When a conducting loop rotating in a magnetic field $B$has an angular velocity $\omega$, the flux through the loop of area $A$ at an angle $\theta = \omega t$, 

$\varphi =\left|B\right|\left|A\right|\cos \theta$

$=B\mathrm{A }\cos \left(\omega t\right)$

The induced emf,

$\epsilon = -\frac{d\varphi }{dt}=BA\omega \sin \left(\omega t\right)$

So,  $\epsilon \mathrm{a}\mathrm{n}\mathrm{d}\frac{d\varphi }{dt} \propto \sin \left(\omega t\right)$

So, the emf is maximum when, $\omega t=\theta =\frac{\pi }{2}.$

Since the emf in the smaller loop will act opposite to that of the larger loop,

${\epsilon }_{Net}= {\epsilon }_{2A}- {\epsilon }_A=B\left(2A\right)\omega \sin \omega t-B\left(A\right)\omega \sin \left(\omega t\right)$

$=B\left(2A-A\right)\omega \sin \omega t=BA \omega \sin \omega t$

So, the amplitude of the maximum net emf is equal to $BA$ which is same as that in the smaller loop.