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 $\text{GH}=\frac{\left(b-vt\right)}{b}\cdot a$
$=a-\left(\frac{a}{b}vt\right)$
$\therefore \text{FG}=2\text{GH}=2\left[a-\frac{a}{b}vt\right]$
Induced emf $e=\text{Bv}\left(\text{FG}\right)=2\text{Bv}\left(a-\frac{a}{b}vt\right)$
$\therefore$ Induced current $i=\frac{\text{e}}{\text{R}}=\frac{2\text{Bv}}{\text{R}}\left[\text{a}-\frac{\text{a}}{\text{b}}\text{vt}\right]$
or $i=k_1-k_{2}t$
Thus, $i-t$ graph is a straight line with a negative slope and positive intercept.