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Physics · PYQ NEET UG 2024
हिन्दीIn a uniform magnetic field of \( 0.049 \text{ T} \), a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is \( 9.8 \times 10^{-6} \text{ kg m}^2 \). If the magnitude of magnetic moment of the needle is \( x \times 10^{-5} \text{ A m}^2 \); then the value of 'x' is :
Correct answer · Option B
\( 1280 \pi^2 \)
Explanation
Time period \( T = \frac{5}{20} = 0.25 \text{ s} \). Using \( T = 2\pi \sqrt{\frac{I}{M B}} \), we get \( M = \frac{4\pi^2 I}{T^2 B} = \frac{4\pi^2 (9.8 \times 10^{-6})}{(1/16) \times 0.049} = 1280\pi^2 \times 10^{-5} \text{ A m}^2 \). Hence \( x = 1280\pi^2 \).
