Physics - Current Electricity Question with Solution | TestHub

Current ($i$) is constant in the conductor.

The resistance of an element is given by: $\mathrm{dR}=\frac{\mathrm{pdx}}{\pi {\mathrm{r}}^2}$

The change in voltage is: $\mathrm{dV}=\mathrm{idR}=\frac{\mathrm{i}\rho \mathrm{dx}}{\pi {\mathrm{r}}^2}$

The electric field $\mathrm{E}$ is: $\mathrm{E}=\frac{\mathrm{dV}}{\mathrm{dx}}=\frac{\mathrm{i}\rho }{\pi {\mathrm{r}}^2}$

${\mathrm{V}}_{\mathrm{d}}=\frac{\mathrm{eE}\tau }{\mathrm{m}}$

Therefore, ${\mathrm{V}}_{\mathrm{d}}\propto \mathrm{E}$.

Since $E\propto \frac{1}{r^2}$, if $r$ decreases, $E$ will increase, and thus ${\mathrm{V}}_{\mathrm{d}}$ will increase.