## In This post you will get answer of Q.

A circular copper disc $10 ,cm$ in diameter rotates at $1800$ revolution per minute about an axis through its centre and at right angles to disc. A uniform field of induction $B$ of $1, Wb, m^{-2}$ is perpendicular to disc, What potential difference is developed between the axis of the disc and the rim?

Solution:

## Here, $l=r=5,cm=5times10^{-2},m$,

$omega=2 pi frac{1800}{60}rad,s^{-1}$

$=60, pi, rad,s^{-1}$,

$B=1,Wb,m^{-2}$

$varepsilon=frac{1}{2}Bl^2omega =frac{1}{2}times1times(5times10^{-2})^{2}times60pi=0.23,V$