# Does magnetic field do any work

1. Sep 2, 2006

### AlbertEinstein

Sorry, but while going through the article on magnetism in wikipedia I read the following which i couldn't not catch it.

When a charged particle moves through a magnetic field B, it feels a force F given by the cross product:
F= q V x B
where is the electric charge of the particle is the velocity vector of the particle is the magnetic field.

Because this is a cross product, the force is perpendicular to both the motion of the particle and the magnetic field. It follows that the magnetic force does no work on the particle; it may change the direction of the particle's movement, but it cannot cause it to speed up or slow down.

This might give you pause: Simple bar magnets seem to be entirely able to pick up small metal objects, which certainly seems to require that they do work on those objects. As David Griffiths points out in his textbook Introduction to Electrodynamics, this law is absolute -the magnetic field doesn't do any work. However, quite like the normal force of an inclined plane, which also can't do work, the magnetic field can redirect the efforts of existing forces, and then those forces can indeed do work in the relevant direction.

I am not able to understand the underlined lines. I do not know about which "existing forces" is he talking about; and the same applies to "those forces".I will really appreciate the help.Thanks in advance.

2. Sep 2, 2006

### quasar987

Hi Albert,

I think I know what he's talking about. An exemple will be best to illustrate the idea.

Consider a charged particle voyaging in an electrostatic field at constant speed v following a trajectory of electrostatic equilibrium* (i.e. E=0, along its path but E$\neq$0 just a little off its path). No work is done on the particle since E=0. But now we turn on a uniform magnetic field in a direction perpendicular to the velocity of the particle. As a result, the particle changes its course, which bring it in a region of non-vanishing E and work is being done on it.

I like this exemple because at first no work is done. Then the magnetic force is applied, which in itself, does no work on the particle, but is the cause of the resulting electrical work done on the particle by the surrounding E field.

The much simpler exemple of a charged particle moving in a constant electric field and constant magnetic field could have been evoqued just as well but I find it less elengant, because there is work done by the electric field in the begining. The the idea remains that the magnetic field cause the particle trajectory to change and thus affect the work done on it by the electrical force.

*A particle voyaging in the middle of a quadrupole lens for instance.

3. Sep 3, 2006

### pseudovector

First, although static magnetic fields can't do work, time varying magnetic fields can. Second, whenever work is done under the influence of a static magnetic field, some external source will have to supply the energy require (as opposed to an electric field that can do work on its own).

Let's analyze the example you gave, about the bar magnetic picking up staples. We'll start with a simpler example: two dipole moments - current loops - pointing in the same way. Both have mass, but one is fixed and the other is directly under the the first one. Let's say the magnetic forces are greater than gravity so the fixed loop pulls the other up. So, obviously, someone did some work against the gravity, but where did the energy come from? the loop itself. While the loop is ascends, the current in both decreases, due to mutual inductance, so the loops serve as some sort of battery.
When you move a magnet near a staple, the staple feels a changing magnetic field that does work and changes the staple's internal energy - it makes many of the current loops in the metal point in the same direction as those in the magnet. After that both magnet and staple behave as magnetic dipoles. (actually some quantum effects such as ferromagnetism make this a bit more complicated, but the basic idea is similar)

Bottom line, time dependent magnetic field can do work, and static magnetic fields can take energy from other sources and use it to move things.

4. Sep 4, 2006

### AlbertEinstein

Thanks, quasar987 and pseudovector.
But again I do not understand one thing said by pseudo
Afterall, why the current in both will decrease if the second loop ascends(actually I am confused betwn mutual induction and self induction).I'll be grateful if you explain it once more.
thanks once again.

5. Sep 4, 2006

### Gokul43201

Staff Emeritus
It's not clear which part of this is from the wiki and which are your own comments.

Nevertheless, you might find this useful: https://www.physicsforums.com/showpost.php?p=997210&postcount=14

6. Sep 5, 2006

### Meir Achuz

I notice that all these posts ignore ferromagnetism, which changes everything.

7. Sep 5, 2006

### Gokul43201

Staff Emeritus
Don't leaving us hanging...please explain.

8. Sep 6, 2006

### Meir Achuz

A permanent magnet does work when it attracts an iron nail.

9. Sep 6, 2006

### Galileo

Very often it will look like magnetic force really does do work and often very difficult to find out what really is responsible for performing the work, but it is never the magnetic force.
The fundamental equation $q(\vec v \times \vec B)$ tells you magnetic forces will never ever do work, whether they are time varying or not.
Look up how a magnetic field exerts forces on dipole moments and see if the real work being done is by the magnetic field.

10. Sep 7, 2006

### Chronos

The myth of free energy from permanent magnets is difficult to dispell. Just keep in mind it takes energy to create a permanent magnet in the first place. Potential energy is not the same thing as 'work'. A boulder rolling downhill looks like free energy too, until you consider the amount of energy it took to place the boulder there to begin with.

