Due to the flow of current in a circular loop of radius `R`, the magnetic induction produced at the centre of the loop is `B`. The magnetic moment of the loop is (`mu_(0)`=permeability constant)

Options

A. `(2piBR^(2))/(mu_0)`
B. `(BR)/(2pimu_0)`
C. `(2piBR^(3))/(mu_0)`
D. `(BR^(2))/(2pimu_0)`

Answer

C

Explanation

For a circular charged loop, `B_(center)=(mu_0)/(4pi)((2pii)/R)=(mu_0i)/(2R)therefore i=(2BR)/(mu_0) therefore ` Magnetic moment `M- I xx A=I xx piR^(2)=(2BR)/(mu_0)(piR^(2))therefore M=(2piBR^(2))/(mu_0)`.