work on free charges, it can influence the atomic structure of magnets, indirectly affecting energy transfer. The conversation highlights the complexity of magnetic interactions, noting that classical electromagnetic laws may not fully account for the behavior of permanent magnets under certain conditions. Participants argue about the adequacy of existing theories and the need for rigorous proofs to support their claims. Ultimately, the debate reflects ongoing challenges in reconciling classical electromagnetism with quantum mechanics in understanding magnetic work.
  • #121
DaleSpam said:
Yes, of course it ignores the role of quantization. Classical EM is a non-quantum theory. But it isn't crucial here, permanent magnets are accurately modeled by classical EM simply by using an empirically determined constituitive relationship between B and H, as described above and in Jackson and any other EM textbook.
So you keep saying. But see below.
OK, I am glad to learn that I was in error with my mistaken belief that you did not accept the equations of classical EM.
Well I'm formally obliged to accept that as a retraction. Fine - even; thank you. But, being suspicious minded I can't help but see a trap of sorts...
Since you do accept classical EM and since classical EM logically implies that the work done is given by E.j then we must logically be in agreement.
Nice try. If this 'inextricable linkage' is valid, here's my request to you. Consider the following simple scenario. We have two magnetic dipoles of equal moment m and orientation along x axis. One, labelled mc, is a classical, perfectly conducting loop current. The other, labelled mq, is an electron intrinsic magnetic moment. Assume for both some effective enclosed area a and mean circulating current I such that |mc|, |mq| = Ia. Somewhat iffy in the intrinsic moment case, but we have to somehow allow the conception of a possible E.j style work being done on it. A uniform ramping B field Bx = xB0*kt is applied (x the unit vector along x), such that both mc and mq are nominally subjected to the same emf ζ = a(-dB/dt) = -aB0k.

You must know from my frequently stated position what will be the contrasting behavior. mc will act as a shorted perfect inductor - a ramp current I' = I0'k't will be induced such that the enclosed flux ∅ = aBav is invariant wrt time t - diamagnetic behavior. This is consistent with work being done on mc - there is a continual building up of a magnetic field energy owing to the ramp I'. Hence mc' = mc(1+I'a/m) is time varying. I maintain zero E.j style work is done on mq - there is no equivalent ramp current to I' in the classical loop current. So mq is time invariant.

Your position logically insists E.j style work must be done on mq. Explain please - really explain how.
 
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  • #122
Q-reeus said:
Nice try. If this 'inextricable linkage' is valid, here's my request to you. Consider the following simple scenario. ...
The inextricable linkage is valid. Please see:
http://farside.ph.utexas.edu/teaching/em/lectures/node89.html
or (for a treatment involving the "macroscopic" quantities)
http://www.phy.duke.edu/~rgb/Class/Electrodynamics/Electrodynamics/node33.html

If you believe that it is not valid despite the proofs then it is up to you to provide good evidence supporting your belief. It is not up to me to chase down every unsubstantiated claim you can manufacture. My opinions are based on the best evidence I have available, if you have some additional evidence I have overlooked, then provide it.
 
  • #123
Q-reeus, you are picturing an electron as an actual ball of something. Perhaps, an infinitely small one. But it's not. Regardless of how you decided to localize an electron, by either considering an electron around a nucleus, or placing it in a cavity, it is a distributed particle. And that distribution will have angular momentum and corresponding magnetic moment and current separate from intrinsic moment of an electron. When you start increasing an applied magnetic field, as you describe above, the electron's state will change, resulting in a current. Simplest case here is hydrogen atom. If you apply an increasing B field, you can cause the electron to transition to higher energy levels. These will have an overall magnetic moment. So in the end, you have exactly the same time-varying magnetic moment. It simply happens to increase in discrete jumps, because, well, Quantum Mechanics. If you do 3D particle in a box, you'll see same behavior. Though, it's a bit more difficult to describe mathematically due to symmetries.

Ultimately, any elementary dipole should be viewed as a particle field. You don't have just the spinor. You also have a spatial distribution of some sort. And whenever any work is being done on your elementary dipole, it will always be done via that spatial distribution. All a bare spinor can do by itself is precess in a magnetic field. No energy is transferred in spinor precession. So any work being done on an elementary dipole is still going to be E.j work done on the particle field.
 
