Question

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

Answer 1

In equilibrium: $2T_o=mg$
or $T_o=\frac{mg}{2}$ ...(i)

Magnetic moment, $M=iA=(\frac{\omega }{2\pi }Q\left)\right(\pi R^2)$
$\tau =MBsin{90}^o=\frac{\omega BQ{R}^2}{2}$

Let $T_1$ and $T_2$ be the tensions in the two strings when magnetic field is switched on $(T_1>T_2)$.
For translational equilibrium of ring is vertical direction,
$T_1+T_2=mg$ ...(ii)

For rotational equilibrium,
$(T_1-T_2)\frac{D}{2}=\tau =\frac{\omega BQ{R}^2}{2}$
or $T_1-T_2=\frac{\omega BQ{R}^2}{2}$ ...(iii)

Solving equations (ii) and (iii) we have
$T_1=\frac{mg}{2}+\frac{\omega BQ{R}^2}{2D}$
As $T_1>T_2$ and maximum values of $T_1$ can be $\frac{3T_o}{2}$, we have
$\frac{3T_o}{2}=T_o+\frac{{\omega }_{max}BQ{R}^2}{2D}$
$\therefore {\omega }_{max}=\frac{D{T}_o}{BQ{R}^2}$