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

1. Aug 16, 2012

### Miyz

Hallo everyone,

Can a magnet do work on another magnet? (I believe it can. Just wanted to make sure.)

What formula's do you use to support you're answer?
Finally,
I know that magnetic field can not do ANY work on a free charge based on Lorentz force so no need to reference that.
Regards,

Miyz.

Last edited: Aug 16, 2012
2. Aug 16, 2012

### Miyz

Re: Can a magnet's magnetic field preform work on another magnet?

Page 8-12.

Miyz,

3. Aug 16, 2012

### Staff: Mentor

Re: Can a magnet's magnetic field preform work on another magnet?

Here is my previous response, with two minor edits:

4. Aug 16, 2012

### Darwin123

Re: Can a magnet's magnetic field preform work on another magnet?

The magnetic field is not doing any work on the electric charge carriers. It is doing work on the matrix of nuclei that keep the electrons in the magnet.
A magnetic is not composed of electric currents alone. It contains the atoms which supplied the electrons that created the electric current. The work done on the magnets includes the work done on the atoms from which the electrons were supplied.
Even when the magnetic field of one magnet does work on the other magnetic, it does not do work on the current. The electric currents can be pictures as d-orbital electrons jumping from nucleus to nucleus. However, the nucleus and the jumping electrons are oppositely charged. There is an electric field from the nucleus that attracts the jumping electrons.
The result is that when the electric current changes direction, the jumping electrons change direction. The jumping electrons then pull the nuclei by its electric field. It is the electric field that does work on the nuclei.
The thing to remember is that "the magnet" is a system. It isn't just electric currents. If there weren't nonmagnetic forces holding the magnet together, the electrons would not stay in the magnet.
Work done on any component of a system is not necessarily the work done on the system. The electric currents are a component of the magnet, not the magnet itself.
A magnetic field can't do work on a current. However, a magnetic field can do work on a system that includes a current.

5. Aug 17, 2012

### Q-reeus

If the almost consensus position in https://www.physicsforums.com/showthread.php?t=621018, from which this follow-on obviously derives were true (dW = E.j dv for any EM system), then uniformly magnetized permanent magnets should interact as though they were perfectly conducting surface current inductors. That is, the so-called surface magnetizing currents Im owing to bulk cancellation of the combo of orbital and spin electronic contributions to magnetization, should perfectly obey Lenz's law and the consequences of the classical Faraday's law curl E = -dB/dt from which Lenz's law derives. So a long straight magnetized rod enclosed within a similarly shaped solenoid should completely demagnetize when the solenoid generates a B field Bs equal in magnitude and of the same sign as that of the magnet's initial B field Bm. [Edit: not quite perfectly, as there is a finite but relatively tiny 'angular KE' contribution owing to the electronic gyromagnetic ratio μ/S ~e/me] This manifestly does not happen. The actual magnetic response is known to be quite complex and material dependent - particularly in the demagnetizing regime when Bs opposes Bm. Assuming the rod is fully magnetized, when Bs has the same sign as Bm, typically there is very little change in the latter regardless of how great Bs is made.

In short, permanent magnets do not obey classical EM in this important respect, and it cannot be maintained that dW = E.j dv covers the situation. I therefore disagree with #3, while #4's picture of magnetization as "d-orbital electrons jumping from nucleus to nucleus" is at best only partially true (orbital contributions are an important contribution in ferrites but not otherwise) and imo misses the real point here. QM 'exchange interactions' stemming from Pauli exclusion principle are intimately tied up with any detailed energy exchanges (includes magnetic domain growth and reorientation), an observation I admit to not being at all qualified to expand upon in any detail. Beside that, there is the electron's intrinsic magnetic moment which clearly cannot be modeled as a tiny classical loop current. If it could, then particularly when Bs has the same sign as Bm, Lenz's law would continue to hold as for a classical perfectly conducting solenoid but does not. I followed only a tiny fraction of the postings in the above linked thread, so pardon please if I am repeating other's arguments already made there.

Last edited: Aug 17, 2012
6. Aug 17, 2012

### Staff: Mentor

Please provide a mainstream reference supporting this or a rigorous derivation from Maxwells equations which supports your point.

Classical EM does not explain how a material is magnetized, but given that it is magnetized, it correctly describes the forces and energy.

