Can magnetic fields do positive work in certain circumstances?

[haruspex]

But (in the right circumstances) a magnetic field is a non-conservative force which can do positive work.

Are you saying there are circumstances in which the work done in a magnetic field depends on the path taken?

 

[Steve4Physics]

Are you saying there are circumstances in which the work done in a magnetic field depends on the path taken?

A magnetic field is conservative is certain situations. But if there are currents (or time-varying electric fields) a magnetic field is generally non-conservative.

How about a simple electric motor. After each complete rotation, the displacement of every part of the motor is zero – the same as if there had been no movement. But the fields have done positive work, e.g. raising a load, during the rotation.

Similarly a non-conservative magnetic field does negative work each complete rotation in a dynamo or magnetic braking system for example.

 

[haruspex]

A magnetic field is conservative is certain situations. But if there are currents (or time-varying electric fields) a magnetic field is generally non-conservative.

How about a simple electric motor. After each complete rotation, the displacement of every part of the motor is zero – the same as if there had been no movement. But the fields have done positive work, e.g. raising a load, during the rotation.

Similarly a non-conservative magnetic field does negative work each complete rotation in a dynamo or magnetic braking system for example.

Ah, you are varying the magnetic field. Could do the same with gravity, or electrostatics...

 

[etotheipi]

From the Maxwell equation$$\nabla \times \mathbf{B} = \frac{1}{c} \left( \mathbf{j} + \frac{\partial \mathbf{E}}{\partial t} \right)$$if $\mathbf{j} = \mathbf{0}$ and $\partial \mathbf{E} / \partial t = \mathbf{0}$, then $\mathbf{B} = \nabla \varphi$ for some $\varphi: \mathbb{R}^3 \rightarrow \mathbb{R}$. This is the sense in which a magnetic field can be described as conservative.

However, magnetic forces never do work, no-matter whether the field is conservative or non-conservative. This is clear from taking the inner product of the magnetic force acting on an element of charge with its velocity$$\mathcal{P} = \delta q[\dot{\boldsymbol{x}} \times \mathbf{B}] \cdot \dot{\boldsymbol{x}} = 0$$In a motor, the work is being done by an electric (Laplace) force, not a magnetic force.

 

[Steve4Physics]

Are you saying there are circumstances in which the work done in a magnetic field depends on the path taken?

Take a current carrying wire..

Ah, you are varying the magnetic field. Could do the same with gravity, or electrostatics...

You can make a motor with a permanent magnet and a constant current (i.e. with constant magnitude magnetic fields). I don’t think you can do the equivalent with constant magnitude electric (or gravitational) fields.

I’ve just seen @etotheipi’s post #8. I agree a magnetic field does no work on a charge. F = qvXB: the force is always perpendicular to the velocity, so the force does no work.

I’m not familiar with the Laplace force. I always thought you could treat the force between two current-carrying conductors as an interaction between their magnetic fields. And I can't see how a field can be described as non-conservative if it always does zero work. I need to do some reading-up on the Laplace force. However, it’s my bedtime here in the UK, so I wish you all a good-night!

 

[etotheipi]

In Lorentz-Heaviside units with $c=1$, the energy density $u$ of the electromagnetic field is$$u = \frac{1}{2} \left(|\mathbf{E}|^2 + |\mathbf{B}|^2 \right)$$Now consider a region $\Omega$ with current density $\mathbf{j}(\boldsymbol{x}) = \rho(\boldsymbol{x}) \mathbf{v}(\boldsymbol{x})$. Given that $\nabla \times \mathbf{B} = \mathbf{j} + \frac{\partial \mathbf{E}}{\partial t}$, the power of the Lorentz force per unit volume is $$\begin{align*}
p = \mathbf{E} \cdot \mathbf{j} &= \mathbf{E} \cdot (\nabla \times \mathbf{B}) - \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t} \\ \\

&= \left( - \mathbf{B} \cdot \frac{\partial \mathbf{B}}{\partial t} - \nabla \cdot (\mathbf{E} \times \mathbf{B}) \right) - \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t} \\ \\

