A circular coil of radius `4 cm` and of `20` turns carries a current of `3` amperes. It is placed in a magnetic field of intensity of `0.5 weber//m^(2)`. The magnetic dipole moment of the coil is

Options

A. 0.15 ampere - `m^(2)`
B. 0.3 ampere - `m^(2)`
C. 0.45 ampere - `m^(2)`
D. 0.6 ampere - `m^(2)`

Answer

B

Explanation

M = NIA
`= 20 xx 3 xx pi (4 xx 10^(-2))^(2) = 0.3` ampere`-m^(2)`