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 $( t \rightarrow \infty)$ 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. $\frac{\pi^2 r^4 B_0^2}{\tau R}$

Answer 1

Answer: (B)

heat equated $=\int\limits_{0}^{\infty} i ^{2}$ Rdt
$=\int\limits_{0}^{\infty} \frac{\varepsilon_{\text {ind }}{ }^{2}}{ Rt }$ Rdt
$=\frac{1}{ R } \int\limits_{0}^{\infty} \varepsilon_{\text {ind }}^{2} dt$
$=\frac{1}{ R } \frac{\pi^{2} r ^{4} B _{0}^{2}}{\tau^{2}} \int\limits_{0}^{\infty} e ^{-2 t / \tau} dt$
$=\left.\frac{\pi^{2} r ^{4} B _{0}^{2}}{ R \tau^{2}} \frac{ e ^{-2 t / \tau}}{-2 / \tau}\right|_{0} ^{\infty}$
$=+\frac{\pi^{2} r^{4} B_{0}^{2} \tau}{2 R \tau^{2}}\{0+1\}$
$\frac{\pi^{2} r^{4} B_{0}^{2}}{2 R \tau}$