  • #124
K^2 said:
Q-reeus, you are picturing an electron as an actual ball of something. Perhaps, an infinitely small one. But it's not. Regardless of how you decided to localize an electron, by either considering an electron around a nucleus, or placing it in a cavity, it is a distributed particle. And that distribution will have angular momentum and corresponding magnetic moment and current separate from intrinsic moment of an electron.
Am I right in thinking you are referring to orbital contribution here? I recognize such does contribute, but in ferromagnetic media only to a relatively quite small degree. If you read back through my postings, I make it clear orbital contribution can be quite significant in ferrites. But there orbital contribution response to an applied B is strongly ferromagnetic in character, far from the diamagnetism expected if classical E.j work is done on a lossless, classical circulating current. if you meant 'dressed' vs 'bare' electron contribution to intrinsic spin, I'm not sure how this translates into explaining the known fact of magnetic saturation - implying rather strongly intrinsic moments are the chief contribution to ferromagnetism. Please elaborate.
When you start increasing an applied magnetic field, as you describe above, the electron's state will change, resulting in a current. Simplest case here is hydrogen atom. If you apply an increasing B field, you can cause the electron to transition to higher energy levels. These will have an overall magnetic moment.
Don't question that transitioning will occur, but to what extent in response to how great an applied B?
So in the end, you have exactly the same time-varying magnetic moment.
Not sure what you mean by that - magnetic susceptibility of hydrogen - say as liquid, is tiny and iirc right, diamagnetic. [oops - that refers to diatomic hydrogen, and you meant atomic. This sounds like spin-orbit coupling which gets tricky.]
It simply happens to increase in discrete jumps, because, well, Quantum Mechanics.
Which if you follow my #45, is a key argument imo against the notion that literal E.j work can account for energy changes, or rather, be equated to the negative of net magnetic field energy change. Pick an interval 'between jumps' and where can E.j work be invoked?
Ultimately, any elementary dipole should be viewed as a particle field. You don't have just the spinor. You also have a spatial distribution of some sort. And whenever any work is being done on your elementary dipole, it will always be done via that spatial distribution. All a bare spinor can do by itself is precess in a magnetic field. No energy is transferred in spinor precession. So any work being done on an elementary dipole is still going to be E.j work done on the particle field.
Doubtless you are far more clued up on QM here than my rudimentary grasp. Just basically, what is the net result then on the intrinsic moment that can be equated to E.j work done? Doesn't it get back to overall susceptibilities? Huge in ferromagnetic media, tiny and of the opposite sign in most other materials. As mentioned above, what I have read suggests intrinsic moment is overwhelmingly the dominant source contribution in ferromagnetic media. Classical loop currents, for which E.j follows readily enough, always predict weak diamagnetism, and it seems absurd to suppose one could ever construct a permanent magnet from such in the first place. But then, I do have much to learn, so fire away please.
 
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  • #125
harrylin said:
This has been discussed before - but perhaps not sufficiently explicit and elaborate?

The kinetic energy of the rocket is not supplied by you pushing the button.
In contrast, earlier we found that the kinetic energy of the magnets must be supplied by their magnetic fields; there is nowhere else that the energy could come from. Any induced currents are induced by those magnetic fields.

My point was that the concept of indirect work was, if not completely absurd, certainly useless.

As another example, consider pushing on an accelerating train car, perpendicular to its motion. Unless you have superhuman strength, the train car is not going to move off the tracks, but will continue along as if you weren't there. You are still transferring energy to the train car by pushing on it, but you are certainly not doing any work on it since the force you apply is perpendicular to its motion.

To me, it is useless to say "the magnetic field does work indirectly".

Much better is to simply say that the magnetic field does no work since it is everywhere perpendicular to the force it applies on a charge or current distribution. The concept of work is well defined. The concept of "indirect work" is not.
 
  • #126
gabbagabbahey said:
My point was that the concept of indirect work was, if not completely absurd, certainly useless.

As another example, consider pushing on an accelerating train car, perpendicular to its motion. Unless you have superhuman strength, the train car is not going to move off the tracks, but will continue along as if you weren't there. You are still transferring energy to the train car by pushing on it, but you are certainly not doing any work on it since the force you apply is perpendicular to its motion.

To me, it is useless to say "the magnetic field does work indirectly".

