# 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!

• – user4552
Aug 6, 2013 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.

• ''However, forces on magnetic dipoles and more general objects don't have the form v⃗ × - they're not perpendicular to v⃗ , so they do work in general.'' Doesn't the force acting on any more general object arises from the Lorentz force? and due to this any net magnetic force on any composite system never does work?
– Lost
Oct 13, 2020 at 20:13
• Thete is always an electric field doing the work. You just have to look for it. Griffiths explain it well in his textbook of electromagnetism.
– nasu
Nov 12, 2020 at 1:00
• "Magnetic moments may be visualized as objects with a forced motion of charges" So the intrinsic magnetic moment of the electron can be seen as moving charges? Nov 15, 2021 at 18:45

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 do work, 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.

• This answer seems correct as opposed to Luboš Motl's.
– Lost
Oct 13, 2020 at 20:15
• No, it's still electric field doing the work in the last instance. Even it may look otherwise.
– nasu
Nov 12, 2020 at 1:02

It is the electric field that does the work, not the magnetic field! When one has current in the loop, it can undergo a voltage drop or rise according to the inductance of the coil. Inductance relates to the electric field and its work.

"The induced electromotive force in any closed circuit is equal to the negative of the time rate of change of the magnetic flux enclosed by the circuit.[14][15]"

If the dipole is still, of course there is no work on it! When it moves, there is a change in the magnetic flux within the loop over the time of its motion resulting in an impulse (electromagnetic force times the time). This is classical physics. The electrical impulse changes the momentum of the current inside the loop, either gaining or losing energy according to the work done or received. If there is zero current, there is zero dipole!

Further even with an electron there is precession in a magnetic field, which results in additional motion of the charge elements of the electron in time. As the field goes from weaker to stronger, the precession of the electron increases precisely to accommodate the work done on it! See: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/larmor.html for detail!

• Inductance happens when there is changing magnetic field but there can be a magnetic force(as we call it) even without having a changing magnetic field. Mar 20 at 7:32

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.

• The counter EMF that you have neglected will cancel the work done by the magnetic field so that the field ends up doing no work. Feb 15, 2021 at 10:09

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.

The key theorem governing work and energy in classical electromagnetism is Poynting's theorem. To properly deal with a magnetic dipole we need to use the "macroscopic" version of Poynting's theorem which is less familiar to many. A good derivation is in section 11.2 here: https://web.mit.edu/6.013_book/www/book.html

$$-\nabla \cdot (E \times H) = \frac{\partial}{\partial t} \left( \frac{1}{2}\epsilon_0 E \cdot E + \frac{1}{2} \mu_0 H \cdot H \right) + E \cdot \frac{\partial P}{\partial t} + H \cdot \frac{\partial \mu_0 M}{\partial t} + E \cdot J$$

In this expression the term on the left and the first term on the right involve only the field. They represent the flow of energy and the storage of energy in the field respectively. The remainder of the terms include matter, so they can be considered different forms of work on matter.

In particular for this question is the term $$H \cdot \frac{\partial }{\partial t}\mu_0 M$$. The magnetization, $$M$$, is the density of magnetic moment. Thus a magnetic dipole is simply a Dirac distribution of $$M$$, so it is indeed possible for the magnetic field $$H$$ to do work on a magnetic dipole.

The reason that this question may be confusing is if one is trying to treat a magnetic dipole using the more well-known microscopic version of Poynting's theorem. It is always best to use the most appropriate laws for a situation rather than try to use awkward tools simply because they are more familiar.

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...