Timeline for Is the Lorentz force proportional to $B$-field in the wire or immediately outside of it?
Current License: CC BY-SA 4.0
10 events
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| May 18 at 14:12 | comment | added | Ján Lalinský | @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. | |
| May 18 at 2:45 | comment | added | Joseph Summerhays | The H field matters only because $B=\mu H$. B is not produced (directly) by currents. H is produced by currents. B is produced by the interaction between H and the permeability of the material. I don't love that you say "only external fields matter". I know what you mean, you mean fields that are local to the wire but produced by currents external to the wire. But it sounds like the field outside the wire, when that's not what you mean (I think). | |
| May 18 at 2:08 | comment | added | Michael Seifert | @JosephSummerhays: I guess I don't see why you should have to think about $\vec{H}$. The Lorentz force on each element of the wire only depends on $\vec{B}$ experienced by the element, not $\vec{H}$. And if you look at it in terms of the $\vec{B}$'s produced by the current and by the external agents, then it's obvious that only the external fields matter. | |
| May 17 at 21:27 | comment | added | Joseph Summerhays | This answer misses the point. I understand that a small segment of wire dl doesn't generate it's own H-field to interact with (or to the extent it does, it cancels itself out). I want to know if the H-field is generated elsewhere, then when we calculate the B, there's B in the wire and outside the wire. I've been told by other physicists that the force uses B immediately outside the wire, which makes no sense to me. | |
| May 17 at 18:49 | comment | added | Michael Seifert | 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. | |
| May 17 at 18:47 | comment | added | Michael Seifert | @JánLalinský: I'm aware of a a few exceptions to Newton's Third Law in the context of magnetic fields, but those all involve time-dependent fields or currents that don't obey $\nabla \cdot \vec{J} = 0$ (which is secretly the same thing.) It would be interesting to do the explicit calculations to confirm that the self-force on a long wire is zero, but I would be shocked if it isn't. Maybe once I've got my grades in. | |
| May 17 at 18:20 | comment | added | Ján Lalinský | > 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. | |
| May 17 at 17:59 | comment | added | Michael Seifert | @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. | |
| May 17 at 17:16 | comment | added | BioPhysicist | 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 | |
| May 17 at 16:30 | history | answered | Michael Seifert | CC BY-SA 4.0 |