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 $(t \rightarrow\infty)$ 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. $\frac{\pi^{2}r^{4}B^{2}_{0}}{ \tau R }$

Answer 1

Answer: (B)

$\frac{\pi^{2}r^{4}B_{0}^{2}}{2 \tau R}$

Answer 2

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, $t$, $\phi=BA=\pi r^{2}\,B_{0}e^{\frac{t}{\tau}}$ Emf induced in the loop, $\varepsilon=-\frac{d\phi}{dt}$ $=\pi r^{2}\, B_{0} \frac{1}{\tau}e^{-\frac{t}{\tau}}$ Net heat generated in the loop $=\int\limits_{0}^{\infty} \frac{\varepsilon^{2}}{R}dt =\frac{\pi^{2}r^{4}B_{0}^{2}}{\tau^{2}R} \int\limits_{0}^{\infty}e^{\frac{2t}{\tau}} dt$ $=\frac{\pi^{2}r^{4}B_{0}^{2}}{\tau^{2}R}\times\frac{1}{\left(-\frac{2}{\tau}\right)}\times\left[e^{-\frac{2t}{t}}\right]_{0}^{\infty}$ $=\frac{-\pi^{2}r^{4}B_{0}^{2}}{2\tau^{2}R}\times\tau\left(0-1\right)$ $=\frac{\pi^{2}r^{4}B_{0}^{2}}{2\tau R}$