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A vibration magnetometer placed in magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of 2 sec in earth's horizontal magnetic field of 24 microtesla. When a horizontal field of 18 microtesla is produced opposite to the earth's field by placing a current carrying wire, the new time period of magnet will be

1. 1 s
2. 2 s
3. 3 s
4. 4 s

Correct answer: 3 s

Solution

The time period of oscillation of a magnetic dipole in a magnetic field is given by T = 2π√(I/MB), where I is the moment of inertia, M is the magnetic moment, and B is the net magnetic field. Initially, B = 24 μT, and when the opposing field of 18 μT is added, the net field becomes B = 24 - 18 = 6 μT. Since T ∝ 1/√B, the new time period is T' = T√(B/B') = 2√(24/6) = 2√4 = 4 s.

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