Answer 1

The angular velocity of the disc is ω. Also, the magnetic field of magnitude B is perpendicular to the disc.

Let us take a circular element of thickness da at a distance a from the centre.

Linear speed of the element at a from the centre, v = ωa

Now,

\[de = Blv\]

\[de = B \times da \times a\omega\]

\[ \Rightarrow e =  \int_0^r \left( B\omega a \right)da\]

\[ e = \frac{1}{2}B\omega r^2\]