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Topic: Torque on magnetic dipole in uniform field· Magnetic Effects of Current and Magnetism

In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is 9.8×10^-6 kg m^2. If the magnitude of magnetic moment of the needle is x×10^-5 Am^2; then the va…

In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is \(9.8\times10^{-6}\) kg m\(^2\). If the magnitude of magnetic moment of the needle is \(x\times10^{-5}\) Am\(^2\); then the value of 'x' is :

magnetic needle suspended in field B

1. \(1280\,\pi^2\)

2. \(5\,\pi^2\)

3. \(128\,\pi^2\)

4. \(50\,\pi^2\)

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Answer: 1. \(1280\,\pi^2\)

Key idea: A magnetic needle in a field oscillates like a torsional pendulum: \(T=2\pi\sqrt{\frac{I}{MB}}\).

Period \(T=\frac{5}{20}=0.25\) s. Squaring, \(M=\frac{4\pi^2 I}{BT^2}\).

\(M=\frac{4\pi^2\times 9.8\times10^{-6}}{0.049\times(0.25)^2}=\frac{4\pi^2\times 9.8\times10^{-6}}{3.0625\times10^{-3}}=1.28\times10^{-2}\pi^2\) A m\(^2\).

Writing this as \(x\times10^{-5}\) gives \(x=1280\,\pi^2\). Option 1.

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