7. Aug 17, 2012

### Q-reeus

The rigorous derivation you require of me and no one else here has been adequately summarized by use of my example in #5 imo. Instead of a terse challenge, how about furnishing your own 'rigorous', or just rational, reconciliation of your viewpoint in #3 with the long-magnetized-rod-in-solenoid scenario I have furnished.
Easy to just say. Then provide your own explanation of just where the total conventionally calculated magnetic energy density comes from when, say, Bs = 100*Bm (hint here: (a+b)2 = a2+b2+2ab, and, even allowing for non-linearity in getting to the b2 bit (non-linear magnetization), it's that 2ab bit that could be a bit worrying for your viewpoint - imho). [Forgot to explicitly add that we assume magnetic saturation early on here, say when Bm = 0.01 Bs (final)] That is, how do you equate net solenoid input energy with the notional net stored energy using only ME's? No problem computing solenoid input energy via integrating over a non-linear rod magnetic susceptibility. But how to explain how those tiny magnetic dipoles somehow resist Lenz's law to give that ~ 2ab part. Moral - don't try putting me on the back-foot pal! Looks like our relationship is back to the usual situation. :tongue:

Last edited: Aug 17, 2012
8. Aug 17, 2012

### vanhees71

I don't know, how often I posted this trivial thing now, but you can describe permanent magnets within usual classical Maxwell electrodynamics by using the effective magnetization current in the Ampere-Maxwell Law, $\vec{j}_{\text{mag}}=c \vec{\nabla} \times \vec{M}$, where $\vec{M}$ is the magnetization density of the magnet, in addition to the usual convection currents of moving charges, $\vec{j}_{\text{convection}}=\rho \vec{v}$. The conduction current is usually described in linear-response approximation to the electric field by Ohm's law, $\vec{j}_{\text{cond}}=\sigma \vec{E}.$ These are the usual simplified equations assuming bodies at rest and the velocity of all charges much smaller than $c$, i.e., the non-relativistic approximation for the description of matter. Note that this approximation can be misleading. E.g., the description of the homopolor generator always needs a fully relativistic treatment. It's one of the most prolific direct application of relativistic effects in engineering :-).

I do not know of any example where Maxwell electromagnetics is proven wrong within the range of applicability of the classical approximation to full quantum electrodynamics, which is of course the most comprehensive theory we have about electromagnetics. This holds for both, "vacuum QED", where one considers scattering events of a few particles (usually two particles) mediated by the electromagnetic interaction, and "in-medium QED" like plasma physics and condensed-matter physics.

9. Aug 17, 2012

### Staff: Mentor

Fair enough. See here for a rigorous proof based on established mainstream science demonstrating that the power density delivered by an EM field is given by E.j in all cases: http://farside.ph.utexas.edu/teaching/em/lectures/node89.html

You are the one attempting to overthrow established mainstream science with nothing more than a handwaving assertion of some problem with no evidence to support it, either theoretical or experimental. It isn't up to me to prove you wrong, it is up to you to prove yourself right.

I don't know what you mean by "the back-foot", but if you cannot justify your assertion here then there is nothing left to do, the default position (particularly on PF) is that the mainstream scientific theory stands in the absence of compelling evidence to the contrary (which you certainly have not provided).

Last edited: Aug 17, 2012
10. Aug 17, 2012

### Q-reeus

Flux quantization in superconducting circuits, as experimentally verified in 1961 by Fairbanks & Dever, Doll & Nabauer, means that a closed circuit supercurrent can only respond to a time-changing flux - and thus an E = -dA/dt, in discrete jumps. In between, no change occurs and in that interval Lenz's law fails totally, and holds only as an average over a periodic interval. An electron can in a way be thought of as the ultimate in supercurrent miniaturization - with the added restriction there are no flux jumps at all. Hard then to see how any time-changing E fields involving relative motion of fully magnetized magnets can be made to 'do work' on each other - save for the usual resistive eddy currents which are not the main concern here. As we are merely talking about a redistribution of energy within a system, seems a whole lot more sensible to me to treat the situation in terms of magnetic energy only, at least for slow motions in the frame of interest.

Last edited: Aug 17, 2012
11. Aug 17, 2012

### Q-reeus

I will give it a look over. [Edit: Now had a look; this is just a pretty standard derivation of Poynting's theorem. Not at all in conflict with #5 and #10 imo.]

As far as this bit goes:
Overthrowing mainstream science? Putting things a little dramatically there. Well as I say, why not just give your own explanation. Still, if it comes down to a case of quoting esteemed authority figures, I just managed to find the following, which might give pause for thought: http://physics.stackexchange.com/qu...field-do-work-on-an-intrinsic-magnetic-dipole A range of opinions, but I side with Lubos Motl there.

Last edited by a moderator: May 6, 2017
12. Aug 17, 2012

### Dotini

To a non-physicist, it seems intuitively obvious that a magnet does work, as when picking up a paperclip. I was easily able to find the following link to confirm that is true.

Question

Since magnetic forces can do no work, what force IS doing the work when a bar magnet causes a paper clip to jump off a table and stick to the magnet?

