For this topic, I recommend to carefully read the Feynman Lectures vol. II. As I've written earlier (I thought even in this thread, perhaps it's one of the deleted postings, although I don't know, why it should be off-topic). So here it is again:
What's basic are the local Maxwell equations, among the Faraday's Law (using Heaviside-Lorentz units)
\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.
The problem in many textbooks is that this fundamental law, which is always correct, is put into integral form not carefully enough, when it comes to moving surfaces and their boundaries. The correct derivation is pretty simple. The first step is to integrate the equation over an arbitrary surface F with boundary \partial F and using Stokes's Theorem, leading to
\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \int_F \mathrm{d}^2 \vec{F} \cdot \partial_t \vec{B}.
The tricky point is now to put the time derivative on the right side of the equation out of the integral, and there many textbooks omit the change of the surface with time if it is moving. Doing everything correctly, you can easily show (using also Gauss's Law for the magnetic field, i.e., \vec{\nabla} \cdot \vec{B}=0 that
\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}-\frac{1}{c} \int_{\partial F} \mathrm{d} \vec{r} \cdot (\vec{v} \times \vec{B}).
Here, \vec{v}=\vec{v}(t,\vec{r}) is the velocity of the boundary curve as a function of time and the location of the point on the surface. All integrals have to be taken at one fixed time, t, of course. Finally we combine the line integrals on both sides of the equation to get the correct form of Faraday's Law in integral form
\int_{\partial F} \mathrm{d} \vec{r} \cdot \left (\vec{E} + \frac{\vec{v}}{c} \times \vec{B} \right )=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B}.
The homopolar generator, however, is easier studied using a more microscopic picture of the situation, than just using Faraday's Law. Suppose we have a long cylindrical bar magnet and assume homogeneous magnetization in its interior (for simplicity). Then also \vec{B} is homogeneous in the interior of the magnet,
\vec{B}=B \vec{e}_z.
Now rotate the magnet around the z axis and wait a little until a stationary condition is established. Then the total force on the conduction electrons in the magnet must be 0, i.e.,
-e n \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right)=0,
i.e.,
\vec{E}=-\frac{\vec{v}}{c} \times \vec{B}.
Thus there is an electric field due to the Hall effect: The electrons drift within the magnet and build a surface-charge density on its surface such that the electric and magnetic forces are compensated as in our ansatz.
Now we have
\vec{v} = \vec{\omega} \times \vec{r} \; \Rightarrow \; \vec{E}=-\frac{B \omega}{c} (\vec{e}_z \times \vec{r}) \times \vec{e}_z = -\frac{B \omega \rho}{c} \vec{e}_{\rho}.
Here \rho is the usual radial cylinder coordinate.
Thus a voltmeter connecting the center of the cylinder with it's rim (using, e.g., coal brushes) reads
U=\frac{B \omega R^2}{2c}.
Of course, there is no contradiction with Faraday's Law, because the magnetic flux is constant in time in this situation and the same is true for the line integral, because \vec{E}+\vec{v}/c \times \vec{B} =0.