Torque required to hold a small circular coil of 10 turns, 2 × 10 - 4 m 2 area and carrying 0 . 5 A current is the middle of a long solenoid of 10 3 turns per meter carrying 3 A current, with its axis perpendicular to the axis of the solenoid is

Step-by-step solution

Dipole moment of the coil, M = N A I = 10 × 2 × 10 - 4 × 0 . 5 ⇒ M = 10 - 3   A   m 2 Magnetic field due to solenoid, B = μ 0 n I = 4 π × 10 - 7 × 10 3 × 3 ⇒ B = μ 0 n I = 12 π × 10 - 4   T Torque = M B sin θ = 12 π × 10 - 4 × 10 - 3 Torque = 12 π × 10 - 7   N   m