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

Options

A. `0.15Am^(2)`
B. `0.3Am^(2)`
C. `0.45Am^(2)`
D. `0.6Am^(2)`

Answer

B

Explanation

`N=20,I=3A,A=pir^(2)=3.14xx10^(-4)xx16m^(2)`
Magnetic dipole moments M = NIA
`thereforeM=20xx3xx3.14xx10^(-4)xx16=0.3Am^(2)`