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Question

A bar magnet takes π/10 second the complete one oscillation in an oscillation magnetometer. The moment of inertia of the magnet about the axis of rotation is 1.2 × 10−4 kg m2 and the earth's horizontal magnetic field is 30 μT. Find the magnetic moment of the magnet.

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Solution

Given:
Time taken by the bar magnet to complete one oscillation, T = π10 s
Moment of inertia of the magnet about the axis of rotation, I = 1.2 × 10−4 kgm2
Horizontal component of Earth's magnetic field, BH = 30 μT
Time period of oscillating magnetometer T is given by
T=2πIMBH
Here,
M = Magnetic moment of the magnet
On substituting the respective values, we get
π10=2π1.2×10-4M×30×10-61202=1.2×10-4M×30×10-6M=1.2×10-4×40030×10-6M=16×102 A-m2M=1600 A-m2

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