Question

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–

• A. $\frac{2g}{1 + \frac{\mu_{0}Q^{2}}{4\pi m\mathcal{l}}}$
• B. $\frac{g}{1 + \frac{\mu_{0}Q^{2}}{4\pi m\mathcal{l}}}$
• C. $\frac{4g}{1 + \frac{\mu_{0}Q^{2}}{4\pi m\mathcal{l}}}$
• D. $\frac{6g}{1 + \frac{\mu_{0}Q^{2}}{4\pi m\mathcal{l}}}$

Answer 1

Answer: (B)

$\frac{g}{1 + \frac{\mu_{0}Q^{2}}{4\pi m\mathcal{l}}}$

Answer 2

As mass come down cylinder will rotate about it axis. Thus charge on cylinder also rotate due to which electric current is produced electrical current will depend on angular speed of cylinder.

mg – T = ma

v = at

w = = $\frac { \mathrm { at } } { \mathrm { R } }$

n = frequency of revolution =$\frac { \omega } { 2 \pi }$=

Effective current i = Qn =

Magnetic field due to i on the axis = m₀ni

B =$\frac { \mu _ { 0 } \times 1 } { \ell }$

=

Electric field due to time varying magnetic field

E = = $\frac { \mu _ { 0 } \mathrm { Qa } } { 4 \pi \ell }$

Torque (t) due to electric field = qER

= Moment of inertia of cylinder is zero

\ Net toque on it should be zero

\ Torque due to tension of the string = Torque

of electric field

TR =

T =

mg – T = ma

ma = mg –

a =$\frac { \mathrm { g } } { 1 + \frac { \mu _ { 0 } \mathrm { Q } ^ { 2 } } { 4 \pi \mathrm {~m} \ell } }$