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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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$\frac{B_{0}^{2} π^{2} r^{4} R}{5}$

$\frac{B_{0}^{2} π^{2} r^{4}}{R}$

$\frac{B_{0} {10r}^{3}}{R}$

$\frac{B_{0}^{2} {πr}^{2}}{R}$

answer is D.

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Detailed Solution

$P = \frac{ε^{2}}{R}; ε = - \frac{d}{dt} (BA) = A \frac{d}{dt} (B_{0} e^{-t}) {= AB}_{0} e^{-t}$

$\Rightarrow P = \frac{1}{R} {({AB}_{0} e^{-t})}^{2} = \frac{A^{2} B_{0}^{2} e^{-2t}}{R}$

At the time of starting t = 0 so $P = \frac{A^{2} B_{0}^{2}}{R}$

$\Rightarrow P = \frac{{({πr}^{2})}^{2} B_{0}^{2}}{R} = \frac{B_{0}^{2} π^{2} r^{4}}{R}$

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