The original assumption that a magnetic field can do no work is incorrect. A magnetic field has an energy density that is equal to the magnetic induction (B) squared divided by twice the permeability (mu sub zero). If you were to sum (integrate) this energy of the magnet over all of its field before it picked up the paper clip and compared it to the same sum after you picked up the paper clip, you would discover that there was a loss of field energy. The paper clip has in effect 'shorted out some lines of magnetic flux'.

How much energy was lost? If you took hold of the paper clip and pulled it out to such a distance that the magnetic pull was insignificant, the work you did in this process would exactly equal the amount of energy lost when the clip was on the face of the magnet. When you picked up the clip with the magnet the clip was accelerated toward the magnet acquiring kinetic energy. This kinetic energy will equal, ignoring air drag, the loss of magnetic energy in the field. This kinetic energy will be dissipated in the form of heat on impact of the clip with the magnet.

For further understanding of the energy in a magnetic field, you may want to study magnetic fields in solenoids. See the Reference below.

Physics, Volume 2 by Halliday and Resnick

Answered by: Robert Gardner, M.S., Retired Physicist

Respectfully submitted,
Steve

13. Aug 17, 2012

### Staff: Mentor

You are specifically claiming that Maxwell's equations predict that putting a bar magnet in a current carrying solenoid will demagnetize it, are you not?

That is clearly contrary to experiment, so if Maxwell's equations actually made such a prediction then they would be overthrown. Indeed, that was your point in suggesting that they make such a prediction. So I think the description is accurate, not dramatic.

However, your claim is also unsubstantiated (either via reference or derivation) and it appears that you have no plans to substantiate it, so it can be safely dismissed.

As I said in in my point 4 both E and j depend on B, so B does work indirectly. In particular, Lubos is talking about the magnetization which is directly linked to j via the equation that vanhees71 posted above.

Last edited: Aug 17, 2012
14. Aug 17, 2012

### Staff: Mentor

The work is entirely accounted for by E.j so the B field only does work via its influence on E and j. See the reference I posted above for a derivation.

15. Aug 17, 2012

### vanhees71

Dotini, there is no contradiction between the quote you give and the fact that magnetic fields don't do work. Of course, if you compare the final state (paper clip attached to the magnet) with the initial state (paper clip separated from the magnet), you of course find that the change in magnetic-field energy is given by the work necessary to pick up the paper clip, attaching it to magnet.

This is precisely the content of Poynting's theorem, nicely explained at DaleSpam's link. This epxlains the the work is done by induced currents and electric fields during the transient (i.e., time-dependent!) situation when picking up the paper clip!

16. Aug 17, 2012

### Dotini

I can create a charged object by rubbing a balloon on my hair. And then do work with it by rolling an aluminum can around on the floor. No magnet is required in this instance, as the work is done by an electric field.

Is this agreeable? Or is the work done by me moving the balloon?

Respectfully,
Steve

17. Aug 17, 2012

### Dotini

Shocking and disconcerting that physicists and perhaps even textbooks are in disagreement!

Respectfully,
Steve

18. Aug 17, 2012

### Q-reeus

Go back and read what I actually said and argued in #5, and then try and not twist it out of shape, as you proceed to do below.
Nonsense. The aim was to show that Faraday's law applied to a permanent magnet modeled as a classical charge/current distribution would indeed necessarily 'discharge' as claimed there. Do you dispute that simple observation? That this is not the case points to the radically different response of a real magnet - owing to QM. Stop twisting what I have been saying!
Is this how the real DaleSpam defends his own position?
Huh? As I said, there are a range of views expressed there, and I back those of Lubos, which is *not* just about force density! Try reading at least all of his entries there.
As Lubos rightly imo points out in that link I gave, electrons cannot be successfully modeled as the limit of a classical loop current. Electrical interactions of the E.j type simply do not and cannot apply in that wholly QM regime. And it carries over to a permanent magnet as a QM glued ensemble of such. You can dismiss this yet again as 'hand-waiving', but I notice you still haven't supplied any kind of coherent rebuttal to either #5 or #10. Just hope I don't get stuck in a useless circular dialogue with you here. Please, be prepared to shift position, however painful it seems at the time. Must go. :zzz:

Last edited by a moderator: May 6, 2017
19. Aug 17, 2012

### Staff: Mentor

I don't think that classical EM (Maxwell's equations and the Lorentz force law) accurately describes the work on the can. Classical EM does accurately describe the work on the paperclip. The two situations are not analogous.

20. Aug 17, 2012

### Staff: Mentor

Tell me how I am twisting this out of shape:
You must have started responding before my edit.