0
$\begingroup$

Does increasing the relative permeability of a wire increase the force felt by the wire (given the current and H field remain the same)?

The answer to this question seemed like an obvious "yes" to me, because the formula for the lorentz force on a wire (assuming B and I are perpendicular) is $F=BIl$. And $B=\mu H$. So $F=\mu HIl$. So $F$ is proportional to $\mu$.

But in answer to other questions I heard it claimed that I was "thinking that the magnetic permeability of the wire changes the force it feels, which [the person answering did] not think is true."

The only way I could think of such that increasing the magnetic permeability would not increase the force was if perhaps the $B$ in $F=BIl$ was the $B$ of the material the wire was passing through, not the wire itself. So I asked another question, where I learned "$B$ refers to external magnetic field $B_{ext}$, due to sources other than those in the element of the wire of length $L$; and it refers to value of this external field at the point of space where the current flows." I.E. inside the wire. I thought this affirmed my original understanding that force was proportional to permeability, until I read these two comments:

"Does this imply that if a current-carrying wire of permeability $\mu_0$ feels a force $F$, then the same wire but of permeability $\mu_r\mu_0$ feels a force $\mu_rF$?"

"No. $\mu_r$ characterizes how well the conductor magnetizes. This does not affect the above magnetic force. It does affect the net force a little, because there is also another force, proportional to product of magnetic moment and gradient of external magnetic field, but this is usually considered negligible when the above force is present."

How on earth does that make any sense?! I feel like every step of logic in my understanding has been affirmed, but the conclusion has been rejected. Does $F=BIl$? yes. Is $B$ where the wire is? yes. Is $B=\mu H$? yes. So $F\propto \mu$? No.

How does this make any sense!?

$\endgroup$

1 Answer 1

1
$\begingroup$

This has ended up being quite the rabbit hole. We will be ignoring ferromagnetic materials the whole time, because they are simply too complicated.

The scenario is treated (somewhat) in Electrodynamics of Continuous Media by Landau and Lifshitz, pp. 127-8:

The calculation of the forces is particularly simple for a linear conductor. Its magnetic properties are of no significance, and, if $\mu=1$ in the surrounding medium, the total force on the conductor is given by the line integral $$\mathbf{F}=J\oint\mathrm{dl}\times\mathfrak{H}/c$$

where $\mathfrak{H}$ is the external $\mathbf{H}$ field.

This ultimately follows from a derivation they begin starting with the stress tensor, assuming a fluid medium with $\mathbf{B}=\mu\mathbf{H}$ and ending with an expression for the force density: $$\mathbf{f}=-\nabla P_0+\frac{1}{8\pi}\nabla\left[H^2\rho(\partial_\rho\mu)_T\right]-\frac{H^2}{8\pi}\nabla\mu+\frac{\mu}{c}\mathbf{J}\times\mathbf{H}$$ They then make a simplification by noting that for $\mu\approx1$, as is common, then the last term in the above is the main contributor, with the others providing only small corrections. As such, we can write $$\mathbf{f}=\mathbf{j}\times\mathbf{H}/c$$So, in the case where we can describe the material as linear and isotropic (basically, the material is paramagnetic or diamagnetic), then it is a valid approximation (at least, given the assumptions) to state that the force on the wire does not depend on the permeability. But, to my reading, it does seem to be just an approximation (albeit an excellent one - I cannot find any paramagnetic or diamagnetic materials with $|\mu_r-1|>0.1$, and the overwhelming majority are much smaller). Later in the same section, though, it is stated that by the conservation of momentum, only the external field can contribute to the net force on a body. My interpretation is that ultimately, the source of the "extra" $\mathbf{B}$-field inside the material comes from the material itself (it is sort of "pinned" to the material, in a sense). A body cannot exert a net force on itself, so there is no way for this "extra" field to result in any additional force. This argument reads as very convincing to me, and makes me feel more inclined that the "approximate nature" of the above expression comes from the nature of the assumptions, such as a fluid medium - I'd need to think more about this.

Zangwill's Modern Electrodynamics addresses this question indirectly in $\S$13.8.2, where the question of the forces on an isolated magnetic body are addressed. In it, Zangwill considers an isolated sample of magnetic matter with magnetization $\mathbf{M}(\mathbf{r})$ in an external magnetic field $\mathbf{B}_0(\mathbf{r})$. He states that it is immaterial whether $\mathbf{M}$ is due to $\mathbf{B}_0$. Finally he gives a form for the force on such a body: $$\mathbf{F}=\int_V\mathrm{d}^3r\,\mathbf{j}_f(\mathbf{r})\times\mathbf{B}_0(\mathbf{r})+\int_V\mathrm{d}^3r\,[\mathbf{M}(\mathbf{r})\cdot\nabla]\mathbf{B}_0(\mathbf{r})$$ which you'll note is equivalent to the sum of the force from a non-uniform magnetic field on a magnetic moment, and the typical Lorentz force $\mathbf{F}=I\int\mathrm{d}\boldsymbol{\ell}\times\mathbf{B}$. Zangwill makes it much less ambiguous that $\mathbf{B}_0$ is the field that would have existed at the location of the body, had it not been there (it is the field due only to the sources) - he notes that the total magnetic field $\mathbf{B}(\mathbf{r})=\mathbf{B}_0(\mathbf{r})+\mathbf{B}_{self}(\mathbf{r})$ is the sum of the external field and the field produced by the magnetic body. Indeed, an object cannot exert a net force on itself (the same conservation of momentum argument), and so we can ignore the self term. We do not have to, though, and it is perfectly reasonable to write the form of the net force in terms of the full field (this forces us to consider the surface current density $\mathbf{K}$): $$\mathbf{F}=\int_V\mathrm{d}^3r\,\mathbf{j}(\mathbf{r})\times\mathbf{B}(\mathbf{r})+\int_S\mathrm{d}S\,\mathbf{K}(\mathbf{r}_S)\times\mathbf{B}_{avg}(\mathbf{r}_S)$$Zangwill makes an important note (emphasis mine):

The bilinear character of the integrands in this section raises the same questions about the macroscopic validity of our force formulae as arose in the Lorentz averaging discussion of Section 2.3.1. ...it is necessary to treat the validity of formulae like [the above] as logically independent assumptions of the macroscopic theory, subject to verification by experiment.

Where the discussion on Lorentz averaging is referring to the process by which we derive the macroscopic fields from the true microscopic fields (essentially removing the atomic-scale fluctuations through taking an average over a spherical region much larger than the scale of atoms). The potential issue here has to do with the fact that in general, $\langle\rho\mathbf{E}\rangle\neq\langle\rho\rangle\langle\mathbf{E}\rangle$, etc. We simply assume that the microscopic forms of things such as the Lorentz force law remain valid when interpreted macroscopically. This is perhaps a bit of a tangent, but interesting context around the whole thing, in my opinion.

My personal thoughts as to the answer to your question as to why the increased permeability would not cause an increased in force has to do with the fact that the bound currents in the material feel equal contributions in all directions due to uniformity of the field. There is simply no mechanism to unbalance these forces other than through a non-uniform field, but this leads to the magnetic moment interaction discussed previously which depends on the gradient of the field. Another solution, from a different perspective, could be the conservation of momentum argument. An increased force on the wire would be equivalent to the wire somehow exerting a (net) force on itself, which cannot happen. There is an interesting question I think left to resolve regarding the fact that L&L's derivation makes this seem (at least, it does to me) approximate, when I suspect that it is actually exact (that is, I suspect it is exact that the permeability does not impact the force, specifically because of the conservation of momentum argument). I would be curious to see a full treatment which shows which is really the case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.