For a moving coil galvanometer, the deflection in the coil is 0.05 rad when a current of 10 mA is passes through it. If the torsional constant of suspension wire is 4.0×10⁻⁵N m rad⁻¹, the magnetic field is 0.01T and the number of turns in the coil is 200, the area of each turn (in cm²) is :

Solution

Physics sir · 4 steps

The coil rests when magnetic torque NBIA equals the wire's torque kθ, so solve for A.

1. Balance torques. The current in the coil feels a magnetic torque NBIA. As the coil turns, the suspension wire twists and pushes back with a torque kθ. The deflection stays steady when the two are equal.

   NBIA = kθ ⇒ A = kθ/NBI

2. Put in numbers. Use SI units, so I = 10 mA = 0.01 A. The top is kθ = 4.0 × 10⁻⁵ × 0.05 = 2.0 × 10⁻⁶ N m. The bottom is NBI = 200 × 0.01 × 0.01 = 0.02 N m⁻¹.
3. Convert units. A = (2.0 × 10⁻⁶)/0.02 = 1.0 × 10⁻⁴ m². Since 1 m² = 10⁴ cm², this is 1.0 cm².
4. So. The area of each turn is 1.0 cm², which is option (d).

The idea to keep. Every moving coil galvanometer question starts from NBIA = kθ. Put every quantity in SI first, and convert the answer to the unit asked only at the end.

Answer from the NTA final answer key. Solution by PrepWiser.