The magnetic flux through a stationary loop with resistance R varies during interval of time ? as $\mathrm{\Phi }=\mathrm{at}(\mathrm{T}-\mathrm{t})$. The heat generated during this time neglecting the inductance of loop will be

Given that, $\mathrm{\Phi }=\mathrm{at}(\mathrm{T}-\mathrm{t})=\mathrm{atT}-{\mathrm{at}}^2$
$\therefore \mathrm{e}=\frac{\mathrm{d\Phi }}{\mathrm{dt}}=\mathrm{aT}-2\mathrm{at}=a(\mathrm{T}-2\mathrm{t})$
Let heat generated in time dt be dH Then
$\mathrm{dH}={\mathrm{i}}^2\mathrm{Rdt}=\frac{{\mathrm{e}}^2}{{\mathrm{R}}^2}\times \mathrm{Rdt}=\frac{{\mathrm{e}}^2\mathrm{dt}}{\mathrm{R}}$
Total heat generated can be obtained by integrating above expression, i,e.,
$\begin{array}{l}{\int }_0^{\mathrm{T}} \mathrm{dH}={\int }_0^{\mathrm{T}} \frac{{\mathrm{e}}^2\mathrm{dt}}{\mathrm{R}}\\ \mathrm{H}=\frac{1}{\mathrm{R}}{\int }_0^{\mathrm{T}} {\mathrm{e}}^2\mathrm{dt}=\frac{1}{\mathrm{R}}{\int }_0^{\mathrm{T}} {\mathrm{a}}^2(\mathrm{T}-2\mathrm{t}{)}^2\mathrm{dt}\\ =\frac{{\mathrm{a}}^2}{\mathrm{R}}{\left[\left({\mathrm{T}}^2\mathrm{t}+\frac{4{\mathrm{t}}^3}{3}-2{\mathrm{t}}^2\mathrm{T}\right)\right]}_0^{\mathrm{T}}\\ =\frac{{\mathrm{a}}^2}{\mathrm{R}}\left[{\mathrm{T}}^3+\frac{4{\mathrm{T}}^3}{3}-2{\mathrm{T}}^3\right]=\frac{{\mathrm{a}}^2{\mathrm{T}}^3}{3\mathrm{R}}\end{array}$