A potential difference is applied between the ends of a conductor of uniform radius `2 mm`. It results in an electric field of strength `250 V//m` along the length of conductor. Find the resistivity of the material if current in the conductor is `2 pi A`.

Explanation

`j = (i)/(A) = (2 pi)/(pi(2 xx 10^-3)^2) = (10^6)/(2) A//m^2`
=`sigma E = (E)/(rho)`
`rho = (E)/(j) = (250 xx 2)/(10^6) = 5 xx 10^-4 Omega-m`.