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

`mu = NiA = 20 xx 3 xx pi xx 16 xx 10^(-4) = 0.3 A-m^(2)`