In a uniform magnetic field of 0.049 T , a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is $9.8 \times 1 0^{- 6} kg m^{2}$ . If the magnitude of magnetic moment of the needle is $x \times 1 0^{- 5} Am^{2}$ , then the value of ' $x$ ' is :
NEET 2024

Question

TestHub · Accepted Answer

Time period of Oscillation, T = 2 π MB I ​ ​ ​ ⇒ 4 1 ​ = 2 π M × 0.049 9.8 × 1 0 − 6 ​ ​ ⇒ 16 1 ​ = 4 π 2 × M × 49 × 1 0 − 3 9.8 × 1 0 − 6 ​ ⇒ M = 49 × 1 0 − 3 4 π 2 × 9.8 × 1 0 − 6 ​ × 16 = 49 4 π 2 × 9.8 × 16 × 1 0 − 3 ​ = 12.8 π 2 × 1 0 − 3 × 1 0 − 2 × 1 0 2 = 1280 π 2 × 1 0 − 5 Am 2 ​

Physics - Magnetism Question with Solution | TestHub

Options:

Doubts & Discussion