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(B) According to Faraday's Law of electromagnetic induction, the induced electromotive force (Emf) is given by: $\epsilon = -\frac{d\phi}{dt}$.Since t

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$0.05 	ext{ C}$

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(B) According to Faraday's Law of electromagnetic induction, the induced electromotive force (Emf) is given by: $\epsilon = -\frac{d\phi}{dt}$.Since the circuit has a resistance $R$, the induced current $i$ is: $i = \frac{\epsilon}{R} = -\frac{1}{R} \frac{d\phi}{dt}$.We know that current $i = \frac{dq}{dt}$, where $dq$ is the small amount of charge flowing in time $dt$.Substituting this, we get: $\frac{dq}{dt} = -\frac{1}{R} \frac{d\phi}{dt}$.Integrating both sides, we find the total charge $q$ that flows:$q = \int dq = -\frac{1}{R} \int d\phi = \frac{\Delta\phi}{R}$.Given $\Delta\phi = 5 	ext{ Wb}$ and $R = 100 \Omega$, the charge $q$ is:$q = \frac{5}{100} = 0.05 	ext{ C}$.

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. Suddenly, the magnetic flux linked with the circuit changes by $5 \text{ Wb}$. What amount of charge will flow in the circuit as a result?

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