Q. 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
A) $\frac{B_{0}^{2}\pi r^{2}}{R}$
B) $\frac{B_{0}10r^{3}}{R}$
C) $\frac{B_{0}^{2}\pi^{2}r^{4}R}{5}$
D) $\frac{B_{0}^{2}\pi^{2}r^{4}}{R}$
Explanation
(d) $P = \frac{e^{2}}{R};e = - \frac{d}{dt}(BA) = A\frac{d}{dt}\left( B_{o}e^{- t} \right) = AB_{o}e^{- t}$
$\Rightarrow P = \frac{1}{R}\left( AB_{o}e^{- t} \right)^{2} = \frac{A^{2}B_{o}^{2}e^{- 2t}}{R}$
At the time of starting $t = 0$ so $P = \frac{A^{2}B_{o}^{2}}{R}$
$\Rightarrow P = \frac{\left( \pi r^{2} \right)^{2}B_{o}^{2}}{R} = \frac{B_{o}^{2}\pi^{2}r^{4}}{R}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.