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A moving coil galvanometer has a rectangular wire coil of enclosed area $0.001 \mathrm{~m}^2$ and 500 turns. The coil operates in a radial magnetic field of $0.2 \mathrm{~T}$ and carries a current of $6 \pi \times 10^{-8} \mathrm{~A}$. If the torsional spring constant is $6 \times 10^{-7} \mathrm{~N}-\mathrm{m} / \mathrm{rad}$, then the angular deflection of the coil in radians is
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Verified Answer
The correct answer is:
$\frac{\pi}{100}$
Angular deflection of a moving coil galvanometer is given by
$$
\begin{aligned}
\theta & =\left(\frac{N B A}{k}\right) i \\
& =\frac{500 \times 0.2 \times 0.001}{6 \times 10^{-7}} \times 6 \pi \times 10^{-8} \\
& =\frac{5 \times 0.2 \times 10^{-1} \times 6 \pi \times 10^{-8}}{6 \times 10^{-7}} \\
\theta & =\frac{\pi}{100}
\end{aligned}
$$
$$
\begin{aligned}
\theta & =\left(\frac{N B A}{k}\right) i \\
& =\frac{500 \times 0.2 \times 0.001}{6 \times 10^{-7}} \times 6 \pi \times 10^{-8} \\
& =\frac{5 \times 0.2 \times 10^{-1} \times 6 \pi \times 10^{-8}}{6 \times 10^{-7}} \\
\theta & =\frac{\pi}{100}
\end{aligned}
$$
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