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

Explanation

The torque acting on the coil,
`tau=NBI sin theta`
`=30xx1xx6xxpi xx(8xx10^(-2))^(2) sin 60^(@)`
`=3.13N*m`
`:.` If a torque of `3.13N*m` is applied in the opposite direction, there will be no delfection of the coil.