question_answer

An equilateral triangular loop ABC made of uniform thin wires is being pulled out of a region with a uniform speed v, where a uniform magnetic field \[\vec{B}\] perpendicular to the plane of the loop exists. At time t=0, the point A is at the edge of the magnetic field. The induced current (I) vs time (t) graph will be as   

A)        

B)

C)

D)

Correct Answer: B

Solution :

[b] \[I=\frac{\varepsilon }{R}=\frac{\left| d\phi \left. /dt \right| \right.}{R}=\frac{B.\frac{dA}{dt}}{R}\] \[=\frac{B\frac{d}{dt}\left( \frac{1}{2}h.b \right)}{R}\propto \frac{Bb\frac{dh}{dt}}{R}\propto Bbv\] \[\therefore \,\,\,\,b\propto t\Rightarrow I\propto t\]