Induced emf, $\text{e}=-\text{L}\frac{\text{di}}{\text{dt}}$

For   $0\le \text{t}\le \frac{\text{T}}{4}$,

i-t graph is a straight line with positive constant slope.

$\therefore \frac{\text{di}}{\text{dt}}=\text{constant}$

$\Rightarrow \text{e}=-\text{ve}$ and constant For $0\le \text{t}\le \frac{\text{T}}{4}$

For   $\frac{\text{T}}{4}\le \text{t}\le \frac{\text{T}}{2},$

$\because$ i is constant  $\therefore \frac{\text{di}}{\text{dt}}=0$

$\Rightarrow \text{e}=0$    For   $\frac{\text{T}}{4}\le \text{t}\le \frac{\text{T}}{2}$

For   $\frac{\text{T}}{2}\le \text{t}\le \frac{3\text{T}}{4}$,

i-t graph is a straight line with negative constant slope.

$\therefore \frac{\text{di}}{\text{dt}}=\text{constant}$

⇒   e = + ve   and constant     For  $\frac{\text{T}}{2}\le \text{t}\le \frac{3\text{T}}{4}$

For   $\frac{3\text{T}}{4}\le \text{t}\le \text{T}$,

i is zero $\therefore \frac{\text{di}}{\text{dt}}=0$

$\Rightarrow \text{e}=0$  For   $\frac{3\text{T}}{4}\le \text{t}\le \text{T}$

From this analysis, the variation of induced emf with time as shown in the figure below.