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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 ]
 

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\begin{aligned} &M=9.85 \times 10^{-2} \mathrm{~A} / \mathrm{m}^{2} \\ &I=5 \times 10^{-6} \mathrm{~kg} \mathrm{} \mathrm{m}^{2} \\ &T=\frac{5}{10}=0.5 \mathrm{~s} \end{aligned}

\begin{aligned} &T=2 \pi \sqrt{\frac{I}{M B}} \\ &0.5=2 \pi \times \sqrt{\frac{5 \times 10^{-6}}{9.85 \times 10^{-2}} \times B} \\ &0.25=4 \pi^{2} \times \frac{5 \times 10^{-6}}{9.85 \times 10^{-2} \times B} \end{aligned}

\begin{aligned} B &=\frac{20}{0.25} \times 10^{-4} . \\ &=80 \times 10^{-4} \\ B &=8 \times 10^{-3}T=8\; mT \end{aligned}

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vishal kumar

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