This is a followup question to this one.
I already found some similar questions in PhysicsSE (as this one and this one) but I didn't found the answer I was looking for.
In a conducting loop wire moving inside a constant magnetic field the charges move, thus a potential difference is induced. Anyway the charges just accumulate to one side of the loop and, after the initial movement of electrons, no current is flowing in the loop. For instance, see the first example at page 128 of these notes.
(Of course things are different if the loop in entering or exiting the region with magnetic field).
The definition of the electromotive force is $$ \mathcal{E}=\oint\textbf{f}_s\cdot\text{d}\textbf{l}, $$ where $\textbf{f}_s$ is a non-conservative force per unit charge.
Using the integral form of Faraday's law, is it correct to state that in the loop moving inside a constant magnetic field $$ \mathcal{E}=-\frac{\text{d}}{\text{d}t}\int_\Sigma \textbf{B}\cdot \text{d}\textbf{A}=0? $$
Moreover, online I found some examples of derivations of the emf when there is no loop involved: 1-dimensional example, 3-dimensional example. But I don't get along which path they are integrating. In the examples involving loops, in order to have $\mathcal{E}\neq 0$, we should have that whether $\textbf{B}$, or the dimension of the loop, or the angle between $\textbf{B}$ and the surface (delimited by the loop) are varying (right?). But with no loop?
Moreover$^2$, the second point of the answer to this question states that emf inside an open ended (non-looped) conductor is $$ \mathcal{E}=\int_A^B \textbf{E}_s\cdot \text{d}\textbf{l},$$ where $\textbf{E}_s$ is the electrostatic field, and the integral doesn't seem to be around a circuit (otherwise it would be zero).
How is this true? Are we integrating over a loop which extends outside the conductor so that $\textbf{E}_s$ contributes only in one direction? What does this result prove? Am I mistaking the definition of emf?