# Does a magnetic field do work on an intrinsic magnetic dipole?

When you release a magnetic dipole in a nonuniform magnetic field, it will accelerate.

I understand that for current loops (and other such macroscopic objects) the magnetic moment comes from moving charges, and since magnetic fields do no work on charges (F perpendicular to v) it follows that the work done on the dipole (that caused its gain in kinetic energy) must have come from somewhere other than the magnetic forces (like electric forces in the material).

However, what about a pure magnetic moment? I'm thinking of a particle with intrinsic spin. Of course such a thing should be treated with quantum mechanics, but shouldn't classical electrodynamics be able to accommodate a pure magnetic dipole? If so, when I release the pure dipole in a nonuniform B-field and it speeds up, what force did the work? Is it correct to say that magnetic fields DO do work, but only on pure dipoles (not on charges)? Or should we stick with "magnetic forces never do work", and the work in this case is done by some other force (what?)?

Thanks to anyone who can alleviate my confusion!

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–  Ben Crowell Aug 6 '13 at 4:36

Yes, of course that if a field - magnetic field - is able to make a bar magnet rotate or move, it is doing work. The statement that magnetic fields don't do any work only applies to point-like pure electric charges.

Magnetic moments may be visualized as objects with a forced motion of charges (solenoids have the same magnetic field as bar magnets), and if something is moving, the magnetic force is becoming a force that does work.

In terms of formulae, the magnetic force on a charge is $q\vec v\times \vec B$ which is identically perpendicular to $\vec v$ and that's why it does no work. However, forces on magnetic dipoles and more general objects don't have the form $\vec v\times$ - they're not perpendicular to $\vec v$, so they do work in general.

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Thanks! That makes sense if it's true. But I thought that maybe since the formula for the force on a dipole $\vec F=\vec\nabla (\vec \mu \cdot \vec B)$ is usually derived by considering some limit of a charge/current distribution, it would somehow inherit the "no work" property of the Lorentz force law. How do you derive the forces on the "more general objects" you mentioned? Is there something more fundamental or different than the Lorentz force law? –  Joss L May 30 '11 at 4:33
@Lubos Motl: I disagree. You can construct a pure magetic dipole from a vanishingly small current loop. And because of the Lorentz law, no work can be done. Also, because no magnetic monopoles have been discovered, all magnetic sources in our world have to come from moving charges on which no work can be done. –  Kasper Meerts May 30 '11 at 21:00
Dear @kmm, what you write is silly, indeed. The objects known as "magnets" - such as iron, you know, they are ferromagnets, some children have already heard about it - have most of their magnetism coming from the spin of the electron that is not moving at all. Again, it is nonsense that one cannot do work with magnetic field. Magnetic moments in magnetic fields are clearly doing work. What matters is a nonzero $\vec j$, and it can be nonzero with arbitrarily small loops, including "vanishingly small" ones, as the spin example shows. –  Luboš Motl Jun 1 '11 at 9:36
@Luboš Motl: Could you be a little less condescending, it's very childish. Spin is a purely quantum mechanical phenomenon and because we were talking about work, I presumed everything was classical. Work isn't well defined at quantum level so I still stand by my point, it's ridiculously easy to prove that magnetic fields can never do work. –  Kasper Meerts Jun 1 '11 at 18:44
@lubos doesn't the work come from the internal energy in the bar magnet? –  John McVirgo Jun 3 '11 at 16:50

I'm going to take a risk and try to answer this, even though my answer is different to Lubos's and he does have a reputation that is overwhelming right compared to mine.

Static Magnetic fields don't don't do work, period (us), full stop (uk). So the work comes from the magnetic dipole itself whose internal energy is affected by the external force that positioned it in the static magnetic field in the first place.

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Your answer might be different, but that's good because you're right and he's wrong. –  Kasper Meerts May 30 '11 at 20:52
An electron has a magnetic dipole moment and, because of that, can be accelerated by a non-uniform static magnetic field. Then, if an initially static electron acquires a kinetic energy using this method, are you suggesting it comes from its internal energy (i.e. rest mass)? –  mmc May 31 '11 at 3:30
@mmc, I was thinking more of a a loop of wire carrying a current where internal electric fields are created to maintain its orientation in a static magnetic field, from an applied force. It doesn't apply to an electron because it's not seen as a spinning charge and hence current loop in the classical sense. –  Larry Harson May 31 '11 at 10:59
Someone actually told me that the rest mass of an electron will change when it's placed in a B field, and this is where the kinetic energy comes from. He was in a hurry and I'm going to talk to him later, but he's a professor and seems like he knows what he's talking about. –  Joss L May 31 '11 at 14:46
"So the work comes from the magnetic dipole itself whose internal energy is affected by the external force that positioned it in the static magnetic field in the first place." I can't make any sense out of this. What form of internal energy would this be? –  Ben Crowell Aug 6 '13 at 4:35

Is it correct to say that magnetic fields DO do work?

