A circular coil of radius 4 cm has 50 turns. In this coil a current of `2 A` is flowing. It is placed in a magnetic field of `0.1 "weber"//m^(2)`. The amount of work done is rotation it through `180 ^(@)` from its equilibrium position will be

Options

A. `0.1 J`
B. `0.2 J`
C. `0.4 J`
D. `0.8 J`

Answer

A

Explanation

The magnetic moment of current carrying loop
`M=niA=ni(pir^(2))`
Hence the work done in rotating it through `180^(@)`
`W=MB(1-costheta)=2MB=2(nipir^(2))B`
`=2xx(50xx2xx3.14xx16xx10^(-4))xx0.1=0.1J`