Suppose a thin conducting disc of radius R is rotated anticlockwise, about its axis, in a plane perpendicular to a uniform magnetic field of induction \(\vec{B}\) (see the figure in the above Note for reference). \(\vec{B}\) points downwards. Let the constant angular speed of the disc be ω.

Consider an infinitesimal element of radial thickness dr at a distance r from the rotation axis. In one rotation, the area traced by the element is dA = 2πrdr. Therefore, the time rate at which the element traces out the area is

\(\frac{d A}{d t}\) = frequency of rotation x dA = fdA

where f = \(\frac{\omega}{2 \pi}\) is the frequency of rotation.

.’. \(\frac{d A}{d t}=\frac{\omega}{2 \pi}\) (2 πr dr) = ωr dr

The total emf induced between the axle and the rim of the rotating disc is

\(|e|=\int B \frac{d A}{d t}=\int_{0}^{R} B \omega r d r=B \omega \int_{0}^{R} r d r=B \omega \frac{R^{2}}{2}\)

For anticlockwise rotation in \(\vec{B}\) pointing down, the axle is at a higher potential.