If we have a constant magnetic field, $B$, perpendicular to a rotating metal disk (essentially Faradays's disk), with some angular frequency, $f$ . Then there will be an emf across the ends of the disk, intuitively, if we consider an electron inside the disk, it would be in motion with the disk then we see from Flemings LHR that it experiences a force (towards the center of the disk). Short proof, WLOG, Consider an electron moving to the right (and the magnetic field going out the plane), the conventional current is directed to the left $\implies$ the electron will experience an upward force (referred to as a Lorentz force). As such the electrons will gather towards the inner part of the disk, and this side will be more negatively charged compared to the outer rim. As such, there will an EMF $\epsilon$ induced across the disk. Alternatively, we can just use Faraday's law (Clearly the amount of magnetic flux cut by the disc in one second is $ \pi r^{2} f B $ (This is when $d$ is taken to be negligible).
For our initial disk(annulus essentially - cause I'm no longer considering $d$ to be negligible) lets' say it has an inner diameter of $d$ and outer diameter $D$. If we were to consider the outer end to the inner end, we see the EMF induced is $ \epsilon = \frac{\pi}{4} \times (D^{2} - d^{2}) \times fB $ Now, what happens as $d$ increases drastically (note that $d <D $ always), such that we have a loop. Will there still be emf? My intuition tells me, yes, but I'm not certain, it'll be great if I can get a response!