Yes! I show this quantitatively:

Each charged particle experiences action of magnetic force. This force is transmitted to a conductor in which the charges move. As a result, the magnetic field acts with a certain force on the current-carrying conductor. Let the volume charge density, (electrons in a metal, for example) is equal to $\rho$. Let distinguish a mental element of volume $dV$ of the conductor. There is a charge equal to $\rho dV$. Then the force acting on the element $dV$ of the conductor can be expressed by the Lorenz formula $\overrightarrow{F}=q(\overrightarrow{v}\times\overrightarrow{B})$ in the form:

$$\overrightarrow{dF}=\rho (\overrightarrow{v}\times\overrightarrow{B})dV$$

Since $\overrightarrow{j}=\rho\overrightarrow{v}$ where $\overrightarrow{j}$ is the current density vector we can write:$$\overrightarrow{dF}=(\overrightarrow{j}\times\overrightarrow{B})dV$$ If the current flows through a thin conductor, then the following holds:$$\overrightarrow{j}dV= \overrightarrow{dl}I$$ where I is a current in a thin conductor(wire) and $\overrightarrow{dl}$ is the vector of an element of the wire in direction of the current. Thus:

$$\overrightarrow{dF}=I(\overrightarrow{dl}\times\overrightarrow{B})$$ This is nothing more than Ampere's force. So the resulting Ampere's force acting on the contour of the current (current loop) in the magnetic field is determined as a line integral along the current loop:

$$\overrightarrow{F}=I\oint(\overrightarrow{dl}\times\overrightarrow{B})$$ If the magnetic field is nonuniform, then the integral is generally different from zero.

A conclusion:

It follows directly from the Lorenz law that magnetic fields do work on a current loop.

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Since Newton's third law doesn't hold for magnetic forces, how do you show the sum of the internal forces for the loop sum to zero? If you can't do this, then I don't see how you can sum the external forces as the force on the centre of mass of the current loop. –  John McVirgo May 31 '11 at 12:14
This seems to contradict Griffith's book. The formulas are fine, but your conclusion that magnetic forces do work on conductors is the opposite of what he says, I think. I'll have to take at his book again later and think about this some more. –  Joss L May 31 '11 at 14:42
@John McVirgo: The loop is made of a conductor. We are interested in forces applied to the conductor. Newton's third law applies. –  Martin Gales Jun 1 '11 at 6:42
@Joss L: In an electric motor, what other than the magnetic forces do work on current-carrying conductors? –  Martin Gales Jun 1 '11 at 6:43
@John McVirgo: Consider an extremely trivial example: A current(an electron) moves along a straight, long, thin, superconducting stationary wire with initial velocity $v_0$. Magnetic field is off. Now a static magnetic field, perpendicular to the direction of the wire is switched on for a while and then switched off. After this, the electron acquires a velocity $v$ along the wire and the wire acquires a velocity $u$, perpendicular to the direction of the wire. Because the magnetic field will not affect the energy of the electron, its speed(relative to the field) must remain the same, i.e. ... –  Martin Gales Jun 6 '11 at 8:33

Note that the failure of the “rest mass” m to be constant resolves a paradox concerning what one is taught in elementary physics courses: On one hand, one is (correctly) taught that an external magnetic ﬁeld can “do no work” on a body, so a body moving in an external magnetic ﬁeld cannot gain energy. On the other hand, one is (also correctly) taught that a magnetic dipole released in a non-uniform external magnetic ﬁeld will gain kinetic energy. Where does this kinetic energy come from? Equation (B6) shows that it comes from the rest mass of the body.

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Larry Harson's answer, although short, is right, to my point of view. How I explain it doesn't fit in a comment so I make a separate answer.

Consider a simple magnetic moment, familiar to those who study space plasmas: the electron, trapped in earth's magnetosphere, precisely in the Van Allen Radiation belt. Let B_0>0 be a static homogenous field oriented towards the z direction. If you put an electron with initial speed Vx>0, Vy and Vz being zero, the lorentz force will make the electron circle in the magnetic field, and if you look to the trajectory of the electron from above (z being the vertical, "up", direction, oriented towards the sky), it will rotate counter-clockwise. As the electron has negative charge, this will create a magnetic dipole, oriented against the external magnetic field B_z. Here you come to the result that free electrons in a static homogenous magnetic field are essentially "diamagnetic".

In Earth's magnetosphere, the electrons circle around magnetic flux lines, their speed along these lines being kept constant if the norm of the magnetic field is constant. They "follow" magnetic flux lines.

