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A magnet and a coil move relative to each other. In the frame of reference of the magnet, there is a magnetic field and consequently a force acting on the charges in the coil according to the Lorentz force $F=qv\times B$ but there is no net electric field. In the frame of reference of the coil, there is a magnetic field and also an electric field, induced by the magnet, $E'$ that moves the charges in the coil, producing a current. But, in the first case no work is done on the charges, since the force is perpendicular to the velocity. In the second case, the force $qE'$ does work on the charges. How is this "paradox" resolved in classical electromagnetism?

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This seems like hw. Look up Lorentz transformations... The E in your one frame is $v\times B$ in the other. – Chris Gerig Jan 14 '12 at 16:32
And you're probably not moving relativistically, so $\gamma\approx 1$. – Chris Gerig Jan 14 '12 at 16:34
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Here is a nice explanation. – Omar Jan 14 '12 at 19:23

4 Answers

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When the magnet is moving, the electric field of the magnet is doing the work, pushing the current carriers around the wire. When the magnet is still and the wire is moving, the magnetic field produces a force in the current carriers, but this force does no work, it is the constraint force that keeps the electrons in the wire that is doing the work. The paradox is resolved by noting the the wire is moving, so the constraint is not time-independent.

The constraint force is perpendicular to the surface of the wire pushing on the charge carriers in the direction of motion (because the whole thing is moving). This force is doing the work on the charge carriers in this frame (although it is somewhat strange to think of a constraint force doing work).

The push of the current carriers against the wire's constraint force gives the breaking force on the wire, which slows it down so as to conserve energy, as the resistance gives off heat.

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Sorry, but you copied the misconception of the OP that energy is Lorentz-invariant and has to agree in all reference frames. That has led you to "solve" a paradox that never existed. Also, your detailed comments about "constraint forces" are wrong because in this experiment much like in all non-accelerated ones, constraint forces aren't doing any work. – Luboš Motl Jan 17 '12 at 14:10
@lubos Motl: Constraint forces do do work in inertial frames when the object is moving, precisely because energy (and its flows) are not Lorentz (Galilean really) invariant. The classic example is: I am on a train going 50m/s and I lean on a pole against the direction of motion, so that it pushes against me with 100N in the direction of motion. So the pole is doing 5000W of work on me! This is correct, because the shift of reference frame mixed the momentum/stress flows in my rest frame into energy flows in the station frame. – Ron Maimon Jan 17 '12 at 15:45
I'd give this a +1, but the constraint force isn't generally perpendicular to the wire in this case - surprisingly perhaps. The sum of the external force on the metal ions and the Lorentz force gives rise to a net force that causes the wire to move in a particular way. The static electric field of the metal ions provides the constraining force for the electrons, which summed with their Lorentz force must give the same net force for the ions. Hence the constraining force on electrons in some part of the wire depends upon the magnetic field at, and velocity of, that segment of the wire. – John McVirgo Jan 19 '12 at 1:53
up vote 1 down vote

There is no paradox: the two reference frames have different answers to the question "how much work is being done", indeed. It's because "work being done" is nothing else than energy and energy isn't a relativistic invariant; it is the time component of a 4-vector. According to relativity, various quantities are observer-dependent i.e. relative – a justification of the name "relativity" – and energy is one of them.

From the magnet's viewpoint, the electrons' energy may be conserved but the momentum is not. Because the energy from the coil's perspective is a mixture of the energy and the momentum in the magnet's reference frame, and because the momentum is being changed from the magnet's viewpoint, it follows that energy of the electron is being changed from the coil's viewpoint.

With this being said, one should still emphasize that the increased/decreased energies of individual electrons get averaged out around the coil.

