A conducting square loop is placed in a magnetic field $$B$$ with its plane perpendicular to the field. The sides of the loop start shrinking at a constant rate $$\alpha .$$ The induced emf in the loop at an instant when its side is $$'a'$$ is

Solution :
At any time $$t,$$ the side of the square $$a = \left( {{a_0} - \alpha t} \right),$$   where $${{a_0} = }$$   side at $$t = 0.$$
At this instant, flux through the square :
$$\eqalign{ & \phi = BA\cos {0^ \circ } = B\left( {{a_0} - \alpha {t^2}} \right) \cr & \therefore {\text{emf}}\,\,{\text{induced }}E = - \frac{{d\phi }}{{dt}} \cr & \Rightarrow E = - B.2\left( {{a_0} - \alpha t} \right)\left( {0 - \alpha } \right) = + 2\alpha aB \cr} $$