Question

A conducting disc of radius R is rotating with angular velocity $\omega$.Mass of electron is and charged e. If electrons are the current carries in a conductor, the potential difference between the center and the edge of the disc is:

• A. $\frac{m{\omega }^2{R}^2}{e}$
• B. $\frac{m{\omega }^2{R}^2}{4e}$
• C. $\frac{m{\omega }^2{R}^2}{3e}$
• D. $\frac{m{\omega }^2{R}^2}{2e}$

Answer 1

The correct option is D $\frac{m{\omega }^2{R}^2}{2e}$

Due to rotations, the electron experfence a centripctual force $F_e$ given by :-

$F_e=m{\omega }^{2}r$

Due to the fonce, a radical electric field is

generated which is :-

$E=\frac{F_e}{e}[asF=qE]$

$\Rightarrow E=\frac{m{\omega }^{2}r}{e}$

So, the potential difference generated between the centre

(r = 0) and edge of dise (r = R) is obtained as :-

V = $\int E.dr={\int }_o^R\frac{m{\omega }^{2}r}{e}dr=\frac{m{\omega }^2}{e}{\int }_o^{R}rdr=\frac{m{\omega }^2}{e}{\left[\frac{r^2}{2}\right]}_o^R$

$\Rightarrow V=\frac{m{\omega }^2{R}^2}{2l}$