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

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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 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 $(emf)$ is $E = -\frac{d\phi}{dt}$.$E = -\frac{d}{dt} (B_0 \pi r^2 e^{-t}) = -\pi r^2 B_0 \frac{d}{dt}(e^{-t}) = \pi r^2 B_0 e^{-t}$.At the instant the key $(K)$ is closed,$t = 0$.Therefore,the induced $emf$ at $t = 0$ is $E_0 = \pi r^2 B_0 e^0 = \pi r^2 B_0$.The electrical power $P$ developed in the resistor $R$ is given by $P = \frac{E_0^2}{R}$.Substituting the value of $E_0$,we get $P = \frac{(\pi r^2 B_0)^2}{R} = \frac{B_0^2 \pi^2 r^4}{R}$.

Shown in the figure is a circular loop of radius $r$ and resistance $R$. $A$ variable magnetic field of induction $B = B_0 e^{-t}$ 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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