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Consider a single charge moving only under the influence of Magnetic Field $\vec{B}$. The charged particle moves in a circle and the work done by $\vec{B}$ is 0. Now consider a current element in a uniform magnetic field ($\vec{B`}$). Now the derivation of the expression of force results in the the equation $\vec{F} = i \vec{l} \times \vec{B}$ (for a uniform magnetic field) where $\vec{l}$ is the vector joing the ends of the current element. For simplicity, I will consider a straight conductor. When the two vectors are perpendicular, the current element experiences a net non-zero force which does not cause a torque and causes translational motion. Therefore, in this case the magnetic field does positive work.

How is this contradiction possible? Magnetic field, which did no work on a single charge, now does positive work when the current carrying elements clumped together in a conductor. Is there any implicit assumption in the whole process which causes this? Can someone please give a good explanation?

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  • $\begingroup$ Think about the source of $\vec{B}$. If it is free to move, it will move because of a reaction force imposed on it by the wire, the opposite motion will end up having a net cancelling effect on the total work done. If it is not free to move, then whatever is fixing it is what is actually doing the work. Not the magnetic field. $\endgroup$ – Jim Jul 20 '15 at 16:31
  • $\begingroup$ The source is not disturbed in any way. And unless it experiences any displacement, no work will be done on it by any force. $\endgroup$ – user117913 Jul 20 '15 at 16:35
  • $\begingroup$ related: physics.stackexchange.com/q/67826/23473 $\endgroup$ – Jim Jul 20 '15 at 16:36
  • $\begingroup$ Moreover it would be helpful if you could focus on the system described and show me the solution there $\endgroup$ – user117913 Jul 20 '15 at 16:37
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    $\begingroup$ Recall that the equation you stated is really just a fancy approximation of $F=q\vec v\times\vec B$. Once the wire begins moving in the direction of $\vec l\times\vec B$, the velocity vector of all charged particles will change. That means any net displacement in the wire, $\vec d$, over any amount of time, $t$, gives a component of velocity to $\vec v$. This means the direction of the force on the wire will change as it moves with the magnetic field. At no point will the charged particles ever move parallel to the magnetic force felt because $q\vec v\times\vec B$ is always perpendicular $\endgroup$ – Jim Jul 20 '15 at 16:56
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The claim, "the magnetic force does no work," while technically true, is misleading in so many situations and actually helpful in so few that we probably shouldn't put as much emphasis on it as many textbooks do. The magnetic field stores energy, and a change of the configuration of a system (like moving a current) can convert that stored energy into other forms, so the system will certainly behave as if the magnetic field is doing work, but careful analysis will always show that it's not actually the magnetic force that's doing the work associated with the change of magnetic energy. Usually it's an electric force.

In your example, I think the subtlety lies in assuming that $I\vec{\ell}$ for the current $ = q \vec{v}$ for individual charges. The charges will tend to change trajectory, i.e. $\vec{v}$ changes, due to the magnetic field, but $\vec{\ell}$ maintains its direction. Why? Because the wire (magically? We never really explain it very well in class) constrains the charges to move along its length in spite of their lateral movement produced by the magnetic field. As others have said above, the wire exerts a force on the charges to keep them in a straight line, and so the charges exert a reaction on the wire which moves it.

So technically, the magnetic force isn't doing any work on the current; it's just redirecting the force that would otherwise be pushing the current down the wire so that that force has a component pointing perpendicular to the wire instead.

Now for the bonus question: a magnetic dipole, e.g. a current loop or a small refrigerator magnet, will experience a force in a nonuniform magnetic field (though not in a uniform one). The dipole will then accelerate if this is the only force acting on it, so work is being done on it. If it's not the magnetic force doing the work, then what is it? (Really, as in this case, it gets silly sometimes to insist the magnetic force isn't doing work. For many intents and purposes, you might as well treat it as though it is.)