11. Sep 7, 2006

### leright

ok, let me try to explain this. As Galileo stated, the magnetic field does no work on elements of charge or current, but, as I interpret it, megnetic fields CAN do work.

Let us take two rings of current. One ring of current has current flowing in the clockwise direction and we will say it lies in the x-y plane and the other ring has current flowing in the counterclockwise direction and it is lying in a plane parallel to the x-y plane but translated along the z axis by a certain distance. Both rings have the same diameter. The rings will produce a magnetic field with a magnetic dipole moment given by the direction of the surface defining the loop, which is defined by the direction of the current in the loop. So, both magnetic moments due to both current rings are pointing in the -z direction. Now, keep in mind that the B-field due to the second ring (which is NOT the same thing as the magnetic moment and, unlike the moment, the B-field varies) will exert forces on the the invividual current elements that are not in the direction of the motion you observe between the rings (they will attract), but these forces add up to a net force which causes the attraction between the rings. It is important to realize here that the moment is not the same as the B-field. If the B-field was the same as the moment then there would be no net force as the forces would all be pulling radially outward on the ring, and assuming the ring is rather strong, these forces would balance.

So, the magnetic field from one current ring DOES do work on another current ring, but as Galileo pointed out, a B-field doesn't do work on moving charges flowing in the current loops. However, the sum of the mechanical forces on all of the moving charges in the loop causes a net force to be exerted.

Keep in mind that the field due to a ferromagnetic material (permanent magnet) is caused by current also...it is caused by currents called eddy currents flowing through the magnetic material and the analysis of the two current loops can be easily applied to the a couple of bar magnets.

Does this mean that you get free energy? Well, in the case of the current loops you need to apply an EMF to get the current flowing in the loops in the first place, and as the loops are brought close together there is a net change in flux through each loop, so by faraday's law there is an EMF induced in the loop that opposes the current in the loop, so constant work must be done by the EMF source, even in a superconductor to maintain the force of attraction.

In the case of the bar magnet, as chronos pointed out, it take energy to take iron (or any ferromagnetic material.....a ferromagnetic material is just something with a permanent dipole moment) and make it into a magnet. You must have an initial changing magnetic flux through the iron to produce that moment. You might ask why the permanent magnet maintain its moment and thus its eddy current? It's simply because of faraday's law! If the eddy current was reduced for some reason there would be a reduced B-field through the magnet, and therefore by faraday's law there would be an EMF to counter this reduced eddy current. There is generally no net reduction in eddy current over time. Also, with bar magnets, if you let them attract at some speed the by faraday's law the eddy current is reduced, but then you pull them back apart at some speed and the eddy current is restored! However, you can reduce the eddy current and thus the magnetic moment by having the magnets attract very rapidly, but then pull them away very slowly. If you do this enough you can kill the magnet's dipole moment.

Also, I'd like to note that many people feel that the lorentz force is not accounted for in maxwell's equations. when you look at the differential forms of the equations (in maxwellian form, not hertzian form) Faraday's law DOESN'T take into account the lorentz force because certain parts of the integral form of the equation are ignored when the differential form is derived, but it is taken into account if you look at the integral form of faraday's law.

12. Sep 8, 2006

### Gokul43201

Staff Emeritus
leright, this is....lewrong! (bad joke, sorry)

Ferromagnetism requires no currents whatsoever and eddy currents are not what you think they are.

13. Sep 8, 2006

### leright

All magnetic fields are generated by currents, and permanent magnets are no exception. Except in ferromagnets the "current" is produced by a bunch of nuclei and electrons with quantum spins all aligned in one particular direction, which results in a net "current".

Sure a ferromagnet doesn't behave identically to a ring of current, but the ferromagnet can be modeled in quite the same fashion.

Last edited: Sep 8, 2006
14. Sep 8, 2006

### Meir Achuz

The electron magnetic moment, which is responsible for ferromagnetism, is a relativistic QM effect that is not related to any current or "current".

15. Jun 18, 2008

### burashka

This is an old topic but it seems to be not closed yet. I've seen repeated claims that magnetic field can do work under some conditions.

1) Electron spin and magnetic momen are indeed quantum (and also relativistic) in nature. Yet, there is quantum-mechanical expectation of current associated with a single spin. Derivation is given, for example, in Landau, v.3.

2) Whenever work is done on a spin, it is the work of electric field on the current mentioned in point 1. This, for example, happens if a spin flips in an external H-field by emitting or absorbing a photon.

3) Since the Lorentz force applies to every elementary particle that makes up a macroscopic object, the magnetic field does no work on each particle separately nor on the system as a whole.

4) Whatever work is apparently done can be attributed to the work of electric fields that the particles of a macroscopic object create on the atomic scale, even if the average is zero. This is always done at the expense of the internal energy of the object (as in the example of two current loops which was offered in a previous post).

