Physics - EMI/AC Question with Solution | TestHub

Let$2a$be the side of the triangle and$b$the length$\mathrm{AE}$.



$\frac{\text{AH}}{\text{AE}}=\frac{\text{GH}}{\text{EC}}$

$\therefore \text{GH}=\left(\frac{\text{AH}}{\text{AE}}\right)\text{EC}$

or$\text{GH}=\frac{\left(b-vt\right)}{b}·a$

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

$\therefore \text{FG}=2\text{GH}=2\left[a-\frac{a}{b}vt\right]$

Induced emf$e=\mathit{\text{Bv}}\left(\text{FG}\right)=2\mathit{\text{Bv}}\left(a-\frac{a}{b}vt\right)$

$\therefore$Induced current$i=\frac{\mathit{\text{e}}}{\mathit{\text{R}}}=\frac{2\mathit{\text{Bv}}}{\mathit{\text{R}}}\left[\mathit{\text{a}}-\frac{\mathit{\text{a}}}{\mathit{\text{b}}}\mathit{\text{vt}}\right]$

or$i = k_1- k_{2}t$

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