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. At time t = 0, side DC of the loop is 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 :

Let 2 a be the side of the triangle and b the length AE. \[\frac{AH}{AE}=\frac{GH}{EC},\]\[\therefore \]\[GH=\frac{AH}{AE}CE\] or \[GH=\frac{(b-vt)}{b}.a=a-=\left( \frac{a}{b}vt \right)\] \[\therefore \]\[FG=2GH=2\left[ a-\frac{a}{b}vt \right]\] Induced \[e\,m\,f\,e=Bv(FG)=2B\left( va-\frac{a}{b}vt \right)\] or \[i={{K}_{1}}-{{K}_{2}}tv\] Thus \[i-t\]graph is a straight line. Hence, the correction option is [b].