Question

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 $30\degree$ with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Answer 1

Length of a side of the square coil, $l$ = 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, $θ$ = 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,
$τ = nBIAsinθ$
Where,
A = Area of the square coil
= $l × l = 0.1 × 0.1 = 0.01 m^2$
So, $τ = 20 × 0.8 × 12 × 0.01 × sin30\degree$
= $0.96 N m$
Hence, the magnitude of the torque experienced by the coil is $0.96 N m$.