0
$\begingroup$

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?

$\endgroup$

1 Answer 1

0
$\begingroup$

If only a sector of the disk is in uniform magnetic field, then there can't be equilibrium of charge in the disk, because in the frame of the disk, the charge experiences time-dependent force, due to moving in and out of the magnetic field. The disk will experience electric currents in its frame, and thus dissipation of energy, and thus the disk rotation will slow down.

EMF depends on the exact path, not just on its endpoints.

Value of EMF as you've calculated it is valid only for those paths that are entirely in that sector, where magnetic field is present.

If you place the brush touching the rim far from the magnetic field, then for a path that starts at the axle and goes straight to the brush, motional EMF for that path (due to motion in magnetic field) would be zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.