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 time 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

The correct option is B $\frac{{\pi }^2{r}^4{B}_0^2}{2\tau R}$

Flux through the loop,

$\varphi =\overrightarrow{B}.\overrightarrow{S}$

$\varphi =B_0\pi r^2{e}^{-t/\tau }$
$\epsilon =-\frac{d\varphi }{dt}=\frac{B_0\pi r^2{e}^{-t/\tau }}{\tau }$
Heat $Q={\int }_0^{\infty }\frac{{\epsilon }^2}{R}dt$

$Q={\int }_0^{\infty }\frac{B_o^2{\pi }^2{r}^4{e}^{-2t/\tau }}{R{\tau }^2}$

$={\left[\frac{B_o^2{\pi }^2{r}^4{e}^{-2t/\tau }}{2R\tau }\right]}_{t=0}^{\infty }$

$=\frac{{\pi }^2{r}^4{B}_o^2}{2\tau R}$