Question

A ring of radius $R=50\text{cm}$ having uniformly distributed charge $Q=2\text{C}$ is mounted in the middle of a rod suspended by two identical strings. The tension in strings in equilibrium is $T_o=2\text{N}$. 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 of $3\text{N}$. (Take $D=1\text{m}$ and $B=0.05\text{T}$)

• A. $10\text{rad/s}$
• B. $80\text{rad/s}$
• C. $40\text{rad/s}$
• D. $30\text{rad/s}$

Answer 1

The correct option is B $80\text{rad/s}$

Maximum angular speed with which the ring can be rotated is given by $${\omega }_{max}=\frac{D{T}_0}{BQ{R}^2}$$Substituting the data given in the question,

${\omega }_{max}=\frac{1\times 2}{0.05\times 2\times (0.5{)}^2}=80\text{rad/s}$

Hence, option (b) is the correct answer.

Note: In equilibrium: $2T_o=mg$ $\Rightarrow T_o=\frac{mg}{2}...\left(1\right)$ Magnetic moment, $\mu =IA=\left(\frac{Q}{2\pi /\omega }\right)\left(\pi R^2\right)=\frac{Q\omega R^2}{2}$ Now, torque on the rod due to magnetic moment, $\tau =\mu B\sin {90}^{\circ }=\frac{\omega BQ{R}^2}{2}$ Let $T_1$ and $T_2$ be the tensions in the two strings when the magnetic field is switched on $(T_1>T_2)$. $T_1+T_2=mg....\left(2\right)$ For rotational equilibrium, torque acting on the rod should be balanced. $\therefore (T_1-T_2)\frac{D}{2}=\tau =\frac{\omega BQ{R}^2}{2}$ $\Rightarrow T_1-T_2=\frac{\omega BQ{R}^2}{D}....\left(3\right)$ Solving equations $\left(2\right)$ and $\left(3\right)$, we have $T_1=\frac{mg}{2}+\frac{\omega BQ{R}^2}{2D}$ As $T_1>T_2$ and maximum value of $T_1$ can be $\frac{3T_o}{2}$, we have, $\frac{3T_o}{2}=T_o+\frac{{\omega }_{max}BQ{R}^2}{2D}[\because \frac{mg}{2}=T_o]$ $\therefore {\omega }_{max}=\frac{D{T}_0}{BQ{R}^2}$