Q 12-05-053JEE MainJEE Main 2021 (27 Jul, Shift 1)Medium
In a uniform magnetic field, the magnetic needle has a magnetic moment $9.85\times10^{-2}\ \text{A m}^2$ and moment of inertia $5\times10^{-6}\ \text{kg m}^2$. If it performs $10$ complete oscillations in $5$ seconds then the magnitude of the magnetic field is ______ mT. [Take $\pi^2$ as $9.85$]
Numerical value type. Enter your answer.
Answer: 8
Time period $T = \dfrac{5}{10} = 0.5$ s, and $T = 2\pi\sqrt{\dfrac{I}{mB}}$.
$$B = \frac{4\pi^2 I}{mT^2} = \frac{4\times9.85\times5\times10^{-6}}{9.85\times10^{-2}\times0.25} = 8\times10^{-3}\ \text{T} = 8\ \text{mT}$$
Solution by Sreeraj P, M.Sc Physics