A long insulating cylinder of radius $R$ and length $l$ carries a uniformly distributed surface charge $Q$. A string is coiled around the cylinder from which a block of mass $m$ hangs. The mass is free to move downwards and can rotate the cylinder. Neglecting the moment of inertia of the cylinder, calculate the acceleration of the block.

Angular momentum of the mass $m$ relative to the cylinder is given by
$\overrightarrow{L}=mvR\hat{k}=m{R}^2\omega \hat{k}$
Therefore torque, $\overrightarrow{\tau }=\frac{d\overrightarrow{L}}{dt}=m{R}^2\frac{d\omega }{dt}$
The cylinder has a uniform charge density on its surface. The angular velocity of the cylinder increases as the mass accelerates downward, the electric current caused by revolution of the cylinder, that flows in azimuthal direction, increases. The magnetic field created by the current on the surface increases. The increase in magnetic flux induces an azimuthal electric field, $E_{\theta }$, that exerts an electric force on the surface charges whose torque resists the angular acceleration of the cylinder in accordance with Lenz’s law. The rotating cylinder may be approximated by a tightly bound solenoid carrying a time-dependent current $I\left(t\right)$, the magnetic field $B\left(t\right)$ inside the cylinder is uniform at each moment.

The equation of motion of the block is 
$mg-T=ma \dots \dots \dots \dots \dots ..\left(1\right)$
The velocity of the block at time $t$ is 
$v=at$
The angular velocity of the cylinder is 
$\omega =\frac{v}{R}=\frac{at}{R}$
and the frequency is
$\gamma =\frac{\omega }{2\pi }=\frac{at}{2\pi R}$

The effective current over the surface of the cylinder is $i=Qv.$
The magnetic field on its axis is(5)
$B={\mu }_{0}ni=\frac{{\mu }_{0}Qat}{2\pi Rl}\left[\text{where }n=\frac{1}{l}\right]$
$\frac{dB}{dt}=\frac{{\mu }_{0}qa}{2\pi Rl}\dots \dots \dots \left(2\right)$
The time varying magnetic field induces an electric field at the surface of the cylinder 
$E=\frac{R}{2}\frac{dB}{dt}=\frac{{\mu }_{0}Qa}{4\pi l}\dots \dots \dots \left(3\right)$
This electric field is tangential and causes a torque. The magnitude of this torque is
$\tau =qER=\frac{{\mu }_0{Q}^{2}Ra}{4\pi l}\dots \dots \left(4\right)$
Moment of inertia of the cylinder is zero, therefore net torque on it should be zero. 
Hence, the torque due to tension of the string must balance the torque of the electric field
$TR=\frac{{\mu }_0{Q}^{2}Ra}{4\pi l}$
or, $T=\frac{{\mu }_0{Q}^{2}a}{4\pi l}\dots ..\left(5\right)$
From equations $\left(1\right)$ and $\left(5\right)$, we have 
$mg-\frac{{\mu }_0{Q}^{2}a}{4\pi l}=ma$
or, $a=\frac{g}{1+\frac{{\mu }_0{Q}^2}{4\pi ml}}$

Note that if $Q=0, a=g$ Ans.