Motion of charged particles in complex fields > Motion in a homogeneous magnetic field > Problem 10.1.22

Statement

$10.1.22.$ Prove that the angular momentum increment $∆M$ of the ring in problem $10.1.21$ is proportional to the increment of the magnetic induction flux through the ring $∆Φ$: $∆M = (1/2π)Q∆Φ$, where $Q$ is the electric charge of the ring. To prove this, use the fact that the flux of magnetic induction through the side surface of the cylinder is equal to the difference in fluxes through its ends.

Solution

Consider moving the ring by a distance of $dx= \upsilon dt$. The increment of the angular momentum of the ring is written by definition:

$$\Delta M = N dt$$

Where $N$ is the moment acting on the ring calculated in the previous problem:

$$N = 2\pi \rho B_R R^2 \upsilon$$

Now let's use the hint from the condition, the change in flow through the ring section is the flow through the side surface of the cylinder, which it describes during dt:

$$\Delta \Phi = 2\pi R B_R \upsilon dt$$

Note now that the charge density in the ring is:

$$\rho = Q/2\pi R$$

Well, the final answer:

$$\Delta M = \frac{1}{2\pi} Q \Delta \Phi$$

Answer

Q.E.D