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(B) According to Faraday's law of electromagnetic induction,the induced $emf$ is given by $e = -\frac{d\phi}{dt}$.In the first case,the relative veloc

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$2e$

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(B) According to Faraday's law of electromagnetic induction,the induced $emf$ is given by $e = -\frac{d\phi}{dt}$.In the first case,the relative velocity between the magnet and the coil is $v$. Thus,the rate of change of magnetic flux is $\left(\frac{d\phi}{dt}\right)_1 = k \cdot v$,where $k$ is a constant depending on the magnetic field gradient and coil geometry. So,$e = k \cdot v$.In the second case,both the magnet and the coil move away from each other with speed $v$. The relative velocity between them is $v_{rel} = v + v = 2v$.The rate of change of magnetic flux is proportional to the relative velocity,so $\left(\frac{d\phi}{dt}\right)_2 = k \cdot (2v) = 2(k \cdot v) = 2e$.Therefore,the induced $emf$ in the second case will be $2e$.

In the following figure,the magnet is moved towards the coil with a speed $v$ and the induced $emf$ is $e$. If the magnet and the coil recede away from one another,each moving with speed $v$,the induced $emf$ in the coil will be: (Assume the separation in both cases is the same)

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