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?

Explanation

Length of a side of the square coil, `l = 10 cm = 0.1m`
Current flowing in the coil, `I = 12A`
Number of turns on the coil, `n = 20`
Angle made by the plane of the coil with magnetic field, `theta = 30^(@)`
Strength of magnetic field, `B = 0.80T`
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,
`T = n BIA sin theta`
Where,
`A =` Area of the square coil
`rArr l xxl = 0.1 xx 0.1 = 0.01 m^(2)`
`:. T = 20 xx 0.8 xx 12 xx 0.01 xx sin 30^(@)`
`= 0.96N`
Hence, the magnitude of the torque experienced by the coil is `0.96Nm`.