• # question_answer An equilateral triangular loop ADC of uniform specific resistivity having some resistance is pulled with a constant velocity v out of a uniform magnetic field directed into the paper. At 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 A)        B) C) D)

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