Question

A rectangular coil, of sides 4 cm and 5 cm respectively, has 50 turns in it. It carries a current of 2 A, and is placed in a uniform magnetic field of 0.5 T in such a manner that its plane makes an angle of 60^o with the field direction. Calculate the torque on the loop.

(a) 5 × 10^-4 N m

(b) 5 × 10^-2 N m

(c) 5 × 10^-3 N m

(d) 5 × 10^-5 N m

This question was posed to me in an interview for internship.

My question comes from Solenoid and Toroid in portion Moving Charges and Magnetism of Physics – Class 12

Answer 1

Correct option is (b) 5 × 10^-2 N m

The best explanation: The magnitude of torque experienced by a current carrying coil kept in a uniform magnetic field is given by:

τ = MB sin (θ) = (NIA) × B sin (θ)

M → Magnetic moment of the coil

N → Number of turns

I → Current in coil

A → Area of the coil

θ → Angle between normal to the plane of the coil and the direction of the magnetic field

Given: N = 50; A = 4 × 5 = 20 cm^2 = 20 × 10^-4 m^2; B = 0.5 T; I = 2 A

θ = 90^o – 60^o = 30^o

τ = (NIA) × B sin (θ)

τ = 50 × 2 × 0.5 × 20 × 10^-4 × sin (30^o)

τ = 0.05

τ = 5 × 10^-2 N m