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  4. 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 (i) versus time (t) graph will be as <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-rz1gfhtaadbwygdq.jpg />

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Q. 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

Question

NTA AbhyasNTA Abhyas 2022

Solution:

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

$\frac{ AH }{ AE }=\frac{ GH }{ EC }$
$\therefore GH =\left(\frac{ AH }{ AE }\right) EC$
or $GH =\frac{(b-v t)}{b} \cdot a$
$=a-\left(\frac{a}{b} v t\right)$
$\therefore FG =2 GH =2\left[a-\frac{a}{b} v t\right]$
Induced emf $e=B v( FG )=2 B v\left(a-\frac{a}{b} v t\right)$
$\therefore$ Induced current $i=\frac{e}{R}=\frac{2 B v}{R}\left[a-\frac{a}{b} v t\right]$
or $i=k_1-k_2 t$
Thus, $i-t$ graph is a straight line with a negative slope and positive intercept.