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A rectangular coil of effective area $0.05 \mathrm{~m}^2$ is suspended freely in a radial magnetic field of $0.01 \mathrm{~Wb} / \mathrm{m}^2$. The torsional constant of the suspension fiber is $5 \times 10^{-9} \mathrm{Nm} /$ degree. If a current of $300 \mu \mathrm{A}$ is passed through it, then the angle through which the coil rotates is
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The correct answer is:
$30^{\circ}$
$\begin{aligned} & I=\frac{K \theta}{n B A} \\ & \therefore \theta=\frac{\mathrm{nIAB}}{\mathrm{K}}=\frac{1 \times 300 \times 10^{-6} \times 5 \times 10^{-2} \times 10^{-2}}{5 \times 10^{-9}} \\ & \therefore \theta=30^{\circ}\end{aligned}$
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