# A square coil of side 10 cm consists of 20 turns and carries a current of 12 A.

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 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?

Solution:

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.