Question

A ring of radius R having unifromly distributed charge Q is mounted on a rod suspended by two identical strings. The tension in strings in equilibrium is `T_0.` Now a vertical magnetic field is switched on and 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 `3T_0 //2.`

Answer 1

Correct Answer - B::D
In equilibrium,
`2T_0=mg` or `T_0=(mg)/2`…….i
Magnetic moment `M=iA=(omega/(2pi)Q)(piR^2)`
`tau=MBsin90^@=(omegaQR^2)/2`
Let `T_1` and `T_2` be the tension in the two strings when magnetic field is switched on `(T_1ltT_2)`.
For translation equilibrium of ring in vertical direction.
`T_1+T_2=mg`...........i
For rotationasl equilibrium
`(T_1-T_2)D/2=tau=(omegaBQR^2)/2`
or `T_1-T_2=(omegaBQR^2)/2`.........iii
Solving eqn ii and iii, we have
`T_1=(mg)/2+(omegaBQR^2)/(2D)`
As `T_1gtT_2` and maximum values of `T_1` and `(3T_0)/2`,
We have
`(3T_0)/2=T_0+(omega_maxBQR^2)/(2D)`
`:. omega_max=(DT_0)/(BQR^2)`