JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Electromagnetic Induction

The two loops are in series but with opposite orientations (one clockwise, one counterclockwise when viewed from the same side). At time $t$, the flux through the smaller loop is ${\varphi }_1=BA\cos \omega t$ and through the larger loop is ${\varphi }_2=B(2A)\cos \omega t$. Since they are oppositely oriented, the net flux is ${\varphi }_{net}=B(2A-A)\cos \omega t=BA\cos \omega t$. The induced emf is $\epsilon =-\frac{d{\varphi }_{net}}{dt}=B\omega A\sin \omega t$. The amplitude is $B\omega A$, which equals the amplitude of the emf in the smaller loop alone, confirming option (B). The rate of change of flux $|\frac{d\varphi }{dt}|=B\omega A|\sin \omega t|$ is maximum when $\sin \omega t=\pm 1$, i.e., when $\omega t=\pi /2$ or $3\pi /2$, which corresponds to the plane of the loops being perpendicular to the plane of the paper, confirming option (D).