A moving coil galvanometer has 100 turns and each turn has an area of 2.0 cm². The magnetic field produced by the magnet is 0.01 T and the deflection in the coil is 0.05 radian when a current of 10 mA is passed through it. The torsional constant of the suspension wire is x × 10⁻⁵ N-m / rad. The value of x is _______ .

Solution

Physics sir · 4 steps

At steady deflection the magnetic torque equals the spring's restoring torque, so set NIBA equal to kθ.

1. Balance the torques. The field turns the coil with torque NIAB. The suspension wire twists back with torque kθ. The coil stops where the two are equal.

   NIAB = kθ ⇒ k = NIAB/θ

2. Convert to SI. A = 2.0 cm² = 2.0 × 10⁻⁴ m² and I = 10 mA = 1.0 × 10⁻² A. Also N = 100, B = 0.01 T and θ = 0.05 rad.
3. Put in numbers. NIAB = 100 × 10⁻² × 2 × 10⁻⁴ × 10⁻² = 2 × 10⁻⁶ N m. So k = (2 × 10⁻⁶)/0.05 = 4 × 10⁻⁵ N m rad⁻¹.
4. So. The torsional constant is 4 × 10⁻⁵ N m rad⁻¹, so x = 4.

The common slip: 400. Taking 1 cm² as 10⁻² m² instead of 10⁻⁴ m² makes k a hundred times too large and gives 400, not 4.

The idea to keep. In a moving coil galvanometer, deflection settles where NIAB = kθ. Put every quantity in SI first, especially area in m² and current in amperes.

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