Question

When a potential difference of $3V$ is applied between the two ends of a $60$cm long metallic wire, current density in it is found to be $1\times {10}^{-7}A{m}^{-2}$. Find conductivity of the material of the wire in SI system.

Answer 1

Given: $V=3V$
Length $=60$cm $=0.60$m
$J=1\times {10}^{-7}A/m^2$
$E=\frac{V}{d}$
$=\frac{3}{0.60}=5V/m$
We know, $J=\sigma E$
$1\times {10}^{-7}=\sigma \times 5$
$\sigma =2\times 10-8(\Omega m{)}^{-1}$.