Q. A metal disc of radius a rotates with a constant angular velocity ω about its axis. The potential difference between the center and the rim of the disc is (*m* = mass of electron, *e* = charge on electron)

(a) $\displaystyle \frac{m\omega^2 a^2}{e} $

(b) $ \displaystyle \frac {1}{2} \frac{m\omega^2 a^2}{e} $

(c) $\displaystyle \frac{e\omega^2 a^2}{2m} $

(d) $ \displaystyle \frac{e\omega^2 a^2}{m} $

Ans: (b)

Sol:As metal disc rotates , the free electron also rotates with same angular velocity having acceleration = ω^{2}r . Hence Charged particle must have a force .But charged particle is only influenced by either electric field or magnetic field . Here there is no external magnetic field in question , hence it is electric field .

e E = m ω^{2}r ; Where E = Electric field

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

Potential difference between center & rim $\displaystyle = \int_{0}^{a} \vec{E}.\vec{dr}$

$\displaystyle = \int_{0}^{a}\frac{m \omega^2 r}{e} dr $

$ \displaystyle = [\frac{m \omega^2 r^2}{2 e}]_{0}^{a} $

$ \displaystyle = \frac{m \omega^2 a^2}{2 e} $