# Charged Particle Through Torus

*Most recent answer: 02/10/2015*

- rajesh kumar nayak (age 20)

durgapur,westbengal,india

A nice tough question!

Let's look at this from the torus frame. For simplicity, let's consider the case where the torus is just a loop of a single tube of metal, with resistance high enough so the L/R (inductance/resistance) time is short compared to the time for the particle to move through the torus vicinity. The moving charged particle has a magnetic field looping around its path. As it approaches the torus that makes a net magnetic flux through the torus tube. The changing flux generates an EMF that drives current around the small circle of the tube. That generates a flux in the tube of the opposite sign to the one from the moving charged particle. That flux is also changing, making an EMF for loops around the tube going through the middle of the torus.

Since the particle's path parallels those loops, it experiences a force along the direction of its motion. Without going through the detailed calculation, we know the particle must slow down, because the currents in the tube dissipated some energy via Joule heating.

In other word, even though the magnetic field generated by the currents around the torus tube is confined to within the tube, changes in that field generate an EMF *outside* the torus. At least to this level of approximation, we don't need to consider the radiative field explicitly.

It's interesting to look at the same problem from the frame initially at rest wih respect to the particle, with the uncharged torus moving. Where do the EMF's come from? Here, remember that the electric field from the particle induces an uneven charge distribution on the torus, enough to cancel the electric field inside the torus. Now parts of the torus are charged, so their motion makes a magnetic field. That magnetic field makes a flux through the tube, etc.

I've tried going through a quick calculation to estimate the magnitude of the energy loss. If the particle has charge Q, initial speed v (relative to the torus), and the torus has big radius D and little radius r, thickness b, and resistivity ρ (all in cgs units), the energy loss is some dimensionless number times Q^{2}r^{3}bv^{3}/(D^{4}ρc^{4}), still in cgs units.

Mike W.

*(published on 02/10/2015)*

## Follow-Up #1: forces on moving charged particle

- rajesh kumar nayak (age 20)

durgapur,westbengal,india

Yes, it must be an electrical field because it changes the energy of the charged particle. Magnetic forces are at right angles to the motion, and thus do not do work on simple charged particles.

In principle the calculation could be done without considering Joule heating, but it provides a simple way to keep track of the main energy loss. Notice that my rough calculation was done assuming that L/R << D/v, so that the current in the torus is determined by the instantaneous changing flux from the particle. If L/R >> D/v, then there's little Joule heating because the current doesn't respond much before the particle has passed through. So the maximum Joule heating occurs when ρ is just big enough for L/R to about equal D/v.

What happens if ρ=0, for a superconducting torus with no Joule heating? There's still some energy loss via electromagnetic radiation due to the interaction between the particle and the torus. I made a very crude calculation and got an energy loss ~Q^{2}r^{2}v^{5}/(D^{3}c^{5}). Notice that the ratio of this radiative loss to the loss if the superconductor is replaced with a material with resistivity ρ is ρDv^{2}/rbc, which is much less than 1 for ordinary values of the parameters. So it looks like Joule heating is in fact the key.

Mike W.

*(published on 02/18/2015)*