&= - \frac{1}{2} \frac{\partial}{\partial t} \left( |\mathbf{E}|^2 + |\mathbf{B}|^2 \right) - \nabla \cdot (\mathbf{E} \times \mathbf{B})
\end{align*}$$Define $\mathbf{S} := \mathbf{E} \times \mathbf{B}$ as the Poynting vector, then you obtain Poynting's theorem:$$- \frac{\partial u}{\partial t} = \nabla \cdot \mathbf{S} + \mathbf{E} \cdot \mathbf{j} \quad \implies -\frac{dU}{dt} = - \frac{d}{dt} \int_{\Omega} u d\tau = \int_{\partial \Omega} \mathbf{S} \cdot d\mathbf{A} + \int_{\Omega} \mathbf{E} \cdot \mathbf{j} d\tau$$The energy contained in the electromagnetic field inside a region $\Omega$ decreases if energy flows out of the boundary [the first term] or does work on charges inside $\Omega$ [the second term]; the work done on the charges only depends on $\mathbf{E}$, not $\mathbf{B}$.

Now we consider an example of a wire carrying current $\mathbf{j}$ in a magnetic field $\mathbf{B}$. Surface charges in the wire produce an electric field $\mathbf{E}$. In steady state you may write$$\rho \mathbf{E} + \mathbf{j} \times \mathbf{B} = \mathbf{0}$$Here $\mathbf{j} \times \mathbf{B}$ is the external magnetic force density exerted on the charge carriers and $\rho \mathbf{E}$ is the electric force density exerted by the surface charges on the charge carriers. By Newton's third law, the charge carriers exert an electric force density $- \rho \mathbf{E} = \mathbf{j} \times \mathbf{B}$ on the surface charges. The electric force on the whole wire is$$\mathbf{F}_{\text{electric}} = \int_{\mathcal{V}} \mathbf{j} \times \mathbf{B} d\tau = I\boldsymbol{\mathscr{L}} \times \mathbf{B}$$where $\boldsymbol{\mathscr{L}} = \mathscr{L} \hat{\mathbf{n}}$ and $\mathbf{j} = (I/A) \hat{\mathbf{n}}$. The work done on the wire as it moves through an external magnetic field, is electric in nature.

 

[Steve4Physics]

Now we consider an example of a wire carrying current $\mathbf{j}$ in a magnetic field $\mathbf{B}$. Surface charges in the wire produce an electric field $\mathbf{E}$. In steady state you may write$$\rho \mathbf{E} + \mathbf{j} \times \mathbf{B} = \mathbf{0}$$Here $\mathbf{j} \times \mathbf{B}$ is the external magnetic force density exerted on the charge carriers and $\rho \mathbf{E}$ is the electric force density exerted by the surface charges on the charge carriers. By Newton's third law, the charge carriers exert an electric force density $- \rho \mathbf{E} = \mathbf{j} \times \mathbf{B}$ on the surface charges. The electric force on the whole wire is$$\mathbf{F}_{\text{electric}} = \int_{\mathcal{V}} \mathbf{j} \times \mathbf{B} d\tau = I\boldsymbol{\mathscr{L}} \times \mathbf{B}$$where $\boldsymbol{\mathscr{L}} = \mathscr{L} \hat{\mathbf{n}}$ and $\mathbf{j} = (I/A) \hat{\mathbf{n}}$. The work done on the wire as it moves through an external magnetic field, is electric in nature.

Interesting! Alas, I couldn’t follow all of your analysis, especially the references to surface charges. Did you mean ‘fixed’ charges?

From what I’ve seen, the Laplace force (F = IℓxB) is generally referred to as a ‘magnetic force’. Your analysis indicates this is a misnomer.

Empirically it seems to make sense to refer to the Laplace force a magnetic force simply because it is observed to act on a current-carrying conductor (having no net charge) in a magnetic field.

Perhaps it might be preferable to say something like: ‘the Laplace force arises from the Lorentz force, mediated by the electric force’.

