A charge q is spread uniformly over an insulated loop of radius r. If it is rotated with an angular velocity `omega` with respect to normal axis then the magnetic moment of the loop is :

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)`