5) Magnetic field can do no work without exceptions, either it is constant or time- and spatially-varying, in both classical and quantum theories.

16. Jun 18, 2008

### kanato

How does it take energy to create a permanent magnet? In a ferromagnetic material, ferromagnetism is the ground-state. Domain boundaries which allow different parts of the material to have different magnetisms, and thus, allow the whole chunk of material to have less than its maximum magnetic moment, require energy to create.

The classical picture of having permanent currents in the magnet which are reduced as the magnet does work seems untrue. The "currents" (electron spins) cannot decrease, and it's energetically favorable to align them and have a ferromagnet. It costs energy to break their alignments, and I'm fairly sure that below the Curie temperature, this results in a net increase in free energy. Of course you can think of picking things up with a magnet at zero temperature, there the free energy of the magnet is minimized by maximizing the magnetic moment of the material. So the magnet doing work can't be compensated by decreasing the magnetism in the material, it seems.

17. Jun 18, 2008

### kanato

Ah, I think maybe I understand now. When a magnet picks up an iron nail, the energy to do work on the iron comes from within the iron itself. The spins inside the iron can align with the magnetic field, which lowers the energy inside the nail. That's where the energy to do work comes from.

18. Jun 18, 2008

### TVP45

Suppose I had a pair of coreless electromagnets. And I used one to pick up the other. Would work be done? By which one?

19. Jun 19, 2008

### burashka

Work is always done by a force. The classical expression for a point particle is dW/dt=vF. You can't say that an object such as an electromagnet does work.

In your example, work is done on each elementary particle that has moved from its original equilibrium position to a new one. If you are interested in the work of the magnetic force which is given by the Lorentz formula, it is exactly zero. If you are thinking about the work of some phenomenological force which is not described by the Lorentz formula, it may not be zero. You can then attribute this phenomenological force to the existence of magnetic field. This doesn't mean that magnetic field does work.

One difficulty is that the macroscopic objects can not be understood in purely classical electromagnetic theory without addition of some phenomenological forces that are needed to hold the object together. As is well known, a system of classical charges can not be in stable equilibrium. Therefore, one must assume, sometimes implicitely, the existence of phenomenological forces which are not electromagnetic in nature.

20. Jun 19, 2008

### burashka

21. Jun 19, 2008

### TVP45

Thanks for the reply. Note that I was careful to stay away from permanent magnets. I'm afraid I don't know what a phenomenological force is; can you give me a reference or some examples?

Anyway, my next set of questions would be:
1) Can a coil of wire have a magnetic field?
2) Does it require work to create that field if it exists?
3) Is any part of such work stored as potential energy?
4) If so, is any part of that potential energy available to do work?

22. Jun 19, 2008

### kanato

That's true, but we don't have to consider the nail hitting the magnet. Consider the initial system to be the magnet and nail before the nail begins to move, and the final system to be the magnet and nail where the nail has picked up some kinetic energy. In this case I can set the final time to be well enough before the collision that the internal quantum systems of the two systems are isolated enough from each other to be independent, and the only interaction between them is through a classical B field.

In this case, the internal free energy of the magnet can't decrease; it's already at its lowest value. (I suppose there may be some feedback from the B field caused by spins aligning in the iron nail, but I would imagine that's small compared to the other energy scales in the problem, because the nail doesn't pick up the same magnetization as the magnet.) The iron nail has picked up classical kinetic energy, and the only place that could really come from is from the breakdown of domain walls in the unmangetized iron allowing it to form a net magnetic moment.

It's true, although I don't see how that changes anything. You don't have to consider the collision and what happens afterward for my argument.

It's a good question. You can simplify it a bit; consider two parallel current carrying wires. There will be a force between them which is perpendicular to the B field: http://hyperphysics.phy-astr.gsu.edu/Hbase/magnetic/wirfor.html#c1

I guess you can think of it as the B field causes the charge carriers to "bump" into the edge of the conductor, pushing it in the direction of the force. My intuition tells me that this will result in an increase in the resistance of the wire, so that in order to maintain the same amount of current, the power source has to work harder. Or, if the power source is a constant voltage source, then the current would decrease. Either way, energy for the work done comes from the battery.

23. Jun 19, 2008

### burashka

Within the constraints of the second law of thermodynamics, yes.

24. Jun 19, 2008

### burashka

25. Jun 19, 2008

### shadowpuppet

The fields cannot do work because the charge does not accelerate in the direction of the field, and will not accelerate at all if it is stationary. You fail to realize that moving charges also produce magnetic fields (such as those that produce electromagnets):

The net effect of the Lorentz force that you mentioned from the magnetic fields produced by the first moving charge on nearby charges moving in the same direction is to draw them closer. Two aligned magnets that have their opposite poles in proximity also have internal domain currents moving in the same direction, and so they are drawn closer. I hope that this answers your question.

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