Much better is to simply say that the magnetic field does no work since it is everywhere perpendicular to the force it applies on a charge or current distribution. The concept of work is well defined. The concept of "indirect work" is not.

But in the OP case the magnetic field is acting directly in line with the motion, upwards. E fields are induced once motion is realized, but said E forces do not act in the direction of the motion, but normal to it. Your train car analogy is exactly what I've been trying to illustrate, but with limited success. Since a force acts along the motion, it can be said to have done work. When a force acts normal to motion, it does not do work.

So all we have to do is, hold on, are you ready:

draw a picture!

Sorry to be so repetitive, but until that happens, we will argue in vain. Just so you & I are clear on this, I totally agree with your train car explanation. It is indeed the force acting with the motion that does work. Any force(s) acting normal to motion do no work. We agree at least on that point.

Now if we can only establish the direction of the various force vectors. I drew pics in the motor thread, & my opponents dismissed them w/o offering their own pics. I see little point in submitting pics since nobody will change their mind. Tonight I may be tempted to when I get home. In the meantime, if anybody can make a sketch, I would greatly appreciate it.

Claude
 
  • #127
cabraham said:
But in the OP case the magnetic field is acting directly in line with the motion, upwards.

The magnetic field is parallel to the motion. The magnetic force applied is perpendicular, according to the Lorentz Force Law.
 
  • #128
gabbagabbahey said:
The magnetic field is parallel to the motion. The magnetic force applied is perpendicular, according to the Lorentz Force Law.

Can you please illustrate? The lines of force around the magnet bend as they enter/exit the poles. Also, to compute the Lorentz force, Fm, per uXB, can you describe the orientation of u & B? Thanks.

Claude
 
  • #129
Q-reeus said:
Doubtless you are far more clued up on QM here than my rudimentary grasp. Just basically, what is the net result then on the intrinsic moment that can be equated to E.j work done?
None. When we are talking about one magnet interacting with another, no work is being done on the intrinsic moment. Like I said earlier, all a spinor can do is precess, and there is no work involved in that. The B.m angle remains constant. All of the work is done via actual currents. Yes, these don't contribute much to the magnetic field of a ferromagnetic in a steady state, but that's where all of the work will be done. Induced currents.

Of course, when you look at macroscopic picture, you can't really tell one from the other. So you can still describe it classically as classical magnetic moments with current loops.

So while you are absolutely right that an elementary dipole of a fermion cannot be adequately described as a classical current loop, it doesn't really lead to any violations. B field still doesn't do any work.
 
  • #130
K^2 said:
B field still doesn't do any work.

So you agree the point that the B field creates/generates an electrical field that would do the work?
 
  • #131
Miyz said:
So you agree the point that the B field creates/generates an electrical field that would do the work?
As one possibility. In principle, work could be done entirely by an external E field, with B field merely assisting. Something along the lines of Hall Effect, for example, where there is no induced E field, yet amount of work done depends on whether the B field is present.
 
  • #132
K^2 said:
As one possibility. In principle, work could be done entirely by an external E field, with B field merely assisting. Something along the lines of Hall Effect, for example, where there is no induced E field, yet amount of work done depends on whether the B field is present.

Im not sure what you're saying... Is that a yes or a no :-p. I am guessing that its a maybe.
 
  • #133
I'm saying that B field doesn't have to actually induce an electric field for work to be done in the system. But work will always be done by electric field, and never by the magnetic field. So an E field must be present in order for work to be done. Whether B field is the source of that E field is a different question.
 
  • #134
K^2 said:
I'm saying that B field doesn't have to actually induce an electric field for work to be done in the system. But work will always be done by electric field, and never by the magnetic field. So an E field must be present in order for work to be done. Whether B field is the source of that E field is a different question.

That makes no sense at all. If B cannot do work, only E can do work, then how can you say that B field does not have to induce an E field for work to be done in a system? Your 2 statements are contradictory.

When you say that "work can never be done by the magnetic field", that is the OP question under scrutiny. Why does B field never do work? Please tell me the reason, because the OP wants to know if & why, & so do I. Again, would you draw a picture showing the force that lifts the magnet? I don't believe the force lifting the magnet is Fe (electric component of Lorentz force), but rather Fm (magnetic component of Lorentz force). It is Fm that lifts the magnet.

