 Solution

Choose the loop normal along $+z$. Rotation about $x$ gives $B_z=B\cos \omega t$, so the linked magnetic flux is ${\Phi }_B=B{a}^2\cos \omega t$. ThereforeThe 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.