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 _______. Numerical answer type. Work it out, then open the answer.

Answer: 4. Torque balance: NIAB = kθ ⇒ k = (NIAB)/(θ). k = 100 × 10 × 10⁻³ × 2 × 10⁻⁴ × 0.010.05 k = 2 × 10⁻⁶0.05 = 4 × 10⁻⁵ N-m/rad So x = 4.

A moving coil galvanometer has 100 turns and each turn has an area of… | JEE Main 2024 (1 Feb, Shift 2)

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Moving Coil Galvanometer, Ammeter & Voltmeter

A moving coil galvanometer has 100 turns and each turn has an area of $2.0\text{ cm}^{2}$ . The magnetic field produced by the magnet is $0.01\text{ T}$ and the deflection in the coil is 0.05 radian when a current of $10\text{ mA}$ is passed through it. The torsional constant of the suspension wire is $x \times 10^{-5}\text{ N-m/rad}$ . The value of $x$ is _______.

Numerical answer type. Work it out, then open the answer.

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Answer: 4

Torque balance: $NIAB = k\theta \Rightarrow k = \frac{NIAB}{\theta}$ .
$k = \frac{100 \times 10 \times 10^{-3} \times 2 \times 10^{-4} \times 0.01}{0.05}$
$k = \frac{2 \times 10^{-6}}{0.05} = 4 \times 10^{-5}$ N-m/rad
So $x = 4$ .

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