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I know this sounds like a bit of a stupid question and it low key is.

But I know that the magnetic force due to a current carrying wire of radius $R$ at a distance $r$ from the central axis of the wire is given by: $$\vec{B}=\frac{\mu_0 i}{2\pi}\frac{r}{R^2}\hat{e_{\theta}}$$

When $r\leq R$.

Now I'm wondering if this magnetic field causes the charger carriers within the wire to experience a magnetic force due to this created magnetic field.

I know for an individual charge in a reference frame where the magnetic field due to that moving charge is present, it does not experience a force due to its own magnetic field. But since this magnetic field($\vec{B}=\frac{\mu_0 i}{2\pi}\frac{r}{R^2}\hat{e_{\theta}}$) is the result of many charge carriers would this magnetic field have an effect on the charge carriers of the wire?

Using an oversimplified model of a current carrying wire. If the charge carriers DID experience a force due to this field. It would cause them to move helically through the wire.

Now I know ultimately I know this question is slightly redundant since a charge carrying wire isn't nearly as simple as a stream of point charges moving at a constant velocity $\perp$ to $B$.

But still it has been something I've been quite curious about for awhile and I did not find any answers on Google.

Kind Regards.

PH Herman
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1 Answers1

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The electrons inside a current carrying wire do feel magnetic forces, and this causes a radial compressive force on the conductor. That such a force exists follows intuitively from the observation that parallel wires carrying current in the same direction are attracted to each other. If you think of a current-carrying cylindrical wire as infinitely many infinitesimal parallel wires, then they will all attract each other creating a radial pressure gradient.

As given in the answers to Magnetic force in the inside of cylindrical conductor? and Pressure experienced due to magnetic force?, the radial magnetic pressure at a radius $r$ inside a straight cylindrical conductor of external radius $R$ carrying a current $I$ is $$P=\frac{\mu I^2r^2}{8\pi^2 R^4}$$

At normal current densities, the compressive strength of metal wires is easily more than enough to sustain this pressure. A compensating electric field builds up due to the outer metal lattice becoming positively charged as the negative conduction electron move inwards. The situation is different, however, inside gaseous plasmas where magnetic pressure can force the plasma to collapse. This is, for example, how a z-pinch works.

Naively, the electrons will not move helically but radially, since their average motion is along the wire, which is perpendicular to the circular magnetic field. Given that the thermal velocity of the electrons at room temperature is much greater than the drift velocity, however, maybe you are right since charged particles in plasmas do tend to follow magnetic field lines.

David Bailey
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  • the magnetic self-compression is also a (mechanical) stability issue for superconducting magnets. – tobalt Nov 24 '22 at 18:52
  • I must admit the helical thing was a classic moment of not actually doing the math and instead just loosely thinking about it. A rather goofy mistake that I'll blame being a first year student on. But yeah I woKe up hoping no one would notice that mistake lol. I appreciate the full and extensive answer. Definitely helping my understanding, and I'll make sure to look at the attached links. Thanks so much. – PH Herman Nov 25 '22 at 02:40
  • Does the negative conducting electrons moving inwards cause a similar result to the skin effect, where you get a higher effective resistance than you would otherwise expect? Can we continue to assume a uniform current density if the electrons are being pulled towards the center? – Calum McConnell Feb 26 '24 at 15:26