
(d) As the magnetic field directed into the paper is increasing at a constant rate, therefore, induced current should produce a magnetic field directed out of the paper. Thus current in both the loops must be anti-clock-wise.
[image:7_image50.emf]
As area of loop on right side is more, therefore, induced emf o right side of loop will be more compared to the emf induced on the left-side of the loop
$\left\lbrack \therefore e = \ - \frac{d\phi}{dt} = - A\frac{dB}{dt} \right\rbrack$