Question

A metal disc of radius 1m is rotating in Earth’s gravitational field with its axis parallel to
the field. The field strength is 0.5 G. What should be the rotation velocity for the potential
difference between the center and the circumference of the disc to be 1 mV.?

Answer 1

According to Faraday's law when a magnetic flux is changed an electromotive force is generated

(1)$E=-\frac{d\Phi}{dt}$

When the conductor moves in a uniform magnetic field the change in magnetic flux is

(2) $d\Phi=B\cdot dS$ where $dS$ is the area that the conductor covers when moving in a unit of time. If the disk is rotating this area is equals the area of the sector

(3) $dS=\frac{d\psi}{2\pi}\cdot \pi R^2=\frac{d\psi}{2}R^2$ where $R=1m$ radius of disc, and $d\psi=w\cdot dt$ . Substitute (2) and (3) to (1) we have

(4) $E=- B\cdot \frac{wR^2}{2}$

A current will begin to flow in the metal disk, and if the circuit of this current is not closed, a potential difference will occur that prevents the movement of carriers under the action of electromotive force. In this case, the current will disappear if the potential difference exactly matches the electromotive force. That is $E=- 1mV=-10^{-3}V$ . For the rotation velocity of disk from (4) we get

(5) $w=\frac{2E}{BR^2}=\frac{2\cdot 10^{-3}V}{0.5\cdot 10^{-4}T \cdot 1 m^2}=40 \frac{rad}{s}$ (Note $1G=10^{-4}T$ ) The plus sign in this expression means that the rotation is performed according to the rule of the right screw relative to the direction of the magnetic field. If we look in the direction of the magnetic field, the rotation should be clockwise.

Answer: The rotation velocity of a metal disc should be $w=40 rad/s$ .

[1] https://en.wikipedia.org/wiki/Homopolar_generator