To go back to the original title of post #1 ‘Is Nonconservative work negative?”, consider the operation of a simple motor, e.g. Michael Faraday’s:
https://www.researchgate.net/profile/Konstantin-Weise-nee-Porzig/publication/263616041/figure/fig2/AS:296550896029702@1447714651457/Faradays-motor-principle.png

I would say positive nonconservative work is done. I don’t know how this could be explained in terms of (conservative) electric fields.

 

[etotheipi]

In a motor, because you have a (moving) current distribution which is producing its own magnetic fields (in addition to the applied external field), I think $\partial \mathbf{B} / \partial t$ will be non-zero and the electric field non-conservative, $\nabla \times \mathbf{E} \neq \mathbf{0}$.

Let's try and think about it in some more detail. In a motor, you have a current loop through which there is a flux$$\Phi(t) = BS \cos{\varphi(t)}$$which means that there is induced an EMF$$\mathcal{E} = - \dot{\Phi} = BS \dot{\varphi}(t) \sin{\varphi(t)}$$The work done by the power supply per unit time to fight against this motional EMF is$$\frac{dW}{dt} = \mathcal{E}I = BIS \dot{\varphi}(t) \sin{\varphi(t)}$$How does this compare to what we'd get with a moment analysis? The moment of the Laplace force on the loop is$$M = 2\cdot [r\sin{\varphi(t)}] \cdot [BIL] = BIS \sin{\varphi(t)}$$where $S = 2rL$, and the rate at which it does work is simply$$M \cdot \dot{\varphi}(t) =BIS \dot{\varphi}(t) \sin{\varphi(t)} $$which is equal to the work done by the power supply.

So, fundamentally, it's the power supply that is providing the rotational energy to the coil. We can think of the magnetic field as "mediating" the transfer of energy, but magnetic forces specifically never do work.

 

[Steve4Physics]

In a motor, because you have a (moving) current distribution which is producing its own magnetic fields (in addition to the applied external field), I think ∂B/∂t will be non-zero and the electric field non-conservative, ∇×E≠0.

.So, fundamentally, it's the power supply that is providing the rotational energy to the coil. We can think of the magnetic field as "mediating" the transfer of energy, but magnetic forces specifically never do work.

That's a good explanation. Thank you.

But I'm still uncomfortable with the idea that "magnetic forces specifically never do work". For example:
- 2 permanent magnets attract and accelerate towards each other;
- a dipole in a non-uniform magnetic field experiences both a force and a torque, each of which can do work;
- a magnetic monopole (if you are prepared to accept one hypothetically) will accelerate in a magnetic field.

Must such processes always be explained in terms of underlying electric fields? I don’t like it! (to be said in the manner/voice of Victor Meldrew’s catchphrase ‘I don’t believe it’, if you are familiar with the old BBC TV series ‘One Foot in the Grave'!)

 

[etotheipi]

If magnetic monopoles $q_m$ exist, then the Lorentz force equation gain an extra term $q_m(\mathbf{B} - \dot{\boldsymbol{x}} \times \mathbf{E})$, so magnetic fields would be able to do non-zero work. However, magnetic monopoles are not consistent with $\nabla \cdot \mathbf{B} = 0$, so I'm not sure if it makes much sense to worry about them.

Two permanent magnets is quite a messy calculation, so might be simpler to focus on the dipole in the magnetic field. A dipole is just a current loop, so the same arguments as for the wire would apply.

As always, the magnetic field does no work; if you look closely enough, you'll find another force doing the work. But pedagogically, I'm not sure how useful it is to worry about these subtleties

 

[Steve4Physics]

All very informative! Thanks for taking the time.

 

[kuruman]

As always, the magnetic field does no work; if you look closely enough, you'll find another force doing the work.

I always thought that when a permanent magnet that is fixed in space attracts a piece of iron which then accelerates, the (non-uniform) magnetic field from the magnet does work on the accelerating piece. How close do I need to look to ascertain that this is not the case and that some other force is responsible?

 

[etotheipi]

I don't know, but it is an electric field, not magnetic field, otherwise you contradict electromagnetism theory. I give up trying to figure out exact mechanism.

 

[kuruman]

I don't know, but it is an electric field, not magnetic field, otherwise you contradict electromagnetism theory. I give up trying to figure out exact mechanism.

