Question

A conducting metal circular-wire-loop of radius $r$ is placed perpendicular to a magnetic field which varies with time as $B=B_0{e}^{\frac{t}{\tau }}$, 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

Answer: (B)

Here, $B=B_0{e}^{\frac{t}{\tau }}$
Area of the circular loop, $A=\pi r^2$
Flux linked with the loop at any time, $t$,
$\varphi =BA=\pi r^2{B}_0{e}^{\frac{t}{\tau }}$
Emf induced in the loop, $\epsilon =-\frac{d\varphi }{dt}$
$=\pi r^2{B}_0\frac{1}{\tau }{e}^{-\frac{t}{\tau }}$
Net heat generated in the loop
$=\underset{0}{\overset{\infty }{\int }}\frac{{\epsilon }^2}{R}dt=\frac{{\pi }^2{r}^4{B}_0^2}{{\tau }^{2}R}\underset{0}{\overset{\infty }{\int }}{e}^{\frac{2t}{\tau }}dt$
$=\frac{{\pi }^2{r}^4{B}_0^2}{{\tau }^{2}R}\times \frac{1}{\left(-\frac{2}{\tau }\right)}\times {\left[e^{-\frac{2t}{t}}\right]}_0^{\infty }$
$=\frac{-{\pi }^2{r}^4{B}_0^2}{2{\tau }^{2}R}\times \tau \left(0-1\right)$
$=\frac{{\pi }^2{r}^4{B}_0^2}{2\tau R}$