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 × 10⁻⁵ N 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 cm ²) is: (a) 1.5 (b) 2.0 (c) 0.5 (d) 1.0

Answer: (d). Equilibrium: NBIA = kθ ⇒ A = (kθ)/(NBI). A = 4×10⁻⁵ × 0.05200 × 0.01 × 10×10⁻³ = 2×10⁻⁶0.02 A = 10⁻⁴ m² = 1 cm² So the answer is (d) 1.0.

For a moving coil galvanometer, the deflection in the coil is 0.05… | JEE Main 2023 (25 Jan, Shift 2)

Q18JEE Main 2023 · 25 JAN · SHIFT 2Medium

Moving Coil Galvanometer, Ammeter & Voltmeter

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 $4.0 \times 10^{-5}\text{ N m rad}^{-1}$ , the magnetic field is 0.01T and the number of turns in the coil is 200, the area of each turn (in cm $^{2}$ ) is:

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Answer: (d)

Equilibrium: $NBIA = k\theta \Rightarrow A = \frac{k\theta}{NBI}$ .
$A = \frac{4\times10^{-5} \times 0.05}{200 \times 0.01 \times 10\times10^{-3}} = \frac{2\times10^{-6}}{0.02}$
$A = 10^{-4}\text{ m}^2 = 1\text{ cm}^2$
So the answer is (d) 1.0.

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