A metal disc of radius ' \(\mathrm{R}\) ' rotates with an angular velocity ' \(\omega\) ' about an axis perpendicular to its plane passing through its centre in a magnetic field of induction ' \(\mathrm{B}\) ' acting perpendicular to the plane of the disc. The induced e.m.f. between the rim and axis of the disc is (magnitude only)

The correct option is (D).
We can imagine the disc to be a collection of thin rods connected in parallel between the center of the disc and the rim. So if we calculate the induced emf on a thin rod rotating about its axis then this should be equal to that of the disc.

The tiny motional emf developed across the element dr can be written as:
\(\mathrm{dE}=\mathrm{Bvdr}\)
Taking the velocity \(\mathrm{v}=\omega \mathrm{r}\),
\(\mathrm{dE}=\mathrm{B} \omega \mathrm{rdr}\)
On integrating across the rod,
\(\left.\mathrm{E}=\int_0^{\mathrm{R}} \mathrm{B} \omega \mathrm{rdr}=\frac{\mathrm{B} \omega \mathrm{r}^2}{2}\right]_0^{\mathrm{R}}=\frac{\mathrm{B} \omega \mathrm{R}^2}{2}\)