Claude
 
  • #135
Power (work per unit time) is given by F.v. The only force in electrodynamics is the Lorentz Force, F=q(E + v x B). So dW/dt = q(E + v x B).v = q E.v + q(v x B).v = E.j + q B.(v x v) = E.j

The term containing B field cancels, because any vector crossed with itself is zero. (I use property a.(b x c) = b.(c x a) = c.(a x b) )

cabraham said:
That makes no sense at all. If B cannot do work, only E can do work, then how can you say that B field does not have to induce an E field for work to be done in a system? Your 2 statements are contradictory.
E that does work can be external. Again, think of Hall Effect as an example. Work is done by E field, but the amount of work done can depend on the B field.
 
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  • #136
K^2 said:
Power (work per unit time) is given by F.v. The only force in electrodynamics is the Lorentz Force, F=q(E + v x B). So dW/dt = q(E + v x B).v = q E.v + q(v x B).v = E.j + q B.(v x v) = E.j

The term containing B field cancels, because any vector crossed with itself is zero. (I use property a.(b x c) = b.(c x a) = c.(a x b) )


E that does work can be external. Again, think of Hall Effect as an example. Work is done by E field, but the amount of work done can depend on the B field.

But I said that the force doing work is Fm, not B. B does indeed cancel, but vXB does not. It should be well understood that B acting in the direction of motion will indeed do no work since the Fm vector points normal to B. Once again, are we defining work as force integrated over path? If so, it is universally known that B itself is not acting along the path of motion, but rather Fm is. B will drop out of the above equation, but Fm will not.

When I say that "B does the work", what I infer is that the force acting in the direction of motion is Fm = qvXB. Then as we know, B itself is normal to the motion. The Fm term does not drop out AFAIK. Am I mistaken? Also, nobody answered my previous question.

Which force lifts the magnet? It has an upward direction. Is it Fe = qE? Is it Fm = qvXB. If it is the former, where does E originate? Thanks in advance.

Claude
 
  • #137
cabraham said:
But I said that the force doing work is Fm, not B.[..]
1. I find the induced current explanation suspect: a pair of magnets doesn't warm up noticeably and doesn't act like a shock damper.

2. At first sight Wikipedia agrees with you:
"In the physically correct Ampère model, there is also a force on a magnetic dipole due to a non-uniform magnetic field, but this is due to Lorentz forces on the current loop that makes up the magnetic dipole."
- https://en.wikipedia.org/wiki/Force_between_magnets

Regards,
Harald
 
  • #138
Now that I thought some more about this, I conclude that the induced field explanation is suspect for another, more fundamental reason. In my book, work is done by a force along the line of force. Work is not done by another, induced force, that replaces the driving force! (How could I not realize that earlier? :redface:).

Moreover, at the start, at v=0, there can only be a magnetic force that drives the magnetic attraction between the two permanent magnets, as there can be no induced electric force at that time. Moreover, the magnetic force can't be instantly replaced by an equal induced electric force at the start of motion either - that's just nonsense.

Thus, the explanation according to which induced electric field does the work instead of the magnetic field cannot possibly be right.
 
  • #139
harrylin said:
Now that I thought some more about this, I conclude that the induced field explanation is suspect for another, more fundamental reason. In my book, work is done by a force along the line of force. Work is not done by another, induced force, that replaces the driving force! (How could I not realize that earlier? :redface:).

Moreover, at the start, at v=0, there can only be a magnetic force that drives the magnetic attraction, as there can be no induced electric force at that time. Moreover, the magnetic force can't be instantly replaced by an equal induced electric force at the start of motion either - that's just nonsense.

Thus, the explanation according to which induced electric field does the work instead of the magnetic field cannot possibly be right.

I said the same earlier. At the time up to the lower magnet ascending, the upward force can be only Fm, not Fe, because prior to motion, no E field has been induced yet. We seem to agree.

One thing I would like to mention is that in addition to laws of physics, we should observe rules of logic as well. I can't understand how anybody can say that the moving magnet's induced E field is the source of its own lifting force. The cause is happening after the effect it seems to suggest. Thanks for your input. BR.

Claude
 
  • #140
K^2 said:
I'm saying that B field doesn't have to actually induce an electric field for work to be done in the system. But work will always be done by electric field, and never by the magnetic field. So an E field must be present in order for work to be done. Whether B field is the source of that E field is a different question.

