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 time $t=0$ , side $DC$ of the loop is at 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 $\text{GH}=\frac{\left(\text{b}-\text{vt}\right)}{\text{b}}\cdot \text{a}$
$=\text{a}-\left(\frac{\text{a}}{\text{b}}\text{vt}\right)$
$\therefore \text{FG}=2\text{GH}=2\left[\text{a}-\frac{\text{a}}{\text{b}}\text{vt}\right]$
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.