question_answer

A conducting square loop of side L and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A magnetic induction B constant in time and space, pointing perpendicular and into the plane at the loop exists everywhere with half the loop outside the field, as shown in figure. The induced emf is

A) Zero                 

B) \[RvB\]

C) \[vBL/R\]                      

D) \[vBL\]

Correct Answer: D

Solution :

The induced emf is

\[e=\frac{-d\phi }{dt}=-\frac{d(\vec{B}.\vec{A})}{dt}=\frac{-d(BA\,\cos \,0{}^\circ )}{dt}\]

\[\therefore \,\,e=-B\frac{dA}{dt}=-B\frac{d(L\times x)}{dt}\]

\[\therefore \,\,e=-BL\frac{dx}{dt}=-BLv\]