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(A) According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ is given by $\varepsilon = -\frac{d\phi}{dt}$.From

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$0.5$

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(A) According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ is given by $\varepsilon = -\frac{d\phi}{dt}$.From Ohm's law,the induced current is $I = \frac{\varepsilon}{R} = -\frac{1}{R} \frac{d\phi}{dt}$.The charge $Q$ flowing through the circuit is given by $Q = \int I dt = -\frac{1}{R} \int_{\phi_1}^{\phi_2} d\phi$.Therefore,the magnitude of the induced charge is $|Q| = \frac{\phi_2 - \phi_1}{R}$.Given: Resistance $R = 100\, \Omega$,initial flux $\phi_1 = 10\, 	ext{Wb}$,and final flux $\phi_2 = 60\, 	ext{Wb}$.Substituting the values: $Q = \frac{60 - 10}{100} = \frac{50}{100} = 0.5\, 	ext{C}$.

The magnetic flux linked with a circuit of resistance $100\, \Omega$ increases from $10\, \text{Wb}$ to $60\, \text{Wb}$. The amount of induced charge that flows in the circuit is (in coulomb):

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