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)       

Correct Answer: B

Solution :

[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.