Q 12-05-007NEETJEE MainEAMCET 2007 (Engineering)Medium
A bar magnet has moment of inertia $49 \times 10^{-2}\ \text{kg m}^2$ and vibrates in a magnetic field of induction $0.5 \times 10^{-4}$ T. The time period of vibration is $8.8$ s. The magnetic moment of the bar magnet is
Answer: (D) $500\ \text{A m}^2$
Concept. For a magnet oscillating in a uniform field, $T = 2\pi\sqrt{\dfrac{I}{MB}}$, so $M = \dfrac{4\pi^2 I}{T^2 B}$.
Step 1. Substitute. With $4\pi^2 \approx 39.5$:
$$M = \frac{39.5 \times 0.49}{(8.8)^2 \times 0.5 \times 10^{-4}} = \frac{19.35}{77.44 \times 0.5 \times 10^{-4}} \approx 500\ \text{A m}^2$$
Answer. $500\ \text{A m}^2$ (Option D).
Solution by Sreeraj P, M.Sc Physics