This paper addresses the issue. I only scanned through its contents and concluded that I do not have the patience, the time or the expertise to ascertain the correctness of its conclusions. The author's final statement is worth quoting for it provides a counterpoint to the 32 or so pages of document: "Magnetic forces can do work. In this paper, we have shown how."

 

[etotheipi]

On the other hand, experiments indicate that many kinds of particles, including electrons, possesses permanent, elementary magnetic dipole moments that do not seem to arise from underlying classical loops of current or as pairs of magnetic monopoles. At a truly fundamental level, these elementary magnetic dipole moments are quantum-mechanical in nature, but, then, so is electric charge, and we obviously still include electric charges as basic sources in Maxwell’s classical theory of electromagnetism. It is therefore worth studying how we might similarly include elementary dipoles as basic sources in the classical theory of electromagnetism, as well as determine from first principles how they should interact with electric and magnetic fields—without assuming the textbook Lorentz force law (19) as one of our starting ingredients

Such "elementary magnetic dipole moments" are not in the framework of classical EM, as far as I know. In classical EM, all magnetic dipole moments must be treated as the limit of a current carrying loop, and there is no such thing as "intrinsic magnetic dipole".

But maybe I will understand him better if I read it.

 

[etotheipi]

especially the references to surface charges. Did you mean ‘fixed’ charges?

By the way, I never replied to this.

I don't mean the metal ions, if that's what you inferred. I'm talking about a surface charge density $\sigma$ due to excess electrons which accumulate at the boundary of the wire, which arises in this case due to the Hall effect.

There's some more about surface charges here:
https://aapt.scitation.org/doi/abs/10.1119/1.18112?journalCode=ajp

 

[PeroK]

I don’t like it! (to be said in the manner/voice of Victor Meldrew’s catchphrase ‘I don’t believe it’, if you are familiar with the old BBC TV series ‘One Foot in the Grave'!)

I went to the local barber's one day and Richard Wilson was in the chair before me! I nearly said "I don't believe it!", but I bit my tongue.

 

[Dale]

However, magnetic forces never do work, no-matter whether the field is conservative or non-conservative.

But I'm still uncomfortable with the idea that "magnetic forces specifically never do work". For example:
- 2 permanent magnets attract and accelerate towards each other;
- a dipole in a non-uniform magnetic field experiences both a force and a torque, each of which can do work;
- a magnetic monopole (if you are prepared to accept one hypothetically) will accelerate in a magnetic field.

I tend to focus on Poynting’s theorem in this sort of discussion. If you look at Poynting’s theorem for Maxwell’s microscopic equations it is:
$$\nabla \cdot \frac{1}{\mu_0}(\vec E \times \vec B)+\frac{1}{2}(\epsilon_0 E^2+\frac{1}{\mu_0}B^2) + \vec E \cdot \vec J=0$$ From this you can see that the work done on matter is $\vec E \cdot \vec J$ so the magnetic field does no work.

But Poynting’s macroscopic theorem is: $$\nabla \cdot ( \vec E \times \vec H) + \frac{\partial}{\partial t}(\frac{1}{2}\epsilon_0 E^2) + \vec E \cdot \frac{\partial}{\partial t}\vec P + \frac{\partial}{\partial t}(\frac{1}{2}\mu_0 H^2) + \vec H \cdot \frac{\partial}{\partial t} \mu_0 \vec M + \vec E \cdot \vec J_f =0$$ So in this case the magnetic field can do work via the term $\vec H \cdot \frac{\partial}{\partial t} \mu_0 \vec M$

 

[etotheipi]