That's just wrong K...

The magnetic field does create/generate an electrical field. A basic contradiction of you're words is a motor, without B, the electrical force that does work on the loop is not generate and no work is done! And based on Faraday-Maxwell's equation that states: "A changing magnetic field creates an electric field".
 
  • #141
cabraham said:
I said the same earlier. At the time up to the lower magnet ascending, the upward force can be only Fm, not Fe, because prior to motion, no E field has been induced yet. We seem to agree.
Yes I just wanted to add that I now noticed that you already said that in post #126 but you beat me to it. Hopefully my stronger post that emphasises physical principles will have more effect :-p
One thing I would like to mention is that in addition to laws of physics, we should observe rules of logic as well. I can't understand how anybody can say that the moving magnet's induced E field is the source of its own lifting force. The cause is happening after the effect it seems to suggest. Thanks for your input. BR.
Claude
Yes, very right - also cause and effect doesn't allow it!
Thank you too. :smile:

Addendum: I did not find the derivation of the magnetic force in Wikipedia, but I found one here (unverified but the result is correct, so maybe the derivation is also correct!):
http://www.physicsinsights.org/force_on_dipole_1.html
 
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  • #142
cabraham said:
But I said that the force doing work is Fm, not B. B does indeed cancel, but vXB does not. It should be well understood that B acting in the direction of motion will indeed do no work since the Fm vector points normal to B. Once again, are we defining work as force integrated over path? If so, it is universally known that B itself is not acting along the path of motion, but rather Fm is. B will drop out of the above equation, but Fm will not.

This is just utter nonsense now. The cross-product \mathbf{v} \times \mathbf{B} is not only perpendicular to \mathbf{B}, but also to \mathbf{v}, the direction of motion. This is why the magnetic force is always perpendicular to the direction of motion and does no work, according to the Lorentz Force Law
 
  • #143
gabbagabbahey said:
This is just utter nonsense now. The cross-product \mathbf{v} \times \mathbf{B} is not only perpendicular to \mathbf{B}, but also to \mathbf{v}, the direction of motion. This is why the magnetic force is always perpendicular to the direction of motion and does no work, according to the Lorentz Force Law

I'd advise that you think before you pronounce something to be "utter nonsense". I've already stated many times that the velocity of the **electron** is such that Lorentz magnetic force cannot do work on said e-. But we're talking about the force on a magnetic dipole, whose velocity is not in the same direction as that of the e-. The "v" in "vXB" must be stipulated. Up to this point we've been discussing the induced E field due to the motion of the lower magnet being lifted upwards. It so happens that the Lorentz magnetic force is in the direction of the velocity of the magnet, not the electrons comprising a current loop.

If this discussion was about "which force accounts for electrons circulating in a current loop, Fe or Fm, the answer would be unanimous. Everybody here that has stated they feel that Fm does work lifting a magnet, not Fe, has also stated that on a moving charge, only Fe can do work, not Fm.

But if the e- is part of a loop, the Fm force acting on it cannot do work on the electron, but it can do work on the loop because Fm aligns with the loop motion. Likewise for a magnetic dipole. We must differentiate between the two. BR.

Claude
 
  • #144
cabraham said:
I'd advise that you think before you pronounce something to be "utter nonsense". I've already stated many times that the velocity of the **electron** is such that Lorentz magnetic force cannot do work on said e-. But we're talking about the force on a magnetic dipole, whose velocity is not in the same direction as that of the e-. The "v" in "vXB" must be stipulated. Up to this point we've been discussing the induced E field due to the motion of the lower magnet being lifted upwards. It so happens that the Lorentz magnetic force is in the direction of the velocity of the magnet, not the electrons comprising a current loop.

If this discussion was about "which force accounts for electrons circulating in a current loop, Fe or Fm, the answer would be unanimous. Everybody here that has stated they feel that Fm does work lifting a magnet, not Fe, has also stated that on a moving charge, only Fe can do work, not Fm.

But if the e- is part of a loop, the Fm force acting on it cannot do work on the electron, but it can do work on the loop because Fm aligns with the loop motion. Likewise for a magnetic dipole. We must differentiate between the two. BR.

