Consider the coordinate axes shown in the figure. The magnetic field is into the paper, i.e., $$\overrightarrow B = B\widehat k$$. Let loop 1 be the loop of area A and loop 2 be the loop of area 2A. At t = 0, both loops are in the plane of the paper. They start rotating about the x-axis with a constant angular velocity $$\overrightarrow \omega = \omega \widehat i$$. Let the area vectors of loop 1 and 2 be $${\overrightarrow A _1}$$ and $${\overrightarrow A _2}$$, respectively. At t = 0, $${\overrightarrow A _1}$$ = $$A\widehat k$$ and $${\overrightarrow A _2} = 2A\widehat k$$ (we chose area into the plane of the paper as positive). In time t, the loops rotate by an angle $$\theta$$ = $$\omega$$t. Their area vectors make an angle $$\theta$$ = $$\omega$$t with z-axis (direction of the magnetic field). Thus, the magnetic fluxes passing through the loop 1 and the loop 2 at time t are given by

$$\phi$$₁ = $$\overrightarrow B .{\overrightarrow A _1}$$ = B A₁ cos$$\theta$$ = B A cos $$\omega$$t,

$$\phi$$₂ = $$\overrightarrow B .{\overrightarrow A _2}$$ = B A₂ cos$$\theta$$ = 2B A cos $$\omega$$t.

By Faraday's law of electromagnetic induction, induced emfs in the loop 1 and in the loop 2 are given by

$${\varepsilon _1} = - {{d{\phi _1}} \over {dt}} = B\,A\omega \sin \omega t$$,

$${\varepsilon _2} = - {{d{\phi _2}} \over {dt}} = 2B\,A\omega \sin \omega t$$.

The rate of change of flux is maximum when $$\omega$$t = $$\pi$$/2 i.e., when the plane of the loops is perpendicular to the plane to the paper.

Let T = 2$$\pi$$/$$\omega$$ be the time period to complete one revolution. The fluxes $$\phi$$₁ and $$\phi$$₂ decrease with time if t $$\in$$ (0, T/4). By Lenz's law, the induced current should oppose reductions in $$\phi$$₁ and $$\phi$$₂. Thus, induced current in both the loops should be clockwise. Hence, the polarity of 'equivalent batteries' are as shown in the figure. These batteries are connected with reverse polarity. Thus, net emf induced due to both the loop is

$$\varepsilon = {\varepsilon _2} - {\varepsilon _1} = B\,A\omega \sin \omega t$$.

Note that the net emf $$\varepsilon $$ is proportional to the difference in areas of two loops and $$\left| \varepsilon \right| = \left| {{\varepsilon _1}} \right|$$.