It's interesting, but there's something I don't understand, specifically aren't these two laws equivalent? In which case I'm not sure if it makes sense to say one form of the equation allows the magnetic field to do work but the other doesn't. We can start, again using Heaviside-Lorentz units along with $c=1$, with the microscopic law$$-\frac{\partial}{\partial t} \left( \frac{1}{2} |\boldsymbol{E}|^2 + \frac{1}{2}|\boldsymbol{B}|^2 \right) = \nabla \cdot (\boldsymbol{E} \times \boldsymbol{B}) + \boldsymbol{E} \cdot \boldsymbol{j}$$Letting $\boldsymbol{j} = \boldsymbol{j}_{F} + \partial_t \boldsymbol{P} + \nabla \times \boldsymbol{M}$ as well as $\boldsymbol{B} = \boldsymbol{H} + \boldsymbol{M}$ we obtain$$\begin{align*}- \frac{\partial}{\partial t} \left( \frac{1}{2} |\boldsymbol{E}|^2 + \frac{1}{2} |\boldsymbol{H}|^2 + \boldsymbol{H} \cdot \boldsymbol{M} + \frac{1}{2} |\boldsymbol{M}|^2 \right) = \nabla \cdot (\boldsymbol{E} \times \boldsymbol{H}) &+ \nabla \cdot (\boldsymbol{E} \times \boldsymbol{M}) \\ &+ \boldsymbol{E} \cdot (\boldsymbol{j}_{F} + \frac{\partial \boldsymbol{P}}{\partial t} + \nabla \times \boldsymbol{M})\end{align*}
$$But we note that $\nabla \cdot (\boldsymbol{E} \times \boldsymbol{M}) = (\nabla \times \boldsymbol{E}) \cdot \boldsymbol{M} - \boldsymbol{E} \cdot (\nabla \times \boldsymbol{M})$, and the previous becomes$$\begin{align*}
-\frac{\partial}{\partial t} \left( \frac{1}{2} |\boldsymbol{E}|^2 + \frac{1}{2} | \boldsymbol{H}|^2 \right) - \boldsymbol{M} \cdot \frac{\partial \boldsymbol{M}}{\partial t} - \boldsymbol{M} \cdot \frac{\partial \boldsymbol{H}}{\partial t} - \boldsymbol{H} \cdot \frac{\partial \boldsymbol{M}}{\partial t} = \nabla \cdot & \, (\boldsymbol{E} \times \boldsymbol{H}) + \boldsymbol{M} \cdot (\nabla \times \boldsymbol{E}) \\ &+ \boldsymbol{E} \cdot \boldsymbol{j}_F + \boldsymbol{E} \cdot \frac{\partial \boldsymbol{P}}{\partial t}
\end{align*}$$From Maxwell III, as well as $\boldsymbol{B} = \boldsymbol{H} + \boldsymbol{M}$, we know that$$\boldsymbol{M} \cdot \left( \frac{\partial \boldsymbol{M}}{\partial t} + \frac{\partial \boldsymbol{H}}{\partial t} \right) + \boldsymbol{M} \cdot (\nabla \times \boldsymbol{E}) = \boldsymbol{M} \cdot \frac{\partial \boldsymbol{B}}{\partial t} - \boldsymbol{M} \cdot \frac{\partial \boldsymbol{B}}{\partial t} = \boldsymbol{0}$$so the equation reduces to$$-\frac{\partial}{\partial t} \left( \frac{1}{2} |\boldsymbol{E}|^2 + \frac{1}{2} | \boldsymbol{H}|^2 \right) = \boldsymbol{H} \cdot \frac{\partial \boldsymbol{M}}{\partial t} + \nabla \cdot (\boldsymbol{E} \times \boldsymbol{H}) + \boldsymbol{E} \cdot \boldsymbol{j}_F + \boldsymbol{E} \cdot \frac{\partial \boldsymbol{P}}{\partial t}$$which is the macroscopic form of the law.

It seems to me like the $\boldsymbol{H} \cdot \partial_t \boldsymbol{M}$ might be better described as just another a term contributing to the energy flow, and not as magnetic work. But, I'm not sure what is the terminology in the literature

 

[hutchphd]

H is not the magnetic field, nor does it appear in the Lorentz force.

 

[Dale]

aren't these two laws equivalent? In which case I'm not sure if it makes sense to say one form of the equation allows the magnetic field to do work but the other doesn't.

Well, they are at different levels of abstraction. If you can keep track of all of the microscopic quantities then you use the microscopic form. If you cannot keep track of all of the microscopic quantities then you have to abstract them away in the $\vec M$ and $\vec P$ vectors.

