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_0`. Now, a vertical magnetic field is switched on and ring is rotated at constant angular velocity co. Find the maximum value of co h which the ring can be rotated if the strings can withstand a maximum tension of `(3T_0)/2`

Answer

`1//4`

Explanation

In equilibrium `2T_(0)=mg` or `T_(0)=(mg)/2` ………..i
Magnetic moment `M=iA=((omega)/(2pi)Q)(piR^(2)), tau=M B sin 90^(@)=(omegaB Q R^(2))/2`
Let `T_(1)` and `T_(2)` be the tensions in the two strings when magnetic field is switched on `(T_(1)gtT_(2))`
For rotational equilibrium
`T_(1)+T_(2)=mg `..........ii
For rotational equilibrium
`(T_(1)-T_(2))D/2=tau=(oemga B Q R^(2))/2` or `T_(1)-T_(2)=(omegaBQR^(2))/2` ............iii
Solving eqs ii and iii we have `T_(1)=(mg)/2+(omega B Q R^(2))/(2D)`
As `T_(1)gtT_(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))`