Question

A solenoid of 0.5 m length with 100 turns carries a current of 5 A. A coil of 20 turns and of radius 0.02m carries a current of 0.6 A. What is the torque required to hold the coil with its axis at right angle to that of solenoid in the middle point of it?

(a) 1.60 N m

(b) 15.89 × 10^-5 N m

(c) 1.893 × 10^-5 N m

(d) 1.893 × 10^-8 N m

I had been asked this question during an online exam.

Query is from Solenoid and Toroid topic in chapter Moving Charges and Magnetism of Physics – Class 12

Answer 1

The correct choice is (c) 1.893 × 10^-5 N m

To explain: Given: μ0 = 4 × 3.14 × 10^-7; number of turns of solenoid = 100; Current passing through the solenoid = 5 A; Number of turns of coil = 20; Current passing through the coil = 0.6A; Length of the solenoid = 0.5 m; Radius of coil = 0.02 m

Magnetic field of solenoid (B) = μ0nI = μ0 × \(\frac {100}{0.5}\) × 5 = 4 × 3.14 × 10^-7 × \(\frac {100}{0.5}\) × 5

Magnetic moment of the coil (M) = I × A × N = 0.6 × (0.02)^2 × 20

Torque (τ) = MBsin (90^o)

τ = 0.6 × (0.02)^2 × 20 × 4 × 3.14 × 10^-7 × \(\frac {100}{0.5}\) × 5        [sin (90^o) = 1]

τ = 1.893 × 10^-5 N m