This question has been asked already, but no satisfying answer came up. All the answers seem to push the problem further, but do not explain clearly what is going on.
I will reformulate it for a specific case, tell you what I have gathered from the answers and what specifically I am expecting.
Setup: I have a powerful ferromagnet, and a paperclip lying on the table. I touch one pole of the ferromagnet to the clip, then lift the whole thing. The paperclip comes up with the ferromagnet.
When I pull on the ferromagnet, I am applying a force on it, thus doing work on it, providing the energy to lift both the magnet and the clip up the potential well.
The problem: The paperclip though experiences the pull of gravity and the pull of not of my hand, but the pull of the magnetic field. It seems to me the magnetic field is doing work on the clip.
However, the Lorentz force caused by a magnetic field on a moving particle is always perpendicular to the motion, so does no work. This is true also for constrained motion. The magnetic field cannot do work on the atoms of the paperclip.
So here we are. A paperclip is made of charged particles: electrons and nuclei. Magnetic fields cannot do work on charged particles. However, there is an undeniable attractive force on the paperclip by the magnet, and this force is capable of lifting it in a potential well, so it does work.
The answer I am looking for:
It seems to me that in the paradigm of classical mechanics + Lorentz forces + point-like charged particles this question has no consistent answer.
Two options:
1) You can prove me wrong, by showing me clearly how can the magnetic field of the magnet do work on an assembly of particles.
2) If I am right, then I would like to see what further notions are necessary $-$ be it treating the electrons as currents and using Maxwell's eqs or introducing quantum spins. I also want to see how these new concepts imply that the field from a magnet can do work on a piece of metal.
NB: you are allowed to treat both the paper clip and the magnet as infinite in extent in two of three spatial dimensions.
Best Answer
The important thing here is that a magnetic dipole, like a permanent magnetic or induced magnetism in ferrous material, produces a nonuniform field.
The potential energy of a magnetic dipole $\vec\mu$ in a magnetic field $\vec B$ is $$ U = - \vec \mu \cdot \vec B .$$ Most frequently (as in anna v's answer) this is used to explain the torque which causes the magnetic moment to want to align with the external field: the energy is minimized if $\vec \mu$ and $\vec B$ are parallel. Let's suppose they are aligned already; we find the force as $$ \vec F = -\nabla U.$$
However the gradient of a scalar product has a surprisingly complicated expansion, which you can verify by expanding all the component terms. $$ \nabla(\vec \mu\cdot \vec B) = (\vec\mu \cdot \nabla) \vec B + (\vec B \cdot \nabla) \vec \mu + \vec \mu \times (\nabla \times \vec B) + \vec B \times (\nabla \times \vec \mu) . $$ We can simplify this by consider $\vec\mu$ constant, so those gradients go away. From Maxwell's equations we have $\nabla\times\vec B=0$. Finally let's define our coordinate system so that $\vec\mu$ (and therefore $\vec B$, since we already assumed they are aligned) point along the $z$-axis. That leaves us with \begin{align} \vec F &= (\vec \mu \cdot \nabla)\vec B = \left(\mu \frac{\partial}{\partial z} \right) B_z \hat z . \end{align} So for a permanent dipole $\vec\mu$ in a field $\vec B$ we find three limiting cases:
If $\vec \mu, \vec B$ are parallel, the dipole will feel a force in the direction of increasing $|B|$
If $\vec\mu, \vec B$ are antiparallel, the dipole will feel a force in the direction of decreasing $|B|$
If $\vec\mu, \vec B$ are not parallel, the dipole will feel a torque that makes it want to align with the field.
This is pretty much my experience with permanent magnets. To get permanent magnets to repel, you have to constrain their rotation somehow; what they like to do is to flip around and attract. Induced magnetism (e.g. paperclips) is the result of many microscopic aligning torques.
You can find the same result by noting that the energy stored in a volume element of magnetic field is $dU = (\vec H \cdot \vec B) d^3x$, and finding the arrangement of magnets which minimizes the volume of strong field. This is relatively intuitive for parallel dipoles aligned end-to-end, which have a strong field in the empty space between the magnets, and also for antiparallel dipoles set side-to-side, where the "return fields" between the dipoles add. To see the repulsive cases, though, you have to do a messy integral over the fringe fields to confirm that the distant-dipole configurations have less stored energy than the near-but-not-overlapping-dipole configurations.
As for the argument that the Lorentz force $$ \vec F = \frac{d\vec p}{dt} = \frac qm \vec p \times \vec B $$ can do no work, because the force is perpendicular the the momentum and therefore cannot change the magnitude of $|p|$: this argument assumes that the field $\vec B$ seen by the particle is uniform. If $\vec B$ varies along the particle's path, the particle (in its rest frame) sees a time-varying $\vec B$ and an electric field which obeys $-\frac{\partial\vec B}{\partial t} = \nabla \times \vec E$. It's the induced electric field that does the work. There's a nice problem in Griffiths's E&M textbook that works through the argument.