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(B) According to Faraday's law of electromagnetic induction,the induced $e.m.f.$ is given by $\varepsilon = -\frac{d\phi}{dt}$.For loop $1$ (radius $R

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$-\frac{d\vec{B}}{dt}\pi r^2$ in loop $1$ and $0$ in loop $2$

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(B) According to Faraday's law of electromagnetic induction,the induced $e.m.f.$ is given by $\varepsilon = -\frac{d\phi}{dt}$.For loop $1$ (radius $R > r$): The magnetic flux $\phi_1$ is linked only with the region of radius $r$ where the magnetic field exists. Thus,$\phi_1 = B \cdot A = B(\pi r^2)$.The induced $e.m.f.$ is $\varepsilon_1 = -\frac{d}{dt}(B \pi r^2) = -\pi r^2 \frac{dB}{dt}$.For loop $2$ (radius $R$): The loop is completely outside the region where the magnetic field exists. Therefore,the magnetic flux linked with loop $2$ is $\phi_2 = 0$.The induced $e.m.f.$ is $\varepsilon_2 = -\frac{d\phi_2}{dt} = 0$.

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