Solution: Angular Velocity of Paper Cylinders
Concept: The problem states the cylinders are “inertia-less” and charges move with the cylinder. This implies the system behaves like a superconductor where the net change in magnetic flux through any closed loop on the cylinder surface is zero. This is because any change in flux would create induced electric field hence torque. Since the cylinder is inertialess net torque should remain zero.
Case 1: Cylinder Inside (\(r_p < r\))The rotating charged cylinder creates a magnetic field \(B_{ind} = \mu_0 K = \mu_0 \frac{Q\omega}{2\pi l}\). Flux conservation requires: $$ \Phi_{initial} = \Phi_{final} $$ $$ (\mu_0 n I) \pi r_p^2 = (\mu_0 n \frac{I}{\eta}) \pi r_p^2 + (\mu_0 \frac{Q\omega}{2\pi l}) \pi r_p^2 $$ Simplifying (removing \(\pi r_p^2 \mu_0\)): $$ nI \left( 1 – \frac{1}{\eta} \right) = \frac{Q\omega}{2\pi l} $$
Here, the flux through the paper cylinder (radius \(r_p\)) is only due to the solenoid core (radius \(r\)) and its own self-field. $$ \Phi_{total} = \Phi_{solenoid} + \Phi_{self} $$ Conservation of flux: $$ (\mu_0 n I) \pi r^2 = (\mu_0 n \frac{I}{\eta}) \pi r^2 + (\mu_0 \frac{Q\omega}{2\pi l}) \pi r_p^2 $$ Rearranging: $$ \mu_0 n I \left( \frac{\eta-1}{\eta} \right) \pi r^2 = \mu_0 \frac{Q\omega}{2\pi l} \pi r_p^2 $$