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  • $\begingroup$ What I'm missing is the description why electrons move in spiral paths in magnetic fields. It has to do with their magnetic diole moment and their intrinsic spin. $\endgroup$ – HolgerFiedler Jul 21 '15 at 6:54
  • $\begingroup$ Very nice answer. I also like this paper for its clear exposition of how it can be that a non-uniform magnetic field does no work on a classical dipole/atomic dipole even though, as you correctly point out, the behaviour is practically indistinguishable from a working field. Where the logic gets fuzzy is in the consideration of the electron spin term; Deissler introduces the 'rotational kinetic' part of the rest-mass energy, admitting that it cannot arise due to rigid rotation. $\endgroup$ – tok3rat0r Jul 21 '15 at 6:57
  • $\begingroup$ It's simply shown that a rotational contribution to the rest energy is not inconsistent with the Dirac equation, but the question of how such a contribution could arise is explicitly avoided. At that point, I think one has gone as far as it's possible to go by insisting on a separation of particles (and their associated dipole moments) and fields acting on these particles. Only fully relativistic quantum field theory would allow one to delve further into the details... $\endgroup$ – tok3rat0r Jul 21 '15 at 7:08
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    $\begingroup$ @tok3rat0r : it isn't fuzzy when you think of a field variation displacing its own path into a closed Dirac's belt path. Then it looks like a standing field. Standing wave, standing field. Note that the Einstein-de Haas effect demonstrates that " spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics." See Goudsmit on the discovery of electron spin. The dipole moment is there for a reason. $\endgroup$ – John Duffield Jul 21 '15 at 7:09
  • $\begingroup$ @JohnDuffield - thanks, that's interesting, and I hadn't heard of that effect before. The point remains though that to get any insight into the underlying physical mechanism (rather than just an experimental observation of the fact) requires QED as mentioned in my second comment (and at the bottom of that Wiki page). $\endgroup$ – tok3rat0r Jul 21 '15 at 7:21
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When electrons moving through a wire experience a magnetic force they change direction right? And when they push in that direction they don't gain any energy. So it must be the battery which gave the electrons their kinetic energy that is doing the work. As long as the current is flowing, the magnetic force is acting, and it is the battery that is sustaining this interaction.

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Therefore, in this case the magnetic field does positive work. How is this contradiction possible?

The force given by the formula

$$ \vec F= i \vec{l} \times \vec B $$

is magnetic force acting on the current forming particles - mobile electrons. The formula is accurate provided the wire does not move so the current forming particles follow the wire element $\vec{l}$. If so, the work done by magnetic forces on the electrons is zero and since the wire does not move, work done by magnetic forces on the wire is also zero.

If the wire element is allowed to move, mobile electrons no longer move along $\vec{l}$, but their velocity is influenced also by the motion of the wire. In such case above formula generally does not give correctly magnetic force on the wire element.

Except for when the motion of the wire is very slow (slower than the average speed of electrons with respect to the wire). Then, the formula gives accurately magnetic force on the electrons. Electrons move each differently and always perpendicularly to the experienced magnetic force, so no work is done by magnetic forces on them. However, in macroscopic description the confinement of the electrons within the wire means any external force $\vec{F}_{e}$ (due to external bodies, magnetic field...) they experience in a small element of the wire is accompanied by internal force $\vec{F}_{i}$ whose component perpendicular to the wire cancels the same kind of component of the external force. This can be assumed to be ordinary mechanical force obeying principle of action and reaction, so the electrons push back on the wire. The result is, that due to magnetic field the electrons themselves make the wire move and work on it. When the wire element $\ell$ gets displaced by $\Delta \vec s$, the expression

$$ i \vec{l} \times \vec B \cdot \Delta \vec s $$

gives net work done on the wire by the electrons. The magnetic field thus enables the moving electrons to give kinetic energy (or work, if done steadily) to the rest of the wire.

The role of magnetic field is similar to the role of normal force of ground, when a human is in the process of standing up. No work is done by the ground; all the work is done by internal forces in the human body. But the normal force of the ground is necessary to make this possible. Similarly, magnetic forces do not work, but they make possible work to be done by the electrons on the rest of the wires.

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How is this contradiction possible? Magnetic field, which did no work on a single charge, now does positive work when the current carrying elements clumped together in a conductor. Is there any implicit assumption in the whole process which causes this? Can someone please give a good explanation?

It's simpler than you think, and there is no contradiction. Start with a current in a wire, like this:

enter image description here

GNUFDL image by Jfmelero, see Wikipedia

Chuck a charged particle past it and it moves in a circular fashion around the magnetic field lines. No work is done. Now wrap the wire into a number of loops until you've got a solenoid, whereupon your magnetic field is reconfigured to resemble that of a bar magnet:

enter image description here

Image courtesy of Rod Nave's hyperphysics

Chuck a charged particle through the middle of the solenoid, and it moves in a circular fashion around the magnetic field lines. No work is done. Now get hold of two bar magnets, and let them attract one another. When this occurs no work is done. Instead you do work on the magnets when you pull them apart. It's rather like gravity. You do work on a brick when you lift it. Gravity doesn't do work on it when it falls down. It merely converts potential energy into kinetic energy, which typically gets dissipated as heat.

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