# Faraday disk only partially in magnetic field

Take the following standard setup and calculation for a Faraday disk:

A metal disk of radius $$a$$ rotates with angular velocity $$\omega$$ about a vertical axis, through a uniform field of magnitude $$B$$, pointing up along the vertical axis. A circuit is made by connecting one end of a resistor to the axle and the other end to a sliding contact, which touches the outer edge of the disk.

The speed of a point on the disk at a distance $$s$$ from the axis is $$v = \omega s$$, so the force per unit charge is $$\mathbf{f}_{mag} = \mathbf{v} \times \mathbf{B} = \omega s B \mathbf{\hat{s}}$$.

The emf is therefore $$\mathcal{E} = \int_o^a f_{mag} ds = \omega B \int_o^a s ds = \frac{\omega B a^2}{2}$$

However what if the magnetic field didn't reach over the entire disk? What if only one slice of the disk was in a uniform magnetic field. My intuition would say that because the negatively charged particles in the disk are only periodically exposed to the magnetic field and thus only periodically exposed to the force per unit charge, they will move to the outside of the disk more slowly. So the emf should be less. But by how much? And where in the calculation does this come in?