Question

A conducting metal circular -wire loop of radius $r$ is placed perpendicular to a magnetic field which varies with time as $B=B_0{e}^{-t/\tau }$, where $B_0$ and $\tau$ are constants, at $t=0.$ If the resistance of the loop is $R$ then the heat generated in the loop after a long time $(t\to \infty )$ is:

• A. $\frac{{\pi }^2{r}^4{B}_0^4}{2\tau R}$
• B. $\frac{{\pi }^2{r}^4{B}_0^2}{2\tau R}$
• C. $\frac{{\pi }^2{r}^4{B}_0^{2}R}{\tau }$
• D. $\frac{{\pi }^2{r}^4{B}_0^2}{\tau R}$

Answer 1

Answer: (B)

heat equated $=\underset{0}{\overset{\infty }{\int }}{i}^2$ Rdt
$=\underset{0}{\overset{\infty }{\int }}\frac{{\epsilon }_{\text{ind }}{}^2}{Rt}$ Rdt
$=\frac{1}{R}\underset{0}{\overset{\infty }{\int }}{\epsilon }_{\text{ind }}^{2}dt$
$=\frac{1}{R}\frac{{\pi }^2{r}^4{B}_0^2}{{\tau }^2}\underset{0}{\overset{\infty }{\int }}{e}^{-2t/\tau }dt$
$={\frac{{\pi }^2{r}^4{B}_0^2}{R{\tau }^2}\frac{e^{-2t/\tau }}{-2/\tau }|}_0^{\infty }$
$=+\frac{{\pi }^2{r}^4{B}_0^2\tau }{2R{\tau }^2}\{0+1\}$
$\frac{{\pi }^2{r}^4{B}_0^2}{2R\tau }$