A magnetic needle has magnetic moment 9.85 times 10^ -2 text A/m ^2 and moment of inertia 5 times 10^ -6 text kg,m ^2. It performs 10 complete oscillations in 5 seconds in a uniform magnetic field. Find the magnitude of the field in mT.(Take pi^2 = 9.85)

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EEJ MAIN Physics QUESTION #7013

Question 1

A magnetic needle has magnetic moment $9.85 \times 10^{-2}\ \text{A/m}^2$ and moment of inertia $5 \times 10^{-6}\ \text{kg\,m}^2$. It performs 10 complete oscillations in 5 seconds in a uniform magnetic field. Find the magnitude of the field in mT.
(Take $\pi^2 = 9.85$)

10 mT✔️
20 mT
5 mT
1 mT

Correct Answer Explanation

Time period: $T = \frac{5}{10} = 0.5\ \text{s}$

Formula for oscillation: $T = 2\pi\sqrt{\frac{I}{MB}}$

$B = \frac{4\pi^2 I}{MT^2} = \frac{4 \times 9.85 \times 5\times10^{-6}}{9.85\times10^{-2} \times (0.5)^2}$

$= \frac{4 \times 9.85 \times 5\times10^{-6}}{9.85\times10^{-2} \times 0.25} = \frac{20\times10^{-5}}{2.4625\times10^{-3}} = 0.01\ \text{T} = 10\ \text{mT}$

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