A square coil of side `10cm` consists of 20 turns and carries a current of 12A. The coil is suspended vertically and normal to the plane of the coil makes an angle of `30^@` with the direction of a uniform horizontal magnetic field of magnitude `0*80T`. What is the magnitude of torque experienced by the coil?

Options

A. 1.6 N m
B. 1.2 N m
C. 1.4 N m
D. 1.8 N m

Answer

A

Explanation

`tau=NIABsin theta`
Here, `N=25, I=10A, B=0.9T, theta=30^(@)`
`A=a^(2)=12xx10^(-2)xx12xx10^(-2)=144xx10^(-4)m^(2)`
`therefore" "tau=25xx10xx144xx10^(-4)xx0.9xxsin 30^(@)="1.6 N m"`