Question

Calculate the electric field in a copper wire of cross-sectional are 2.0 $m{m}^2$ carrying a current of 1 A. The resistivity of copper = 1.7 $\times {10}^{-8}\Omega m.$

Answer 1

$A=2\times {10}^{-6}{m}^2,i=1A,$

$p=1.7\times {10}^{-8}\Omega -m,E=?$

$R=\frac{\rho l}{A}=\frac{1.7\times {10}^{-8}\times l}{2\times {10}^{-6}}$

$V=iR=\frac{1.7\times {10}^{-8}\times l}{2\times {10}^{-6}}$

$E=\frac{dV}{dL}=\frac{V}{l}$

$\frac{1.7\times {10}^{-8}\times l}{2\times {10}^{-6}}$

$=\frac{17}{2}\times {10}^{-3}\frac{V}{m}=8.5m\frac{V}{m}$