The magnitude of torque experienced by a square coil of side 12 cm which consists of 25 turns and carries a current 10 A suspended vertically and the normal to the plane of coil makes an angle of `30^(@)` with the direction of a uniform horizontal magnetic field of magnitude 0.9 T is

Explanation

Length of a side of the square coil, 1=10 cm=0.1 m
Current flowing in the coil, I= 12 A
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=0.80 T
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,
`tau=nBIAsin theta`
Where, A= Area of the square coil
`=1xx1=0.1xx0.1=0.01 m^(2)`
So,`tau=20xx0.8xx12xx0.01xx sin 30^(@)=0.96Nm`
Hence, the magnitude of the torque experienced by the coil is 0.96 N m.