You may know that at the microscopic level work is $\vec E \cdot \vec J$ but that doesn’t help you if you don’t know those microscopic details. At the macroscopic level you may as well say the magnetic field does work. You cannot determine the work done otherwise.

See section 11.2 here for the derivation
https://web.mit.edu/6.013_book/www/book.html

just another a term contributing to the energy flow, and not as magnetic work

It is energy that flows out of the field and into the matter. That is a transfer of energy, and hence it is indeed work.

 

[Vanadium 50]

"Magnetic fields do no work" is one of those tropes that produces more confusion than clarity. If I hold a magnet over a piece of steel, it jumps up, so clearly something happened.

Magnetic fields do no work on electric monopoles (which we call "charges") but would do work on magnetic monopoles, but they seem not to exist. All the work that magnetic fields seem to do on real materials can ultimately be traced to electric forces in these materials (made of atoms, ultimately) but this is rarely useful. It allows you to keep "magnetic fields do no work" but it is not something you calculate with.

Often the best way to understand this is to look at energetics. The energy in a magnetic field goes as B², not H². The magnetization arises to try and cancel B, and the forces are gradients of the energies.

 

[Dale]

One other thing to consider is how inter-related the EM quantities are. The work term $E \cdot J$ includes all the microscopic work done on matter, including but not limited to the Ohmic losses. So I wanted to see if any further analysis of that term could be done to pull out the Ohmic losses.

If a piece of material is moving at velocity $v<<c$ then we can make the following transformations:
$E' = E + v \times B$
$J' = J - \rho v$
where the primes indicate values in the reference frame where the material is at rest.

Substituting those in we have:
$E \cdot J = (E' - v \times B) \cdot (J' + \rho v)$
$=E' \cdot J' + E' \cdot \rho v - (v \times B) \cdot J' - (v \times B) \cdot \rho v$
$=E' \cdot J' + \rho v \cdot (E+ v \times B) - (J-\rho v) \cdot (v \times B)$
$=E' \cdot J' + \rho v \cdot E - J \cdot (v \times B)$

So finally we end with
$E \cdot J=E' \cdot J' + v \cdot (\rho E + J \times B)$

Since $E'$ and $J'$ are in the reference frame of the material, I would interpret the term $E' \cdot J'$ as being Ohmic losses. Then $E \cdot J$ contains not only the Ohmic losses, but also the material velocity times the Lorentz force density, which I would call the Lorentz power density.

I don't have a reference for this, so I don't know if someone else has already done this derivation, but I thought you might like to see it. It indicates that even with the microscopic Poynting’s theorem you can conclude that magnetic fields can do work. If you dig a bit deeper.

 

[etotheipi]

See section 11.2 here for the derivation
https://web.mit.edu/6.013_book/www/book.html

By the way, this looks like a really nice EM textbook!

 

[vanhees71]

Such "elementary magnetic dipole moments" are not in the framework of classical EM, as far as I know. In classical EM, all magnetic dipole moments must be treated as the limit of a current carrying loop, and there is no such thing as "intrinsic magnetic dipole".

But maybe I will understand him better if I read it.

Well, classical models of matter nowadays are understood as approximations of the quantum-theoretical description in the sense of coarse-grained descriptions of macroscopic observables. In this sense there is also spin and intrinsic magnetic moments, which on the other hand in the quantum-field theoretical description are also included in the electromagnetic current.

 

[vanhees71]

One other thing to consider is how inter-related the EM quantities are. The work term E⋅J includes all the microscopic work done on matter, including but not limited to the Ohmic losses. So I wanted to see if any further analysis of that term could be done to pull out the Ohmic losses.

If a piece of material is moving at velocity v<<c then we can make the following transformations:
E′=E+v×B
J′=J−ρv
where the primes indicate values in the reference frame where the material is at rest.

Substituting those in we have:
E⋅J=(E′−v×B)⋅(J′+ρv)
=E′⋅J′+E′⋅ρv−(v×B)⋅J′−(v×B)⋅ρv
=E′⋅J′+ρv⋅(E+v×B)−(J−ρv)⋅(v×B)
=E′⋅J′+ρv⋅E−J⋅(v×B)

So finally we end with
E⋅J=E′⋅J′+v⋅(ρE+J×B)

Since E′ and J′ are in the reference frame of the material, I would interpret the term E′⋅J′ as being Ohmic losses. Then E⋅J contains not only the Ohmic losses, but also the material velocity times the Lorentz force density, which I would call the Lorentz power density.

