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

In equilibrium,
`2T_0=mg`
or `T_0=(mg)/2….(i)`
magnetic moment, `M=iA=(omega/(2pi)Q)(piR^2)`
`tau=MB sin90^@=(omegaBQR^2)/2`
Let `T_1 and T_2` be the tensions in the two strings when magnetic
field is switched on `(T_1gtT_2)`
For translation equilibrium,
`T_1+T_2=mg....(ii)`
For rotational equilibrium
`(T_1-T_2)D/2=tau=(omegaBQR^2)/2`
or `T_1-T_2=(omegaBQR^2)/D....(iii)`
Solving Eqs. (ii) and (iii) we have
`T_1=(mg)/2+(omegaBQR^2)/(2D)`
As `T_1gtT_2` and maximum values of `T_1` can be `3T_0//2`, we have
`(3T_0)/2=T_0+(omega_(max)BQR^2)/(2D) ((mg)/2=T_0)`
`:. omega_(max)=(DT_0)/(BQR^2)`