Question

A copper wire of cross-sectional area $2.0m{m}^2$, resistivity = $1.7\times {10}^{-8}\Omega m$, carries a current of 1 A. The electric field in the copper wire is:

• A. $8.5\times {10}^{-5}V/m$
• B. $8.5\times {10}^{-4}V/m$
• C. $8.5\times {10}^{-3}V/m$
• D. $8.5\times {10}^{-2}V/m$

Answer 1

Answer: (C)

$8.5\times {10}^{-3}V/m$

from given data $A=2m{m}^2$
$\rho =1.7\times {10}^{-8}\Omega m.$
$i=1A$
$V=iR$ ( from ohms law.)
$V=i\times \frac{\rho l}{A}$
$E.l=i\times \frac{\rho l}{A}$
$\Rightarrow E=i\times \frac{\rho l}{A}$
$=1A\times \frac{1.7\times {10}^{-8}\Omega m}{2m{m}^2}$

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

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