I don't have a reference for this, so I don't know if someone else has already done this derivation, but I thought you might like to see it. It indicates that even with the microscopic Poynting’s theorem you can conclude that magnetic fields can do work. If you dig a bit deeper.

In addition in "macroscopic electrodynamics" it's also not unique, how you shuffle terms from the microscopic theory between "fields" ($\vec{E}$, $\vec{B}$, $\vec{P}$, $\vec{M}$) and "matter/charges/currents" ($\rho$, $\vec{j})$).

 

[Steve4Physics]

By the way, I never replied to this.

I don't mean the metal ions, if that's what you inferred. I'm talking about a surface charge density σ due to excess electrons which accumulate at the boundary of the wire, which arises in this case due to the Hall effect.

There's some more about surface charges here:
https://aapt.scitation.org/doi/abs/10.1119/1.18112?journalCode=ajp

Thankyou. I can't access the full article in your link but I'm familiar with the basic Hall effect. So I get the idea.

I'm *not* going to ask what force acts on the electrons, causing them to redistribute, in the Hall effect!

 

[etotheipi]

Well, classical models of matter nowadays are understood as approximations of the quantum-theoretical description in the sense of coarse-grained descriptions of macroscopic observables. In this sense there is also spin and intrinsic magnetic moments, which on the other hand in the quantum-field theoretical description are also included in the electromagnetic current.

Because I'm slightly stupid; what is a coarse-grained description of a macroscopic observable? I haven't learned a lot of quantum

 

[vanhees71]

It's not necessarily quantum. You consider a many-body system and realize that quite often the "macroscopically relevant" degrees of freedom change significantly over much larger spatial and temporal dimensions than the much more fluctuating microscopic ones. Thus you can simply average over "macroscopic small but microscopic large" spatial and/or temporal regions.

Now to consider the reaction of the matter to some external electromagnetic field or, equivalently, the presence of some external charges and currents you can lump some parts of the equation either to the "fields" or the "matter".

E.g., considering a dielectric you usually think in terms of charges bound within atoms and/or molecules making up your macroscopic matter and usually you describe (in linear response approximation) the result of the action of the external em. field in terms of the polarization $\vec{P}(t,\vec{x})$. You can, however, equivalently also put it into effective charge and current sources, $\rho_{\text{pol}}=-\vec{\nabla} \cdot \vec{P}$ and $\vec{j}_{\text{pol}}=\partial_t \vec{P}$.

Now you can write the inhomogeneous Maxwell equations in the form (in Heaviside-Lorentz units)
$$\vec{\nabla} \cdot \vec{E}=\rho_{\text{pol}}+\rho_{\text{ext}}, \quad \vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} (\vec{j}_{\text{pol}} + \vec{j}_{\text{ext}}).$$
Another, more common way is to use $\vec{P}$ instead. Then you have
$$\vec{\nabla} \cdot (\vec{E}+\vec{P})=\vec{\nabla} \cdot \vec{D}=\rho_{\text{ext}}$$
and
$$\vec{\nabla} \times \vec{B} - \frac{1}{c} \partial_t \vec{D}=\frac{1}{c} \vec{j}_{\text{ext}}.$$
So you can shuffle parts of the "matter part" of the equations to the "field part" and vice versa.

The same is of course true for the magnetic "field and matter parts". Which leads to the introduction of $\vec{H}$ lumping parts of the "atomic currents" and the "elementary magnetization" to the fields. Also the magnetization can be counted into the currents. So you get many ways to formulate "macroscopic electrodynamics" in differently doing this split into a field and a matter part. At the end you should of course get the same result for the observable quantities.

