Question

A thin wire of length `L` is made of an insulating material. The wire is bent to form a circular loop, and a positive charge `q` is distributed uniformly around the circumference of the loop. The loop is then set into rotation with angular speed `omega` around an axis through its centre. If the loop is in the region where there is a uniform magnetic field `vec(B)` directed parallel to the plane of the loop, calculate the magnitude of the magnetic torque on the loop.
A. `(q omegaL^(2)B)/(8pi^(2))`
B. `(q omegaL^(2)B)/(4pi^(2))`
C. `(q omegaL^(2)B)/(2pi^(2))`
D. `(q omegaL^(2)B)/(pi^(2))`

Answer 1

Correct Answer - 1
`tau=mBsintheta=IABsintheta,IAB,(as theta=90^(@))`
Further, as ` I=(q)/(T)=(q)/(2pi//omega)=(qomega)/(2pi),`
`A=pir^(2)=pi((L)/(2pi))^(2)=(L^(2))/(4pi)`
`tau=((qomega)/(2pi))((L^(2))/(4pi))B=(qomegaL^(2)B)/(8pi^(2))`