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(D) The magnetic flux $\phi_2$ linked with loop $(2)$ is proportional to the current $i_1$ in loop $(1)$,given by $\phi_2 = M i_1$,where $M$ is the mu

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(D) The magnetic flux $\phi_2$ linked with loop $(2)$ is proportional to the current $i_1$ in loop $(1)$,given by $\phi_2 = M i_1$,where $M$ is the mutual inductance.According to Faraday's law of electromagnetic induction,the induced emf $\varepsilon_2$ in loop $(2)$ is $\varepsilon_2 = -\frac{d\phi_2}{dt} = -M \frac{di_1}{dt}$.The induced current $i_2$ is given by $i_2 = \frac{\varepsilon_2}{R} = -\frac{M}{R} \frac{di_1}{dt}$,where $R$ is the resistance of loop $(2)$.From the given graph,$i_2$ is a positive constant value. This implies that $\frac{di_1}{dt}$ must be a negative constant.Therefore,the current $i_1$ must be decreasing linearly with time $t$ (i.e.,having a negative slope).Looking at the options,the graph where $i_1$ decreases linearly with time $t$ corresponds to option $(D)$.

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