A circular coil of 30 turns and radius `8.0cm` carrying a current of `6.0A` suspended vertically in a uniform horizontal magnetic field of magnitude `1.0T`. The field lines makes an angle of `60^(@)` with the normal of the coild. Calculate the magnitude of the counter torque that must be applies to prevent the coil form turning.

Explanation

Number of turns on the circular coil n=30, Radius of the coil r=8.0 cm =0.08 cm Area of the coil `=pir^(2)=pi(0.08)^(2)=0.0201m^(2)`, Current flowing in the coil `I=6.0A`
Angle between the field lines and normal with the coil surface, `theta=60^(@)`.
The coil experience a torque in the magnetic field. Hence, it turns. The counter torque applied to prevent the coil from turning is given by the relation.
`tau=n I B sin theta`..........i , `=30xx6xx1xx0.0201xxsin 60^(@)=3.133Nm`