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Lubos, this is not a good answer either--- which force is doing the work when the wire is moving? – Ron Maimon Jan 15 '12 at 4:10
Dear @Ron, it's always the electric force $q\vec E$ that is doing work. If $\vec E=0$ in a given frame, then there's no work being done. What's your problem? – Luboš Motl Jan 15 '12 at 8:43
What you wrote is correct, but my problem is that you don't resolve the paradox. The average business you talk about around the coil is only true when the magnetic field is constant and there is no EMF. If the field is changing, the field will do work. The work is actually being done by the constraint forces, not the B field, as explained in my answer. Your answer is suboptimal. – Ron Maimon Jan 15 '12 at 14:39
@RonMaimon: that's what I think, too... but is it actually right? Perhaps there is really no work done in $\mathfrak{M}$, neither by the Lorentz- nor by the constraint force. I don't know how to do the transformations correctly. – leftaroundabout Jan 15 '12 at 19:12
@leftaroundabout: The transformations are done correctly by Lubos, and there is no paradox in the rest frame of the wire. In the moving frame, the constraint force starts doing work, that is all. It is like riding on a subway and leaning on a pole, when you are moving, the force on the pole does work on you(if it is in the direction of motion), balanced by the opposite sign work on your feet. – Ron Maimon Jan 15 '12 at 19:25
up vote 0 down vote

"There is no work because the force is perpendicular to the velocity". I don't think that's right. The v in your formula qv x B is the velocity of the coil, but it is not the velocity of the electrons. The electrons move inside the coil, and they move in the direction that the force is pushing them: perpendicular to the velocity of the coil. So how is there not a force pushing the electrons?

EDIT: In response to Ron's comment about the constraint force, I've worked it out and I think the following picture will show what is happening:enter image description here

The wire is treated as a pipe full of electrons, and the blue electron shown is actually right against the pipe wall: as the pipe moves upwards, the electron rides along the pipe wall as shown by the blue arrow. The two forces are the vxB force, perpendicular to the electron's true motion, and the constraint force, perpendicular to the wall of the pipe. I think the picture shows that the actual work is done by that COMPONENT of the vxB force which is in line with that COMPONENT of the electron's velocity which is along the direction of the pipe.

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The magnetic force on the electrons is perpendicular to the velocity. It is the constraint force from the wire that is doing the work. – Ron Maimon Jan 15 '12 at 14:38
Ron, I'm going to edit my response with a picture in response to your comment. The constraint force is indeed important, but I think you will agree it turns out to be the vxB force which does the work after all... – Marty Green Jan 15 '12 at 18:23
I'm going to eat my words. Having posted the picture, it's pretty obvious that the guy pushing the pipe is doing the work. There is a reaction force from the magnetic field that he is pushing against, and that's why the electrons move to the left, but obviously the magnet isn't providing any energy. – Marty Green Jan 15 '12 at 18:30
up vote -1 down vote

I'm not quite sure about this one, but I would explain it this way: (let $\mathfrak{M}$ the frame of reference of the magnet, $\mathfrak{C}$ 〃〃〃〃〃 the coil)

  1. No work is done initially in $\mathfrak{C}$, either: the electrons are at rest in the beginning. Only once the force has started built up some speed there is a finite power.
  2. In $\mathfrak{M}$, the electrons also build up velocity. Even in the parts of the wire which enter the field perpendicularly (sort of), so that the electrons are initially deflected in the direction of the wire, the force will eventually (when the electrons move in the wire) point in a direction other than the wire alignment. Once this happens, only the force component in wire direction will take action, so the effective force direction is not perpendicular to the electrons' moving direction any more. To put it another way: when the electrons are pushed to one side of the wire, $\rho$ becomes macroscopically nontrivial, resulting in an electric field due to $\nabla\mathbf{D}\propto\rho$.
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This is not the resolution--- the magnetic force does no work even when the electrons are moving. – Ron Maimon Jan 14 '12 at 16:45
@RonMaimon: however as I said there is also going to be an electric field in the wire. It seems to be this field that does the actual work, though I can't quite figure out how. – leftaroundabout Jan 14 '12 at 17:05
Not all force in a wire are electromagnetic, there are also electronic forces. Electrons exert a push-pull type Fermi-exclusion, that's what you feel when you put your hand against a table. In this case, though, you are right that there is an electric field in the wire that does the work--- it is the constraint force. Not my downvote, by the way. – Ron Maimon Jan 15 '12 at 4:09

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