= Solution
For a moving circuit, <Faraday's law> takes the form
$$
\boxed{\mathcal E(t)=\oint_{C(t)}(\mathbf E+\mathbf v\times\mathbf B)\cdot d\boldsymbol\ell=-\frac d{dt}\int_{S(t)}\mathbf B\cdot\mathbf n\,dS.}
$$
Here $C(t)$ is the instantaneous wire <contour>, $\mathbf v$ its local material <velocity>, and $S(t)$ any spanning surface. The <contour> <orientation> and surface <normal vector> obey the right-hand convention. The <integral> on the right is the <magnetic flux>, differentiated with both the field and circuit position allowed to vary. The <electric field> term is the electric force per unit <electric charge> around the wire, and $\mathbf v\times\mathbf B$ supplies the <motional electromotive force>. Their sum is the <electromotive force> driving charge around the moving circuit. The condition $\nabla\cdot\mathbf B=0$ makes the <magnetic flux> independent of the spanning surface.
Choose the initial loop <normal vector> to be $+\mathbf e_y$, so <rotation> about $+\mathbf e_z$ gives $\mathbf n(t)=(-\sin\Omega t,\cos\Omega t,0)$. The spatially uniform <magnetic field> gives
$$
\Phi(t)=\pi a^2B_0\bigl[-\sin\Omega t\cos\omega t+\cos\Omega t\sin\omega t\bigr]=\pi a^2B_0\sin((\omega-\Omega)t).
$$
Using <Ohm's law> for the resistive loop,
$$
\boxed{I(t)=\frac{\pi a^2B_0(\Omega-\omega)}R\cos((\omega-\Omega)t).}
$$
Positive <electric current> is measured in the <contour> <orientation> associated with $\mathbf n(t)$. Reversing that convention reverses the signed answer. When $\Omega=\omega$, the <magnetic flux> is constant and the induced <electric current> vanishes. This uses the usual thin resistive-circuit approximation with <self-inductance> neglected.
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