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(B) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = B_0 \pi r^2 e^{-t/	au}$.According to Faraday's law,the induced electrom

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$\frac{\pi^2 r^4 B_0^2}{2	au R}$

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(B) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = B_0 \pi r^2 e^{-t/	au}$.According to Faraday's law,the induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\phi}{dt}$.$\varepsilon = -\frac{d}{dt} (B_0 \pi r^2 e^{-t/	au}) = \frac{B_0 \pi r^2}{	au} e^{-t/	au}$.The instantaneous power dissipated as heat is $P = \frac{\varepsilon^2}{R} = \frac{B_0^2 \pi^2 r^4}{	au^2 R} e^{-2t/	au}$.The total heat generated $H$ is the integral of power from $t = 0$ to $t = \infty$:$H = \int_{0}^{\infty} \frac{\varepsilon^2}{R} dt = \frac{B_0^2 \pi^2 r^4}{	au^2 R} \int_{0}^{\infty} e^{-2t/	au} dt$.Evaluating the integral: $\int_{0}^{\infty} e^{-2t/	au} dt = \left[ -\frac{	au}{2} e^{-2t/	au} ight]_{0}^{\infty} = 0 - (-\frac{	au}{2}) = \frac{	au}{2}$.Therefore,$H = \frac{B_0^2 \pi^2 r^4}{	au^2 R} \cdot \frac{	au}{2} = \frac{\pi^2 r^4 B_0^2}{2	au R}$.

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