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?

Answer 1

Hey Dear,

◆ Answer -

τ = 0.96 N.m

● Explaination -

# Given -

l = 10 cm = 0.1 m

n = 20 turns

I = 12 A

θ = 30°

B = 0.8 T

# Solution -

Area of the square coil will be -

A = l^2

A = 0.1^2

A = 0.01 m^2

Torque experienced by the coil is calculated by -

τ = n.I.A.B.sinθ

τ = 20 × 12 × 0.01 × 0.8 × sin30°

τ = 1.92 × 0.5

τ = 0.96 N-m

Therefore, torque experienced by the coil is 0.96 N-m.

Thanks dear..