Is the concept of motional EMF consistent with Faraday's Law?

Consider a circular homogeneous metallic coil sliding on a smooth horizontal surface in a region of uniform magnetic field $$B$$ which is perpendicular to the face of the coil. By Faraday's law, the net flux is constant and hence there is no electro-motive force (EMF) induced in the coil.

But now look at individual electrons present in the coil. Due to the Lorentz force, all electrons will experience a net force in the same direction. Since they are constrained to move on the ring, they will move on one of the half side of the ring. Thus, an electric field will be setup and an EMF is induced. Hence, there will be induced motional EMF.

Please explain whether the EMF will be induced in the coil or not with the appropriate reasoning. Thanks!

• Hall effect in establishing the equilibrium. But let's wait for other answers Oct 26, 2022 at 18:16
• @basics Unfortunately, I haven't learnt hall effect. Oct 27, 2022 at 4:29

Faraday's law in this situation predicts $$\epsilon = \oint \left(\vec{E} + \vec{v} × \vec{B}\right) \cdot \vec{dl} = 0.$$ This does not imply that $$\int_{a}^{b} \left(\vec{E} + \vec{v} × \vec{B}\right) \cdot \vec{dl} = 0.$$ The closed line integral of the Lorentz force is equal to zero in your case; however, an open line integral can be non-zero.