Claude

The Lorentz force doesn't act on the loop. Rather, it acts on the charge carriers in the loop. Its direction is perpendicular to each of the charge carrier's motion, so it does no work.
 
  • #145
gabbagabbahey said:
The Lorentz force doesn't act on the loop. Rather, it acts on the charge carriers in the loop. Its direction is perpendicular to each of the charge carrier's motion, so it does no work.

See the thread about work on a current carrying loop. I drew sketches proving otherwise. Then feel free to comment. BR.

Claude
 
  • #146
harrylin said:
1. I find the induced current explanation suspect: a pair of magnets doesn't warm up noticeably and doesn't act like a shock damper.
I find your implicit hypotheses incorrect. An induced current doesn't have to warm up noticeably or act like a shock absorber.
It seems that your picture of the induction of an electric field in a ferromagnetic substance is completely different from my picture of the induction of an electric field in a ferromagnetic process. The induction that I imagine is like precession. The induction that you imagine is like rolling.
You picture the magnetic field creating an electric current loop that didn't previously exist in the ferromagnetic material. In your picture, there is no electric field previous to the approach of another magnet. The magnetic field from the approaching magnet induces a nonzero electric field. The new electric field pushes electric charges in a circle, creating a "new" current loop that didn't previously exist. The electric charge carriers, which were no moving before, bump into atoms which generates heat. The "new" current loop generates heat which can be measured.
In your picture, there are no magnetic domains, so everything occurs in a homogenous bulk material. Therefore, there aren't any centripetal forces that make the electric current travel in a circle. When the electron bumps into an atom, the momentum can be in any direction. So the induced motion of atoms is random.
I picture the magnet field as rotating an electric current loop that previously existed in the magnetic field. There is a previously existing electric field on the surface of a magnetic domain that forces the electrons to move in a circle. The atoms and the electrons have equal and opposite momentum before the other magnetic comes near.
In my picture, the electric current loop had previously been there in the ferromagnetic material. The current loop concentrates near the surface of a magnetic domain, where there is no resistance. The approaching magnetic field rotates the current loop. The magnitude of the current is unchanged, but the axis of rotation is different. There is no extra heating because the electric current always has the same magnitude.
According to my picture, the component of electric force in the direction of the electric current is generated by rotation instead of creation. After the current loop is rotated, the electric charge carriers bump into atoms that had not been previously subjected to an electric current. So the total momentum of the atoms is changed. The atoms don't move, so the magnetic field doesn't apply force to the atoms. The atoms feel a force only because of the electric fields coming from the electrons.
In my picture, of the other magnetic there are many loops of electric current with their axial directions oriented in a certain direction. The magnetic field applies a total torque to the electric charges that rotates the axes of the loops. However, the loops never increase or diminish their intensity. The electric fields associated with these current loops are also rotated. The rotation induces an electric field in the material that is in the same direction as the current.
In my view, the torque that rotates the electrons is orthogonal to the torque on the atoms. Because they are orthogonal, they can not be the same. The torque on the electrons is caused directly by the magnetic field. However, the torque on the atoms is caused directly by the electric field. The work is done by the second torque, not the first torque.
A torque can do work only if the bodies it is acting upon are moving (i.e., velocity) perpendicular to that torque. A torque can't do work if it is moving in the same direction as the body it is acting upon.
Another way to look at it is picturing each magnetic domain as a bicycle wheel. The angular momentum of the bicycle wheel. The magnetic field is like a person trying to twist the bicycle wheel in a new direction. He applies a torque perpendicular to the angular momentum of the wheel. The work isn't done until the wheel starts to precess. The work is done because the axis of rotation is changing.
The person doesn't make the tire spin. The tire spins at the same rate all the time. However, the precession generates a component of force in the direction of the linear velocity of the tire.
The twisting doesn't cause the spin of the tire any more than the magnetic field causes the loop of current. The force on the tire is due to precession, not due to extra spin. A tire that isn't spinning is easy to twist. The precession of the tire doesn't cause heating.
I do not have the software to draw a diagram and post it in a reply. However, it should be easy for someone else to draw given that description. A magnetic field is applying a torque to a electric current loop. The axis rotates without changing the angular velocity of the electric current.
I know that there are such objects on the macroscopic scale that obey "classical" equations. For example, there is super fluid helium 4. A vortex of helium 4 can be rotated without heating because it has no viscosity. Therefore, I can't see what the difficulty is in picturing electrons in a ferromagnetic material acting in an analogous way. I picture the magnetic dipole of a single electron the same way.
An electron is just droplet of superfluid with a current loop on its equator. I know quantum mechanics is involved. I know about wave-particle duality. However, the electron can behave as a semiclassical particle.
In the semiclassical approximation, an electron is merely a droplet of electrically charged super fluid. The laws of Newton apply to this super fluid. The anomalous behavior can be modeled by letting the viscosity of the superfluid be zero. A droplet of super fluid can rotate without violating Newton's Laws.
So that is what I suggest. Imagine the charge carriers in a ferromagnetic material as being in droplet of an electrically charged super fluid. The droplets are spinning. The magnetic field rotates the droplets, but doesn't change the magnitude of their spin.
Again, I can't present a drawing. However, I can imagine those little loops of current precessing. It is precession that you should be thinking of.
 
