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When a conducting rod moves orthogonally through a magnetic field, charges in that conductor experience a force that is perpendicular to both the magnetic field and their velocity. Let's say that because the electric force on this charge is at this point less than the magnetic force on it, the charge moves a little bit due to the magnetic field.

When the charge moved this little bit moved in a direction that the electric force was opposing it to move in. Therefore this little charge had to have work done on it while it moved this little distance. The force doing work on it was the magnetic force. But the magnetic field is perpendicular to the magnetic field, so the magnetic field could not have done the work referred to.

Who did it?

PS. Could it be that BEFORE the little charge makes its displacement in the direction opposing the electric field, the velocity of the charge-bearing conductor FALLS such that the magnetic force due to v [cross] B is equal to the electric field E holding the charge in place? No work would be done since there is no net force, which makes sense since the magnetic field cannot do work on a charge. And one would need an external force to create a potential difference within the conducting rod.

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    $\begingroup$ What force is moving the rod? $\endgroup$ Commented Oct 19, 2020 at 15:43
  • $\begingroup$ @AdrianHoward I think that's exactly what I was omitting. In my example, gravitational acceleration alone would be driving the velocity, and in examples where v is constant mg+F<ext> would be driving velocity, and more generally an external mechanical force would be driving the velocity of the rod. Therefore the work is done by the external force driving the rod. That's okay. I still don't understand exactly what it looks like (in a micro, non-lumped-matter way) for the external force to translate and become effectively equivalent to the magnetic force. $\endgroup$ Commented Oct 19, 2020 at 16:26
  • $\begingroup$ @AdrianHoward The displacement of the charge is also perpendicular to the velocity of the charge. If we assume an external force moving the rod, then the displacement of the charge is perpendicular to that force as well, so that force could not do the work either, could it? $\endgroup$
    – fishinear
    Commented Jan 12, 2022 at 13:29

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The moving charge makes a magnetic field, against the original B so v would not be constant, you need a force along the way, but only for a short time, if you do not have a closed circuit.

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When a conducting rod moves orthogonally through a magnetic field, charges in that conductor experience a force that is perpendicular to both the magnetic field and their velocity.

A bit more detailed, the charges are not only shifted perpendicular to v and B, but all in the same direction. It is strange, however, that an interaction of an external field with an isotropic electron cannot be the reason for such an asymmetric behaviour.

The reason lies in the following. The electron is not only a charge (with its own electric field). It is also a tiny magnet.

... the charge moves a little bit due to the magnetic field. But the magnetic field is perpendicular to the ... (direction of movement), so the magnetic field could not have done the work referred to. Who did it?

The mechanism behind is the following. The electrons have a kinetic energy (since the rod is moved which is the case for example in n electric generator). The electrons magnetic dipole gets aligned with the external B-field. These magnetic dipoles we associate with a spin. And indeed what happens has an analogy, the gyroscopic effect. The end of the story is, that during the alignment and the displacement the electron emits photons and its magnetic dipole gets disaligned again. The game starts again: alignment - displacement - energetic loss in the form of heat - disalignment.

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