Ring of radius $R$ having uniformly distributed charge $Q$ is mounted on a rod suspended by two identical strings as shown in the given figure. The tension in strings in equilibrium is $T_0$. Now a vertical magnetic field is switched on and the ring is rotated at constant angular velocity $\omega$. Find the maximum $\omega$ with which the ring can be rotated if the strings can withstand a maximum tension $\frac{3T_0}{2}$.

 

Initially $2T_0=mg ...\left(i\right)$

If $\omega$ be the frequency corresponding to the breaking of the string, then current in the ring

$i=\frac{Q}{T}=\frac{Q}{{\displaystyle \frac{2\mathrm{\pi }}{\mathrm{\omega }}}}=\frac{Q\omega }{2}$

Magnetic moment of the loop

$M=iA$

The torque experiences now

$\zeta =MB \sin \theta$

$=iAB$

If $T_1$ and $T_2$ are the tensions in the strings now , then

$\left(T_1-T_2\right)\frac{D}{2}=iAB$

or $T_1-T_2=\frac{2iAB}{D} ...\left(ii\right)$

Also $T_1+T_2=mg \left(iii\right)$

Solving equations $\left(ii\right)$ and $\left(iii\right)$, we get

$T_1=\frac{mg}{2}+\left(\frac{iAB}{D}\right)$

and $T_2=\frac{mg}{2}-\left(\frac{iAB}{D}\right)$

As $T_1>T_2$, $\therefore T_1=\frac{3T_0}{2}$    (Given)

Hence $\frac{3T_0}{2}=\frac{2T_0}{2}+\frac{\omega Q}{2\mathrm{\pi D}}\times {\mathrm{\pi R}}^2\times \mathrm{B}$

or $T_0=\frac{\omega QB{R}^2}{D}$

or $\omega =\frac{D{T}_0}{QB{R}^2}$