A square loop (side a ) rotates in a uniform magnetic field B...

The magnetic flux through the loop at any instant is given by $\Phi = \vec{B} \cdot \vec{A} = B A \cos\theta$, where $\theta$ is the angle between the magnetic field and the normal to the loop area.

Since the loop rotates with an angular velocity $\omega$, we have $\theta = \omega t$.

The area of the square loop is $A = a^2$.

The induced EMF is given by Faraday's law of induction:

$e = -\dfrac{d\Phi}{dt} = -\dfrac{d}{dt}(B a^2 \cos\theta) = B a^2 \omega \sin\theta$

The maximum induced EMF occurs when $\sin\theta = 1$.

The first angle where this occurs is \theta = 90^{}^{\circ}.

The maximum induced EMF is e_{\text\max} = B a^2 \omega.

Answer: $Ba^2\omega$; \theta = 90^{}^{\circ}