What you of course also need then are the constitutive relations, as in linear-response approximation, $\vec{P}=\chi_e \vec{E}$ (or equivalently $\vec{D}=\epsilon \vec{E}$, $\epsilon=1+\chi_e$) and (unfortunately for historical reasons in a sense inconsistent convention) $\vec{M}=\chi_m \vec{H}$ and $\vec{B}=\vec{H}+\vec{M}=\mu \vec{H}, \quad \mu=1+\chi_m$.

Of course this more or less naive classical picture has its limitations (particularly for the magnetic part), but qualitatively it's not so bad to have this picture in mind. A very nice discussion can be found in the Feynman lectures:

https://www.feynmanlectures.caltech.edu/II_32.html

and the following sections.

 

[Charles Link]

In a very simple description of the energy of a magnetic $\vec{m}$ moment in a magnetic field, there is a torque $\tau =m \times B$. The result when integrating this over $d \theta$ is that energy $E=-m \cdot B$. I suppose it could be argued that the energy comes from the change in flux through the current loop, in the case where $m=IA$. In any case it does appear that such a system is capable of doing work.

 

[vanhees71]

Indeed, you can derive also this from a classical picture (a la Ampere). The idea is that an atom ("particle") may be somehow composed of moving charges with some "molecular current". Then the force on this current in a magnetic field is
$$\vec{F}=\int_{\mathbb{R}^3} \mathrm{d}^3 x \vec{j}(\vec{x}) \times \vec{B}(\vec{x}).$$
Now you can assume that $\vec{j} \neq 0$ only for the very small extent of this "atom". So you can expand the magnetic field around the place of the atom (assumed to be at $\vec{x}_0=0$). This gives
$$\vec{F}=\int_{\mathbb{R}^3} \mathrm{d}^3 x \vec{j}(\vec{x}) \times [\vec{B}(0)+\vec{x} \cdot \vec{B}(0)].$$
Now, because of $\vec{\nabla} \cdot \vec{j}=0$ one can derive that
$$\int \mathrm{d}^3 x \vec{j}=0.$$
The 2nd term can be transformed to
$$\vec{F}=\vec{\nabla} (\vec{\mu} \cdot \vec{B})$$
with the magnetic moment
$$\vec{\mu}=\frac{1}{2} \int \mathrm{d}^3 x \vec{x} \times \vec{j}.$$
So indeed the potential of the force is
$$V=-\vec{\mu} \cdot \vec{B}.$$
You can also derive the torque in a similar way
$$\vec{\tau}=\int \mathrm{d}^3 x \vec{x} \times (\vec{x} \times \vec{B})=\cdots=\vec{\mu} \times \vec{B}.$$
From the definition of the magnetic moment you also get the relation to (orbital) angular momentum, rewriting
$$\vec{j}=q n \vec{v}.$$
Here $\vec{n}$ is the particle-number density and $q$ the charge of the particles making up the molecular current. With that we have
$$\vec{\mu}=\int \mathrm{d}^3 x \frac{1}{2} n q \vec{x} \times \vec{v} = \int \mathrm{d}^3 \frac{1}{2} \frac{n q}{m} \vec{x} \times \vec{p}=\frac{q}{2m} \vec{L},$$
where $\vec{L}$ is the angular momentum of the particles making up the molecular current.

Today we know that the magnetic moment is not only made up by such classical currents of moving charges with the corresponding orbital angular momentum, but also by the spin of the particles making up the current, e.g., the electrons around a nucleus in an atom. There is, however, another factor, the Lande or gyromagnetic factor. For an electron this factor is about 2. From the above classical model with the orbital angular momentum of Amperian currents the gyro factor comes out to be 1. Famously Einstein and de Haas thought to have confirmed this in their famous experiment, although de Haas indeed found also values for the gyro factor larger than 1. Not much later other physicists found the gyro factor of ferromagnets to be rather closer to 2, which was of course not understandable in this time (around 1915) . Today we understand it as the empirical evidence that most of the magnetization of a piece of iron is due to electron spins.

For more complicated particles like protons or neutrons and other hadrons the gyro factors' origin are very complicated and even not completely understood today. Here the elementary entities making up the particles are quarks and gluons, and how the spin and magnetic moment is "made up" of these constituents is very complicated and under ungoing research.