= Solution
Choose the loop normal along $+z$. Rotation about $x$ gives $B_z=B\cos\omega t$, so the linked <magnetic flux> is $\Phi_B=Ba^2\cos\omega t$. Therefore
$$
\boxed{\mathcal E(t)=Ba^2\omega\sin\omega t.}
$$
The sign uses the contour orientation associated with $+z$. For the small open gap, this is the corresponding open-circuit terminal emf, with the stated terminal ordering. The tangential induced <electric field> is nonconservative, so its circulation is not the integral of a single-valued electrostatic potential gradient.
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