Magnetic attraction / repulsion from the Lorentz force

  • #26
SamRoss
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...and then infer the force of this magnetic field acting on the equivalent magnetization current of the other magnet.

Very good, but one comment is for uniform magnetization, the equivalent currents are not on the endfaces, but rather along the outside cylindrical surface, as if the magnet were a solenoid.

Thanks, @alan123hk ! This seems to be EXACTLY what I was looking for!

A couple of quick questions to make sure I'm interpreting everything correctly...
1. The black arrows representing the equivalent magnetization current would be the v in the F = qv x B formula (and, of course, the red arrows would be the B and the blue arrows would be the F). Is that correct?
2. There is not really any current in the magnet (at least, none that contributes significantly to the magnetic field). The current depicted in the diagram is the result of the spins of the electrons in the magnet cancelling out on the inside, resulting, as @Charles Link said, in a current which is only on the outside of the magnet as in a solenoid.
1604518038407.png

Of course, the electrons are not literally spinning either but each of them produces a magnetic field as if they were. Is all of this correct?
 
  • #27
Charles Link
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Of course, the electrons are not literally spinning either but each of them produces a magnetic field as if they were. Is all of this correct?
Yes. It might interest you that the equivalent magnetic surface current per unit length ## K_m=\vec{M} \times \hat{n} ## in units where ## B=\mu_o(H+M)##. See also https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/
and also see https://www.physicsforums.com/insights/permanent-magnets-ferromagnetism-magnetic-surface-currents/
 
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  • #28
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I want to comment on @alan123hk 's post 22. Very good, but one comment is for uniform magnetization, the equivalent currents are not on the endfaces, but rather along the outside cylindrical surface, as if the magnet were a solenoid

Thank you for pointing out the problem with the diagram.
I plan to fix this when I have the opportunity to upload it to the internet next time :rolleyes:

Thanks, @alan123hk ! This seems to be EXACTLY what I was looking for!
I am happy if you find this useful :biggrin:
 
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  • #29
Charles Link
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See also https://www.physicsforums.com/threa...perature-relationship-in-ferromagnets.923380/
I don't want to break PF rules about hijacking a thread, but I think the OP @SamRoss might find the above thread of interest. The standard textbook calculations are somewhat lacking for the forces between two permanent magnets, (it is more complex than simple electrostatic type forces), but he may be interested in seeing some calculations involving permanent magnets that are rather straightforward.
 
  • #30
SamRoss
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...but I think the OP @SamRoss might find the above thread of interest.

All new knowledge is of interest. Thank you :)
 
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  • #31
SamRoss
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@alan123hk @Charles Link

I was thinking about this thread again and one nagging thought occurred to me. I understand (hopefully) the equivalent current as the current that would produce the same magnetic field as is seen around the magnet and is in reality produced by the spin of the electrons. This current, then, is fictitious. I'm okay with imagining a fictitious current in order to "explain" the magnetic field, but are we then justified in using the right-hand rule to predict the force on imagined moving charges that aren't really there?
 
  • #32
Ibix
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are we then justified in using the right-hand rule to predict the force on imagined moving charges that aren't really there?
Perhaps a better way to think of it is, for any magnet, you could replace it with a custom designed solenoid and get the same field. Then obviously you can use the usual rules for currents and fields for that equivalent solenoid.
 
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  • #33
SamRoss
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Perhaps a better way to think of it is, for any magnet, you could replace it with a custom designed solenoid and get the same field. Then obviously you can use the usual rules for currents and fields for that equivalent solenoid.

I guess the jump from two things producing the same field to those two things reacting in the same way to a nearby object is not so obvious to me. After all, the forces are supposed to act on charges, not fields. To make a weird analogy - if two people happen to draw the same waves in a Japanese-style sand garden, it doesn't mean they will each react the same way to some nearby music.
 
  • #34
Charles Link
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In the magnetization currents, there is no actual charge transport, but the magnetic field that is calculated from them using Biot-Savart is, in fact, precisely correct. In Griffith's E&M book, he does a derivation to compute the magnetic vector potential ## A ## from an arbitrary distribution of microscopic magnetic moments, and the system acts as if there were bulk currents given by current density ## J_m=\nabla \times M ## and surface current per unit length ##K_m=M \times \hat{n} ##.
To answer another question, for the system that creates the magnetic field, at least for magnetostatics, e.g. between two magnets, the force that is experienced will indeed be the same if one or both are replaced by the equivalent solenoid. This is because they will experience equal and opposite forces.
 
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  • #35
vanhees71
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The problem is that the magnetization arising from fundamental dipole moments of particles like in a permanent magnet, where the dipole moments of the electrons are aligned in macroscopic domains, cannot be fully understood within classical electrodynamics, because they are generic quantum phenomena related to the spin of the particles. However, quantum theory tells you that the magnetic moments of the particles and also the resulting macroscopic magnetization are equivalent to the magnetization current discussed above. This follows from the Hamilton operator describing the interaction of the particles, taking into account both their charge and their magnetic moment. The same Hamiltonian of course also describes the interaction of the particles with the electromagnetic field and the magnetization current leads precisely to the same Lorentz force law as usual currents of moving charges.
 
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  • #36
Charles Link
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@SamRoss It may be worth mentioning as a side item that it is very important that the magnetic currents involve no charge transfer. The reason is this: For a transformer, because there is a changing magnetic flux, (originating from the currents in the primary coils and enhanced by the iron/magnetic surface currents), that passes through the transformer, (it cycles at 60 Hz), the result is an EMF, (from Faraday's law), that creates real (eddy) currents in the iron that generate reverse magnetic fields that would almost completely negate the transformer's operation. The solution to block these eddy currents is simple: the transformer has layers of iron that are separated by plastic laminations. The eddy currents can be almost completely blocked, while the magnetic surface currents continue to persist in a computational sense=the magnetic surface currents are virtually unaffected by the laminations. The tranformer operates in a most ideal fashion with the laminations.
 
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