A metal disc of radius 25 cm rotates with a constant angular velocity 130 rad s^-1 about its axis. Find the potential difference in nV between the centre and rim of the disc if the external magnetic field is abscent ?

Hi,

As the metal disc rotates, the free electron must have an acceleration ω²r towards the centre of the disc. As the external magnetic field is absent. Therefore, we can say electric field will be responsible for making the electron move towards the centre of the disc.

$$\begin{gathered} \mathrm{Let} E \mathrm{be} \mathrm{the} \mathrm{electric} \mathrm{field} \mathrm{at} \mathrm{a} \mathrm{distance} r \mathrm{from} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{the} \mathrm{disc}. \mathrm{From} \mathrm{the} \mathrm{newton}\text{'}\mathrm{s} \\ \mathrm{second} \mathrm{law}, \mathrm{we} \mathrm{have} \\ {F}_{\mathrm{n}}=m{a}_{\mathrm{n}} \\ \Rightarrow eE=m{\omega }^{2}r \\ \mathit{\Rightarrow }E\mathit{=}\frac{m{\omega }^{\mathit{2}}r}{e} \\ \mathrm{Now}, \mathrm{potential} \mathrm{difference} \mathrm{between} \mathrm{the} \mathrm{rim} \mathrm{and} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{disc} \mathrm{is} \\ {V}_R-V_C=∆V={\int }_0^{\mathrm{a}}E\mathit{.}dr \\ \Rightarrow ∆V={\int }_0^{\mathrm{a}}\frac{m{\omega }^{2}r}{e}dr \\ \Rightarrow ∆V=\frac{m{\omega }^2{a}^2}{2e} \\ \Rightarrow ∆V=\frac{9.1\times {10}^{-31}\times \left({130}^2\right)\times {\left(0.25\right)}^2}{2\times 1.6\times {10}^{-19}} \\ \Rightarrow ∆V=3\times {10}^{-9}=3 \mathrm{nV} \end{gathered}$$