question_answer

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

A)     

B)

C)     

D)

Correct Answer: B

Solution :

[b] Let 2a be the side of the triangle and b be the length AE. \[\frac{AH}{AE}=\frac{GH}{EC}\Rightarrow GH=\left( \frac{AH}{AE} \right)EC\] Or \[GH=\frac{(b-vt)}{b}.a=a-\left( \frac{a}{b}vt \right)\] \[\therefore FG=1GH=2\left[ a-\frac{a}{b}vt \right]\] \[\therefore \]Induced emf, \[e=Bv(FG)=2Bv\left( a-\frac{a}{b}vt \right)\] \[\therefore \]Induced current, \[i=\frac{e}{R}=\frac{2Bv}{R}\left[ a-\frac{a}{b}vt \right]\] Or \[i={{k}_{1}}-{{k}_{2}}t\] Thus i-t graph is a straight line with negative slope and positive intercept.