A long round conductor of cross – sectional area S is made of material whose resistivity depends only on distance r from the axis of the conductor  as  $\rho =\frac{\alpha }{r^2}$, where  $\alpha$  is a constant. Find : 

Consider a cylindrical element of radii between r and (r+dr). Its resistance
  
 $dR=\frac{\rho l}{2\pi rdr}\left(or\right)\frac{1}{dR}=\frac{2\pi rdr}{\rho l}\mathrm{..............}\left(1\right)$
$\therefore \frac{1}{R}=\underset{0}{\overset{a}{\int }}\frac{1}{dR}=\underset{0}{\overset{a}{\int }}\frac{2\pi }{\rho l}=rdr$  (where a is the radius of the conductor)
  $=\underset{0}{\overset{a}{\int }}\frac{2\pi rdr}{\left(\frac{\alpha }{r^2}\right)l}=\frac{2\pi }{\alpha l}\underset{0}{\overset{a}{\int }}{r}^{3}dr=\frac{2\pi }{\alpha l}\left(\frac{a^4}{4}\right)=\frac{{\left(\pi a^2\right)}^2}{2\pi \alpha l}=\frac{S^2}{2\pi \alpha l}$
$R=\frac{2\pi \alpha l}{S^2}$               ………(ii)
The resistance per unit length of wire  $R=\frac{2\pi \alpha }{S^2}$
b) Equation (ii) can be written as $R=\left(\frac{2\pi \alpha }{S}\right)\left(\frac{l}{S}\right);$   compare with $R=\frac{\rho l}{S}$ , we get  $\rho =\frac{2\pi \alpha }{S}$
By Ohm’s law  $E=j\rho =\frac{i}{S}\times \frac{2\pi \alpha }{S}=\frac{2\pi \alpha i}{S^2}$