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A bar magnet with moment of inertia 9 × 10⁻⁵ kg m² is suspended in a vibration magnetometer and performs oscillations in a uniform magnetic field of 16π² × 10⁻⁵ T. If it completes 20 oscillations in 15 s, what is the magnetic moment of the bar magnet?

1. 3 Am²
2. 2 Am²
3. 5 Am²
4. 4 Am²

Correct answer: 4 Am²

Solution

T = 15/20 = 0.75 s. From T = 2*pi*sqrt(I/(M*B)): M = 4*pi^2*I/(T^2*B) = 4*pi^2*9e-5/(0.5625*16*pi^2*1e-5) = 4 A*m^2.

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