Physics - EMI/AC Question with Solution | TestHub

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





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

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

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

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

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

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

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

or    i = k₁- k₂t

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