A right angled triangular loop (as shown in the figure) enters into uniform magnetic field (at right angle to the boundary of the field) directed into the paper. The
loop moves always with constant speed. Draw the graph between induced emf e and the distance along the perpendicular to the boundary of the field (say x) along
which loop moves.

$\text{ For }0\le x\le a\text{ The flux enclosed }\varphi =\frac{1}{2}B{x}^2$

$\text{ Hence induced emf, }e=-\frac{d\varphi }{dt}=-Bxv$

$\begin{array}{l}\text{ At }x=0,e=0\\ \text{ At }x=\frac{a}{2},e=-\frac{Bav}{2}\\ \text{ At }x=a,e=-Bav\end{array}$

Now considering for $a<x\le 2a$, The flux enclosed with triangle will be constant hence no induced emf

Finally for $2a\le x\le 3a$

$\varphi =B\left[\frac{a^2}{2}-\frac{(x-2a{)}^2}{2}\right]$

$e=-\frac{d\varphi }{dt}=\frac{B}{2}\left[2\right(x-2a\left)v\right]=Bvx-2Bav$

Hence graph for this part is straight line with positive slope but negative intercept with y-axis.