Question

The magnetic flux through a stationary loop, with resistance $R$, varies with time $t$ as $\varphi =at(T-t)$. Here, $a$ is a constant and $T$ is the time period, with appropriate units. The heat generated in the loop during one time period, neglecting the inductance of the loop will be,

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

Answer 1

The correct option is A $\frac{a^2{T}^3}{3R}$

E.M.F generated in the loop is given by:

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

$\Rightarrow E=-(aT-2at)$

Power generated is,

$P=\frac{E^2}{R}=\frac{(aT-2at{)}^2}{R}$

$\therefore$ Heat gnerated during time $T$ is,

$H={\int }_0^{T}Pdt={\int }_0^T\frac{(aT-2at{)}^2}{R}dt$

$={\int }_0^T\frac{(a^2{T}^2+4a^2{t}^2-4a^{2}tT)}{R}dt$

$={\left[\frac{a^2{T}^{2}t}{R}\right]}_0^T+{\left[\frac{4a^2{t}^3}{3R}\right]}_0^T-\left[\frac{4a^{2}T{t}^2}{2R}\right]$

$\Rightarrow H=\frac{a^2{T}^3}{3R}$

Hence, option $\left(A\right)$ is the correct answer.

Why this Question? Tip: Heat generated is given by, $H=\int Pdt$