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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Solution

The correct option is D
$P = \frac{e^{2}}{R}; e = - \frac{d}{d t} (B A) = - A \frac{d}{d t} (B_{o} e^{- t}) = A B_{o} e^{- t}$
$\Rightarrow P = \frac{1}{R} (A B_{o} e^{- t})^{2} = \frac{A^{2} B_{o}^{2} e^{- 2 t}}{R}$
At the time of starting t = 0 so $P = \frac{A^{2} B_{o}^{2}}{R}$
$\Rightarrow P = \frac{(π r^{2})^{2} B_{o}^{2}}{R} = \frac{B_{o}^{2} π^{2} r^{4}}{R}$

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