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When considering the force on a current carrying wire in a magnetic field, is $B$ in the Lorentz force $F=IlB$ the field inside the wire where the current is flowing, or is it the field immediately outside the wire?

To put it another way, is the Lorentz force greater on a piece of iron than a piece of copper (because iron has a high permeability), given the same current and H field? And if not, then does a wire experience a greater force when the entire circuit is in air rather than submerged in water (because air has a higher permeability than water)?

If the Lorentz force depends on $B$ immediately outside of the wire, how does this not violate localism? Why are the electrons being pushed to the side by the magnetic field outside the wire when they're inside the wire?

To be absolutely clear, I'm not talking about the production of the H field due to the motion of electrons in the wire. You may assume the H field is uniform everywhere. The B field only differs as the permeability of the substance differs.

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In the formula for magnetic force on current-conducting wire element of length $L$:

$$ \mathbf F = I\mathbf L \times \mathbf B, $$ $\mathbf B$ refers to external magnetic field $\mathbf B_{ext}$, 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. Usually, most of the current flows inside the wire (as opposed to its surface), thus $B$ in the formula refers to external magnetic induction inside the wire. However, since $\mathbf B_{ext}$ is continuous in the region where the wire element is present, and source of the external magnetic field is usually quite distant (compared to length $L$), value of this external magnetic field at the surface is almost the same as inside the wire, so usually the difference is negligible.

If the wire is thick or the source of the external magnetic field is very close, so $\mathbf B_{ext}$ varies throughout the cross-section of the wire, then we can use the integral formula taking this into account:

$$ \mathbf F = \int \mathbf j \times \mathbf B~ d^3\mathbf x. $$ Here $\mathbf B$ is assumed to be a function of position, and integral goes over all points of space where current density $\mathbf j$ is non-zero. Thus macroscopic magnetic force on a wire can be expressed as sum of forces on all current elements in the wire (due to moving charge), and is proportional to product of current density and external magnetic field at the position of the element.

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  • $\begingroup$ As ridiculous as this sounds, when I said "immediately outside the wire", I meant the air or free space surrounding it, and not the surface of the wire, which I recognize with have the same or nearly the same permeability. And I know, it makes no sense to say the force is proportional to the magnetic field outside the wire, but I had two physicists whom I respect claim this, so I had to double check. $\endgroup$
    – Joseph Summerhays
    Commented May 17 at 23:53
  • $\begingroup$ Thanks for the answer. This is exactly what I wanted to know. $\endgroup$
    – Joseph Summerhays
    Commented May 17 at 23:54
  • $\begingroup$ Maybe there was just a miscommunication/misunderstanding. There are other formulae for magnetic force, and one of them, using the Maxwell stress tensor, expresses net force on a body using electric and magnetic field on any imaginary surface containing the body. Thus field "just outside the wire" can be used there. The resulting force will be the same as using the above formula. $\endgroup$
    – Ján Lalinský
    Commented May 17 at 23:55
  • $\begingroup$ 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 $\mu_rF$? $\endgroup$
    – Riley Scott Jacob
    Commented May 18 at 13:51
  • $\begingroup$ @RileyScottJacob 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. $\endgroup$
    – Ján Lalinský
    Commented May 18 at 14:03
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Neither. The magnetic field in the Lorentz force equation is the magnetic field due to all current sources other than the wire itself.

One could in principle look at the forces on each microscopic element of the wire. Each element of the wire would feel forces from two sources: from external fields, and from the currents (free and bound) inside the wire. Wires made of different materials will certainly have different amounts of bound currents. But as far as the net force on the wire goes, the bound currents don't matter; because when we sum up all the microscopic forces on the elements of the wire to obtain the net force, all of the forces internal to the site will cancel out (thanks, Newton!) and only the forces from the external fields will matter.

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    $\begingroup$ I think the OP is asking how you handle the magnetic field due to other current sources if the field changes due to material properties in the wire $\endgroup$
    – BioPhysicist
    Commented May 17 at 17:16
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    $\begingroup$ @BioPhysicist: That was my point about the bound currents, etc. Even if the total magnetic field at any point inside the wire differs due to the wire's material properties, the resulting differences in the forces will automatically cancel out when we figure out the net force on the wire. $\endgroup$
    – Michael Seifert
    Commented May 17 at 17:59
  • $\begingroup$ > as far as the net force on the wire goes, the bound currents don't matter; because when we sum up all the microscopic forces on the elements of the wire to obtain the net force, all of the forces internal to the site will cancel out -- Magnetic forces, in general, do not obey Newton's third law. In general, bound currents may affect net external force on any wire element. It is not clear that these effects sum up to zero for the given wire or circuit. $\endgroup$
    – Ján Lalinský
    Commented May 17 at 18:20
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    $\begingroup$ That said, I suppose that if you have an external field that's non-uniform, the distribution of bound currents in the interior of the wire could affect the net force. The idea that the field is uniform is implicit in the formula $F = I l B$, though; if $B$ is non-uniform then the force has to be written as an integral over the length of the wire instead. $\endgroup$
    – Michael Seifert
    Commented May 17 at 18:49
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    $\begingroup$ @JosephSummerhays it's misleading to think of $H$ as produced by only currents. When magnetized body is present, $H$ is affected by its magnetization as well, irrespective of presence of currents. None of this matters for the force formula above, because in this formula, only $B$ appears. $\endgroup$
    – Ján Lalinský
    Commented May 18 at 14:12

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