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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 $9.8 \times 10^{-6} \mathrm{~kg} \mathrm{~m}^2$. If the magnitude of magnetic moment of the needle is $x \times 10^{-5} \mathrm{Am}^2$, then the value of ' $x$ ' is :
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Verified Answer
The correct answer is:
$1280 \pi^2$
Time period of Oscillation, $T=2 \pi \sqrt{\frac{I}{M B}}$
$\begin{aligned}
& \Rightarrow \quad \frac{1}{4}=2 \pi \sqrt{\frac{9.8 \times 10^{-6}}{M \times 0.049}} \\
& \Rightarrow \frac{1}{16}=4 \pi^2 \times \frac{9.8 \times 10^{-6}}{M \times 49 \times 10^{-3}} \\
& \Rightarrow \quad M=\frac{4 \pi^2 \times 9.8 \times 10^{-6}}{49 \times 10^{-3}} \times 16 \\
& =\frac{4 \pi^2 \times 9.8 \times 16 \times 10^{-3}}{49} \\
& =12.8 \pi^2 \times 10^{-3} \times 10^{-2} \times 10^2 \\
& =1280 \pi^2 \times 10^{-5} \mathrm{Am}^2 \\
&
\end{aligned}$
$\begin{aligned}
& \Rightarrow \quad \frac{1}{4}=2 \pi \sqrt{\frac{9.8 \times 10^{-6}}{M \times 0.049}} \\
& \Rightarrow \frac{1}{16}=4 \pi^2 \times \frac{9.8 \times 10^{-6}}{M \times 49 \times 10^{-3}} \\
& \Rightarrow \quad M=\frac{4 \pi^2 \times 9.8 \times 10^{-6}}{49 \times 10^{-3}} \times 16 \\
& =\frac{4 \pi^2 \times 9.8 \times 16 \times 10^{-3}}{49} \\
& =12.8 \pi^2 \times 10^{-3} \times 10^{-2} \times 10^2 \\
& =1280 \pi^2 \times 10^{-5} \mathrm{Am}^2 \\
&
\end{aligned}$
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