Question:
In a uniform magnetic field, the magnetic needle has a magnetic moment $9.85 \times 10^{-2} \mathrm{~A} / \mathrm{m}^{2}$ and moment of inertia $5 \times 10^{-6} \mathrm{~kg} \mathrm{~m}^{2}$. If it performs 10 complete oscillations in 5 seconds then the
magnitude of the magnetic field is___________ $\mathrm{mT}$.
[Take $\pi^{2}$ as $9.85$ ]
Solution:
$T=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}}$
$\mathrm{B}=80 \times 10^{-4}=8 \mathrm{mT}$
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