A wheel of radius ` R` having charg `Q`, uniformly distributed on the rim of the wheel is free to rotate about a light5 horizontal rod. The rod is suspended by light inextensible strings and a magnetic field `B` is applied as shown in the figure. The initial tensions in the strings are `T_(0). If the breaking tension of the strings are `(3T_(0))/(2)`, find the maximum angular velocity `omega_(0)` with which the wheel can be rotated.

Answer

`omega = (dT_(0))/(QR^(2)B)`

Explanation

Initially `2T_(0) =mg` ..(i)
if `omega` be the frequency corresponding to the breaking of the string then current in the ring
` I = (Q)/(T) = (Q)/(2pi//omega) = (Qomega)/(2pi)`
Magnetic moment of the loop
`M =iA`
The torque experinces now
`tau = MB sin 90^(@)`
if `T_(1)` and `T_(2)` are the tensions in the string now, then
`(T_(1) -T_(2)) (d)/(2) = iAB`
or `T_(1) -T_(2) = (2iAB)/(d)...(i)`
also `T_(1) + T_(2) =mg...(iii)`
Solving equations (ii) and (iii) we get
`T_(1) = (mg)/(2) +((iAB)/(d))` and `T_(2) =(mg)/(2) -((iAB)/(d))`
As `T_(1) gt T_(2)`
` :. T_(1) = (3T_(0))/(2)` (Given) Hence
`(3T_(0))/(2) = (2T_(0))/(2) + (omegaQ)/(2pid) xx piR^(2) xx B`
or `T_(0) = (omegaQ BR^(2))/(d)` or `omega = (dT_(0))/(QBR^(2))`
.