NEET 2024 · PhysicsPrevious Year Question
In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds. The moment of inertia of the needle is 9.8 × 10⁻⁶ kg m². If the magnetic moment of the needle is x × 10⁻⁵ A m², the value of x is:
Answer: (C) 1280π². Answer: (C) Solution: Period T = 5/20 = 0.25 s.
- A.128π²
- B.50π²
- C.1280π²✓
- D.5π²
Correct Answer
(C) 1280π²
Solution & Explanation
Answer: (C) Solution: Period T = 5/20 = 0.25 s. A dipole in a uniform field executes SHM: T = 2π√(I/mB) ⟹ m = 4π²I/(B T²). m = 4π² × 9.8×10⁻⁶ / (0.049 × 0.0625) = 4π² × 9.8×10⁻⁶ / 3.0625×10⁻³ = 0.0128 π² = 1280π² × 10⁻⁵ A m². Hence x = 1280π².
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