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In a closed circuit there is only a coil of inductance $L$ and resistance $100 \Omega$. The coil is situated in a uniform magnetic field. All on a sudden, the magnetic flux linked with the circuit changes by $5 \mathrm{Weber}$. What amount of charge will flow in the circuit as a result?
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$0.05 \mathrm{C}$
Emf $=-\frac{\mathrm{d} \phi}{\mathrm{dt}}($ Faraday's Law $)$
$\mathrm{i}_{\text {induced }}=\frac{\mathrm{Emf}}{\mathrm{R}}=-\frac{\mathrm{d} \phi}{\mathrm{dt}} \times \frac{1}{\mathrm{R}}$
$\Rightarrow \frac{\mathrm{dq}}{\mathrm{dt}}=-\frac{1}{\mathrm{R}} \frac{\mathrm{d} \phi}{\mathrm{dt}} \Rightarrow \mathrm{dq}=-\frac{\mathrm{d} \phi}{\mathrm{R}} \Rightarrow \int \mathrm{dq}=-\int \frac{\mathrm{d} \phi}{\mathrm{R}}$
$\Rightarrow \mathrm{q}_{\text {flown }}=\frac{\phi_{\mathrm{i}}-\phi_{\mathrm{f}}}{\mathrm{R}}=\frac{5 \text { weber }}{100}=0.05 \mathrm{C} .$
$\mathrm{i}_{\text {induced }}=\frac{\mathrm{Emf}}{\mathrm{R}}=-\frac{\mathrm{d} \phi}{\mathrm{dt}} \times \frac{1}{\mathrm{R}}$
$\Rightarrow \frac{\mathrm{dq}}{\mathrm{dt}}=-\frac{1}{\mathrm{R}} \frac{\mathrm{d} \phi}{\mathrm{dt}} \Rightarrow \mathrm{dq}=-\frac{\mathrm{d} \phi}{\mathrm{R}} \Rightarrow \int \mathrm{dq}=-\int \frac{\mathrm{d} \phi}{\mathrm{R}}$
$\Rightarrow \mathrm{q}_{\text {flown }}=\frac{\phi_{\mathrm{i}}-\phi_{\mathrm{f}}}{\mathrm{R}}=\frac{5 \text { weber }}{100}=0.05 \mathrm{C} .$
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