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

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

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(D) The magnetic flux $\phi$ through the loop of area $A = \pi r^2$ is given by $\phi = B \cdot A = (B_0 e^{-t}) (\pi r^2)$.According to Faraday's law of induction,the induced electromotive force $(e)$ is given by $e = -\frac{d\phi}{dt}$.$e = -\frac{d}{dt} (B_0 \pi r^2 e^{-t}) = B_0 \pi r^2 e^{-t}$.The electrical power $(P)$ developed in the loop is given by $P = \frac{e^2}{R}$.$P = \frac{(B_0 \pi r^2 e^{-t})^2}{R} = \frac{B_0^2 \pi^2 r^4 e^{-2t}}{R}$.Right after closing the switch,the time $t = 0$.Substituting $t = 0$ into the power equation,we get $P = \frac{B_0^2 \pi^2 r^4 e^0}{R} = \frac{B_0^2 \pi^2 r^4}{R}$.

Shown in the figure is a circular loop of radius $r$ and resistance $R$. $A$ variable magnetic field of induction $B = B_0e^{-t}$ is established inside the coil. If the key $(K)$ is closed,the electrical power developed right after closing the switch is equal to

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