  • #147
cabraham said:
See the thread about work on a current carrying loop. I drew sketches proving otherwise. Then feel free to comment. BR.

Claude

I've no need to look at sketches in this case. Notice the q in the Lorentz Force law? Where there is no charge, there is no Lorentz force. Where there is no moving charge/current, there is no magnetic Lorentz force (qvxB). The Lorentz force acts directly on the charge carriers in the loop, not on the loop itself.

Some mechanism within the loop (in the case of a physical loop connected to a battery, that mechanism is the electric force generated by the battery) keeps the charge carriers moving around the loop and results in the loop as a whole moving (possibly) in the presence of the external magnetic field. This motion gives an additional component to the motion of the charge carriers in the loop.
 
  • #148
Darwin123 said:
I find your implicit hypotheses incorrect. An induced current doesn't have to warm up noticeably or act like a shock absorber. [..]
I know; and that was mere intuition - it's nothing compared to the disproof of the electric field hypothesis that I presented next (and I wasn't the first).
You picture the magnetic field creating an electric current loop that didn't previously exist in the ferromagnetic material. [...]
That was the model that I distrusted by intuition. However, I finally rejected the possibility of any electrical induction as the driving force on other grounds, which I presented next. Moreover, Ampere's model is that we explain the magnetic field by the existence of electric current loops inside the material; I agree that that's the model that we should use - at least in this classical physics forum. That picture leads to the derivation of the magnetic force to which I finally referred with links. I find that description in Wikipedia satisfying and consistent with my textbooks. In that description, the driving force between magnets is basically magnetic, as it is due to the Lorentz force gradient on current loops (more precisely: the Lorentz force that is due to the gradient of the magnetic field at the current loops).
[...] In my picture, the electric current loop had previously been there in the ferromagnetic material. The current loop concentrates near the surface of a magnetic domain, where there is no resistance. The approaching magnetic field rotates the current loop. The magnitude of the current is unchanged, but the axis of rotation is different. There is no extra heating because the electric current always has the same magnitude. [...] the torque on the atoms is caused directly by the electric field. The work is done by the second torque, not the first torque. [...] So that is what I suggest. Imagine the charge carriers in a ferromagnetic material as being in droplet of an electrically charged super fluid. The droplets are spinning. The magnetic field rotates the droplets, but doesn't change the magnitude of their spin.
Again, I can't present a drawing. However, I can imagine those little loops of current precessing. It is precession that you should be thinking of.
That's interesting; can you likewise give a reference to show that your picture is established physics? I made a reference to standard physics; if there is doubt about it, we could try to do an order of magnitude calculation to verify the classical magnetic force equation (and likely we can save that trouble and find an example in a textbook).

Here again implicitly the answer on the OP's question (it should hardly surprise that the simple answer is the correct one: the force between the magnets and which pulls them together... is magnetic, as it results from the Lorentz forces in an inhomogeneous magnetic field):
https://en.wikipedia.org/wiki/Force_between_magnets

and once more a derivation:
http://www.physicsinsights.org/force_on_dipole_1.html
 
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  • #149
Why is this going way out of hand? Seriously. You all are going way to deep.

I ask all of you now to just answer my OP question please.

I think the most suitable answer is that the magnetic field's generate/create electrical fields that would do the work. Based on Maxwell's equations.

