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At what rate a single conductor should cut the magnetic flux so that current of $1.5 \mathrm{~mA}$ flows through it when a resistance of $5 \Omega$ is connected across its ends?
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The correct answer is:
$7.5 \times 10^{-3} \frac{\mathrm{wb}}{\mathrm{s}}$
$\begin{aligned} & \mathrm{I}=\frac{\mathrm{e}}{\mathrm{R}} \text { or } \mathrm{e}=\mathrm{IR} \\ & \mathrm{e}=\frac{\mathrm{d} \phi}{\mathrm{dt}} \therefore \frac{\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{IR}=1.5 \times 10^{-3} \times 5 \\ & =7.5 \times 10^{-3} \mathrm{~Wb} / \mathrm{s}\end{aligned}$
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