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A circular ring of diameter $20 \mathrm{~cm}$ has a resistance of $0.01 \Omega$. The charge that will flow through the ring if it is turned from a position perpendicular to a uniform magnetic field of $2.0 \mathrm{~T}$ to a position parallel to the field is about
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$6.3 \mathrm{C}$
As $\begin{aligned} q & =\frac{-d \phi}{R}=\frac{B A\left(\cos 0^{\circ}-\cos 90^{\circ}\right)}{R} \\ & =\frac{B \pi r^2(1-0)}{R}=\frac{B \pi r^2}{R} \\ & =\frac{2 \times 3.14 \times\left(10^{-1}\right)^2}{0.01} \\ & =6.28 \mathrm{C} \simeq 6.3 \mathrm{C}\end{aligned}$
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