Now does a magnet do any kind of work on another magnet? Yes.
What force is the cause of work? Electric Force/field.
What caused the electric force? The MAGNETIC FIELDS.
Then what did the magnetic field/force do? Well, it created the electrical field that eventually does the work, did it do work or not? Well that depends on how people accept it... It created the force that does work...

Miyz,
 
  • #150
harrylin said:
That's interesting; can you likewise give a reference to show that your picture is established physics? I made a reference to standard physics; if there is doubt about it, we could try to do an order of magnitude calculation to verify the classical magnetic force equation (and likely we can save that trouble and find an example in a textbook).
A lot of articles treat the electrons in atoms as current loops that rotate in a magnetic field. Magnetic domains are a little more complicated since they can change shape, grow and rotate. However, the point is always that the current loops at the macroscopic scale rotate.
Here are some links and corresponding quotes.

http://www.physics.sfsu.edu/~lea/courses/ugrad/360notes15.PDF
“In a ferromagnetic material, the atomic dipoles tend to align with each other,
even in the absence of applied fields. the dipoles align in regions called domains.
Normally the domains are oriented randomly. When an external field is applied,
the dipoles at the edge of a domain feel an additonal torque, and the net effect is
to cause the domains with magnetization parallel to the applied field to grow. If
the applied field is strong enough, the other domains actually rotate to align with
the applied field. “

Maybe this is your heating. The hysteresis losses cause heating in the ferromagnetic material.
http://ecee.colorado.edu/~mcleod/teaching/EandM3400/Lab%20Book/Chp_13.pdf
“In magnetic terms, atoms and molecules inside matter resemble tiny current loops. If a
piece of matter is situated in a magnetic field, the moment of magnetic forces partly aligns
these loops, and we say that the substance is magnetized. The magnetic field produced by
the substance is due to these aligned current loops, known as Ampere's currents. A substance in the magnetic field can therefore be visualized as a large set of oriented elementary current loops situated in a vacuum. These oriented loops can be replaced by equivalent macroscopic currents situated in a vacuum, known as the magnetization currents.

In ferromagnetic materials, groups of atoms (Weiss' domains) are formed as small saturated magnets. Magnetization of ferromagnetic materials is obtained by aligning these domains, which is accompanied by hysteresis losses.”http://hyperphysics.phy-astr.gsu.edu/hbase/solids/magpr.html
“All atoms have inherent sources of magnetism because electron spin contributes a magnetic moment and electron orbits act as current loops which produce a magnetic field. In most materials the magnetic moments of the electrons cancel, but in materials which are classified as paramagnetic, the cancellation is incomplete.”http://www.cientificosaficionados.com/libros/aceleradores2.pdf
“On a macroscopic scale, when two fixed adjacent current loops have the same orientation, the magnetic forces act to rotate the loops to opposite polarity (Fig. 5.9a). This is a consequence of the fact that when the magnetic moments are aligned antiparallel, magnetic fields cancel so that the field energy is minimized. With no applied field, atomic currents are oriented randomly, and there is no macroscopic field.”

http://whites.sdsmt.edu/classes/ee692gwmm/notes/Lecture3.pdf
“Consequently, this loop will rotate if free to do so.”

Note: All the above references show that magnetic dipoles can be modeled as current loops which can rotate.

Just two more references dealing, not with permanent magnets, but with the concept of electrons as classical balls of charged fluid.
http://www.usd116.org/lbeuschlein/physics/PowerPoint/magnetism.pdf
“Note: Electrons are not actually little balls that rotate and revolve like planets, but imagining them this way is useful when explaining magnetism without quantum mechanics.”

This this not a link. I believe that there is a link to an online copy of the following book, but I forgot where it is. In this book, the writer really does model an electron as an ball of electrically charged fluid, held together by some nonmagnetic force. He did not include spin. However, there is nothing in his model that contradicts the idea of spinning electrons. One can give his electron a spin. The model used by Lorentz would work quite well. Further, his model is consistent with Maxwell's equations.
"The Theory of Electrons" by H. A. Lorentz (1915).

At your request, I gave you some references. Now I have three questions. Please give references and links, if possible.
1) How can a current loop exist without a centripetal force?
2) What centripetal force keeps the electrons in an atom moving in a circle?
3) Is there always an electric field between an electron and the nucleus of an atom?
 

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