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 mI 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

A solenoid of 0.5 m length with 100 turns carries a current of 5 A. A

1.	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 mI 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» 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

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