Question

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. $\frac{B_o^2\pi r^2}{R}$
• B. $\frac{B_{o}10r^3}{R}$
• C. $\frac{B_o^2{\pi }^2{r}^{4}R}{5}$
• D. $\frac{B_o^2{\pi }^2{r}^4}{R}$

Answer 1

The correct option is D $\frac{B_o^2{\pi }^2{r}^4}{R}$

$P=\frac{e^2}{R};e=-\frac{d}{dt}\left(BA\right)=-A\frac{d}{dt}\left(B_o{e}^{-t}\right)=A{B}_o{e}^{-t}$
$\Rightarrow P=\frac{1}{R}(A{B}_o{e}^{-t}{)}^2=\frac{A^2{B}_o^2{e}^{-2t}}{R}$
At the time of starting t = 0 so $P=\frac{A^2{B}_o^2}{R}$
$\Rightarrow P=\frac{(\pi r^2{)}^2{B}_o^2}{R}=\frac{B_o^2{\pi }^2{r}^4}{R}$