A square coil of side 10 cm has 20 turns and carries a current of 12 A. The coil is suspended vertically. The normal to the plane of the coil makes an angle of `30^@` with the direction of a uniform horizontal magnetic field. If the torque experienced by the coil equals 0.96 N - m, find the magnitude of the magnetic field.

Explanation

Here side of square `l = 10 cm = 0.1 m`, hence area of coil `A = (0.1)^(2) = 0.01 m^(2)`, current `I = 12 A`, torque `tau = 0.96 N-m, theta = 30^@` and number of turns in coil N = 20.
From relation `tau = NIAB sin theta,` we have
`B = (tau)/(NIA sin theta) = (0.96)/(20 xx 12 xx 0.01 xx sin 30^(@)) = 0.8 T`.