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 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
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 = n BIA sinθ
Where,
A = Area of the square coil
$\Rightarrow \mid \times 1=0.1 \times 0.1=0.01 \mathrm{~m}^{2}$
$\therefore T=20 \times 0.8 \times 12 \times 0.01 \times \sin 30^{\circ}$
= 0.96 N m
Hence, the magnitude of the torque experienced by the coil is 0.96 N m.