An equilateral triangular loop ADC of uniform specific resistivity...

Let $2a$ be the side of the triangle and $b$ the length $AE$.

$∆ AHG$ and $∆ AEC$ are similar triangles.

$\dfrac{AH}{AE} = \dfrac{GH}{EC}$

$∴ GH = \left(\dfrac{AH}{AE}\right)EC$

$= \dfrac{b - vt}{b} . a = a - \left(\dfrac{a}{b}\right)vt$

$∴ FG = 2GH = 2\left[a - \dfrac{a}{b}vt\right]$



Induced e.m.f., $e = Bv\left(FG\right) = 2Bv\left(a - \dfrac{a}{b}vt\right)$

$∴$ Induced current, $I = \dfrac{e}{R} = \dfrac{2Bv}{R}\left[a - \dfrac{a}{b}vt\right]$

or $I = {k}_{1} - {k}_{2}t$

Thus, the $I - t$ graph is a straight line with a negative slope and positive intercept.