Question

A magnetic flux through a stationary loop with a resistance $R$ and time $t$ varies during the time interval $\tau$ as $\varphi =at(\tau -t)$, where, $a$ is a positive constant. Find the amount of heat generated in the loop during that time. (Neglect the inductance of the loop.)

• A. $\frac{2a^2{\tau }^3}{3R}$
• B. $\frac{a^2{\tau }^3}{3R}$
• C. $\frac{3a^2{\tau }^3}{2R}$
• D. $0$

Answer 1

The correct option is B $\frac{a^2{\tau }^3}{3R}$

Given:

$\varphi =at(\tau -t)$

The $\text{emf}$ generated in the loop is,

$E=-\frac{d\varphi }{dt}=\frac{-d\left(at\right(\tau -t\left)\right)}{dt}$

$=-\left[a\right(\tau -t)+at(-1)$

$=2at-a\tau$

$\therefore$Power generated in the loop is,

$P=\frac{E^2}{R}=\frac{[2at-a\tau {]}^2}{R}$

$\therefore$ Heat produced in the loop in time $\tau$ is,

$H={\int }_o^{\tau }P\cdot dt$

$={\int }_o^{\tau }\frac{[2at-a\tau {]}^2}{R}dt$

$={\int }_o^{\tau }\left(\frac{4a^2{t}^2+a^2{\tau }^2-4a^{2}t\tau }{R}\right)dt$

$=\frac{1}{R}[\frac{4a^2{\tau }^3}{3}+a^2{\tau }^3-\frac{4a^2{\tau }^3}{2}]$

$=\frac{1}{R}\left[\frac{a^2{\tau }^3}{3}\right]$

$\therefore H=\frac{a^2{\tau }^3}{3R}$

Hence, option $\left(B\right)$ is correct.