Now, if the electrons try to get too close to Earth, the magnetic field will intensify if you follow a magnetic flux line, because of the general shape of the magnetic flux lines of a dipole, that get closer to each other the closer you get to the source. To sketch this from the point of view of perturbation theory, imagine that, together with the static B_0 field defined later, you add a centripetal field, B_r<0. Imagine that you are approaching earth's south pole from outer space. Now, the electron will undergo two forces. If it has no V_z speed, then the B_r field will create a force oriented towards -Oz, that will superpose to the centripetal force of the B_0 field. "down", towards outer space. This force is what keeps the electrons from messing with our atmosphere too much: only the most energetic particles reach earth's atmosphere, those that came with a big enough V_z speed that the converging magnetic field could not stop. This is what creates Auroras at earth's poles.

Then, let us consider the process of this "brake" of electrons when approaching the south pole. Now, they have a non-negligible V_z speed. Look closely: because of the B_r component, it will speed their rotation up! Their V_z speed is only converted to V_x/V_y speed. The same process will perform in reverse, when the electron is finally reflected to outer space: they will lose rotational speed, and gain longitudinal speed. Of course, their kinetic energy is conserved.

If you look now to what happens for magnetic dipoles in a magnet, it will be a little bit different. In space, electrons had to be diamagnetic, as their motion was determined by the magnetic field. But if you look at, say the orbital momentum of an electron around an atom, then it is the electric force that makes it circle: it can circle clockwise or counter-clockwise, now. But the magnetic field will add a force, either centrifugal or centripetal, that will shift the energy levels. As we enter now the domain of quantum mechanics, I cannot explain formally without mistake what happens, but the remark about the B_r component of the field slowing down the "rotation" still seem to hold. If you consider a ferromagnetic atom as having its dipolar momentum oriented towards the magnetic field ("clockwise motion" in my model), then the magnetic force is centrifugal, and compensates a part of the electric field. For a given "radius", the equilibrium speed of the electron is thus lower (in the Rutherford model), which would tend to correspond to a "lower energy level". But what happens exactly in the shifts I do not know, one would need to look at these energy level shifts from the point of view of spectroscopy, to measure them and see indeed that, the higher the magnetic field when it is aligned with a dipole, the lower the intrinsic energy of the dipole.

So, finally, when you pull on a magnet, you are just compensating for all the atomic energy levels shifting up, the number of atoms being kept constant the overall energy of your magnet rises. Where to count the rise in the integrated energy of the magnetic field itself, I do not know for certain (as when two magnets are apart, the integral of B^2/(2*µ0) is usually higher than when they are "how they want to be", glued to one another).

One last thing about macroscopic loops of current: you need an electromotive force to make a motor move, the current tends to decrease by itself when a current loop moves according to Laplace's force, just as it increases by itself if it moves against this force (induction, NdFeB magnet falling in an almost adjusted copper tube, very funny experiment). If you see electrons as "bouncing on the walls of your copper wire", it is all about momentum conservation: if the electron gives momentum to the cristalline structure of the copper wire grain, it loses momentum itself. It will need to be reaccelerated by the electric field, to give this momentum again. So now, it is about the speed, gained from the electric field, that is converted to momentum of the macroscopic wire. The magnetic field is just "the man in the middle" that makes possible the transaction.

I hope what I said is clear...

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It's not true that you may represent magnets via orbital angular momentum. Both ferromagnetism and paramagnetism depend on the existence of the spin - the internal angular momentum that, according to quantum mechanics, is carried even by point-like particles - and the spin can't be explained by any charges' velocities. The spin would be exactly zero in classical physics. So there exists no accounting in which the motion of the magnet would come from the change of the internal energy: the state is exactly the same. After all, the spin is quantized and can't be "raised a bit". –  Luboš Motl May 30 '12 at 10:42
Yes, about spin precisely this can't hold "as is". However, although the quantum of spin itself, and hence the quantum of magnetic field generated by one electron, do not change, maybe something in the internal energy of the electron itself changes. I mean: why would magnetic fields work on NOTHING but spins...just because we can't model it down enough, and see how we can fraction the energy of an electron...I agree, this spin problem is a big one, and unfortunately I don't know enough in quantum theory to find any answer to that. –  MrBrody May 30 '12 at 19:26
Nothing can change about the "internal energy" of the electron. Every electron is always the same and fully described by its momentum $\vec p$ and the spin $\vec s$ and the energy of the electron only depends on the former, the momentum. The magnetic moment is $\vec\mu = C\vec s$ with a fixed constant $C$. This "preservation" of the full magnetic moment isn't necessarily a special quantum mechanical feature: a permanent magnet is made out of many spins like that and the total magnetic moment of a bar magnet is also preserved in those exercises, unless you work hard to damage or demagnetize it. –  Luboš Motl Sep 19 '12 at 11:53