Question

An equilateral triangular loop $ADC$ having some resistance is pulled with a constant velocity $v$ out of a uniform magnetic field directed into the screen. At the time $t=0$ , side $DC$ of the loop is at the edge of the magnetic field. The induced current $\left(i\right)$ versus time $\left(t\right)$ graph will be as

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

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 $\text{GH}=\frac{\left(b - v t\right)}{b}\cdot a$
$=a-\left(\frac{a}{b} v t\right)$
$\therefore \text{FG}=2\text{GH}=2\left[a - \frac{a}{b} v t\right]$
Induced emf $e=\textit{Bv}\left(\text{FG}\right)=2\textit{Bv}\left(a - \frac{a}{b} v t\right)$
$\therefore $ Induced current $i=\frac{\textit{e}}{\textit{R}}=\frac{2 \textit{Bv}}{\textit{R}}\left[\textit{a} - \frac{\textit{a}}{\textit{b}} \textit{vt}\right]$
or $i=k_{1}-k_{2}t$
Thus, $i-t$ graph is a straight line with a negative slope and positive intercept.