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(B) When a copper rod is moved in a magnetic field,an electromotive force $(emf)$ is induced.According to Faraday's law of electromagnetic induction,t

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rate of change of magnetic flux

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(B) When a copper rod is moved in a magnetic field,an electromotive force $(emf)$ is induced.According to Faraday's law of electromagnetic induction,the induced $emf$ $(e)$ is given by $e = -\frac{d\phi}{dt}$.Since the rod has a resistance $(R)$,the induced current $(i)$ is given by $i = \frac{e}{R} = -\frac{1}{R} \frac{d\phi}{dt}$.We know that current is the rate of flow of charge,so $i = \frac{dQ}{dt}$.Substituting this into the equation,we get $\frac{dQ}{dt} = -\frac{1}{R} \frac{d\phi}{dt}$.Integrating both sides,we find that the total charge $Q$ developed is proportional to the change in magnetic flux $\Delta\phi$.However,the instantaneous rate of charge flow is proportional to the rate of change of magnetic flux,$\frac{d\phi}{dt}$.

$A$ copper rod is moved in a magnetic field. The charge developed across its ends will be proportional to

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