A circular insulated copper wire loop is twisted to form two loops...

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 $\theta = \omegat$, 

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

$= BAcos \left(\omegat\right)$

The induced emf,

$ε = - \dfrac{dϕ}{dt} = BA\omega sin \left(\omegat\right)$

So, $εand\dfrac{dϕ}{dt} ∝ \sin \left(\omegat\right)$

So, the emf is maximum when, $\omegat = \theta = \dfrac{\pi}{2} .$

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

${ε}_{Net} = {ε}_{2A} - {ε}_{A} = B\left(2A\right)\omega sin \omegat - B\left(A\right)\omega sin \left(\omegat\right)$

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

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