1. 4.55 μ J 2. 2.3 μ J 3. 1.15 μ J 4. 9.1 μ J

Answer: 4. 9.1 μ J Key idea: work done in rotating a coil from θ=0 to 180^° is W=MB(cos0-cos180)=2MB , with M=NIA . A=2.1×10⁻²×1.25×10⁻²=2.625×10⁻⁴ m ² . M=250×85×10⁻⁶×2.625×10⁻⁴≈5.6×10⁻⁶ A m ² . W=2×5.6×10⁻⁶×0.85≈9.5\ μ J, nearest to 9.1 μ J.

Topic: Torque on current loop, moving coil galvanometer, conversion to ammeter and voltmeter· Magnetic Effects of Current and Magnetism

A 250 - Turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 μA and subjected to a magnetic field of strength 0.85 T. Work done for rotating the coil by 180^° against the torque is :

1. 4.55 \(\mu\)J

2. 2.3 \(\mu\)J

3. 1.15 \(\mu\)J

4. 9.1 \(\mu\)J

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Answer: 4. 9.1 \(\mu\)J

Key idea: work done in rotating a coil from \(\theta=0\) to \(180^\circ\) is \(W=MB(\cos0-\cos180)=2MB\), with \(M=NIA\).

\(A=2.1\times10^{-2}\times1.25\times10^{-2}=2.625\times10^{-4}\) m\(^2\).

\(M=250\times85\times10^{-6}\times2.625\times10^{-4}\approx5.6\times10^{-6}\) A m\(^2\).

\(W=2\times5.6\times10^{-6}\times0.85\approx9.5\ \mu\)J, nearest to 9.1 \(\mu\)J.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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A 250 - Turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 μA and subjected to a magnetic field of strength 0.85 T. Work done for rotating the coil by 180^° against the torque is :

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