Question

The figure shows a circular loop of radius $r$ and a resistor of resistance $R$. A variable magnetic field, $B=B_0{e}^{-t}$ where $B_0$ is a constant and $t$ is time, is established inside the loop. If the key $K$ is closed at $t=0$, the electrical power developed just after closing the switch is -

• A. $\frac{B_0^2{\pi }^{2}r}{R}$
• B. $\frac{B_0^2{\pi }^2{r}^2}{R}$
• C. $\frac{B_0^2{\pi }^2{r}^3}{R}$
• D. $\frac{B_0^2{\pi }^2{r}^4}{R}$

Answer 1

The correct option is D $\frac{B_0^2{\pi }^2{r}^4}{R}$

Magnetic flux linked with the coil is,

$\varphi =B.A=B_0{e}^{-t}\left(\pi r^2\right)=B_0\pi r^2{e}^{-t}$

Now, emf induced in the loop is,

$E=-\frac{d\varphi }{dt}=-\frac{d\left(B_0\pi r^2{e}^{-t}\right)}{dt}$

$=-B_0\pi r^2\frac{d\left(e^{-t}\right)}{dt}$

$=B_0\pi r^2{e}^{-t}$

At $t=0$,

$E=B_0\pi r^2\times e^0=B_0\pi r^2$

Now, electric power developed in the loop at $t=0$ is,

$P=\frac{{\epsilon }^2}{R}$

$=\frac{(B_0\pi r^2{)}^2}{R}=\frac{B_0^2{\pi }^2{r}^4}{R}$

Hence, option $\left(D\right)$ is the correct answer.