A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.8 T. What is the magnitude of the torque experienced by the coil?

Explanation

N=20 , `A=100 xx 10^(-4) m^2. 1=12 A`
R=0.8 T `theta = 30^@` Torque `tau = NI( bar A xx bar B)`
`= 20 xx 12 xx 100 xx 10^(-4) xx 0.8 xx sin 30^2@`
`= 20 xx 12 xx 100 xx 10^(-4) xx 0.8 xx 1/2 = 0.96 Nm`