A charge `q` is uniformly distributed on a non-conducting disc of radius `R`. It is rotated with an angular speed `omega` about an axis passing through the centre of mass of the disc and perpendicular to its plane. Find the magnetic moment of the disc.

Options

A. `1/2 q omega r^(2)`
B. `q omega r^(2)`
C. `3/2 q omega r^(2)`
D. `4/3 q omega r^(2)`

Answer

A

Explanation

Magnetic moment is given by `M=IA=q/t(pir^(2))`
`=1/((2pi//omega))(pir^(2))[:'t=T=(2pi)/(omega)]=1/2qomega r^(2)`