Question

Consider a silver wire with a cross-sectional area of $\mathbf{1}\mathbf{}{\mathbf{mm}}^{\mathbf{2}}$carrying$\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{A}$of current. The conductivity of silver is$\mathbf{6}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}\raisebox{1ex}{$\left(A/m^2\right)$}\!\left/ \!\raisebox{-1ex}{$\left(V/m\right)$}\right.$. Calculate the magnitude of the electric field required to drive this current through the wire.

Short answer

$4.76\times {10}^{-3}\mathrm{V}/\mathrm{m}$

Step-by-step solution

Given Data

$$\begin{gathered} \mathrm{Area}A=1{\mathrm{mm}}^2 \\ \mathrm{Current}I=0.3\mathrm{A} \\ \mathrm{Conductivity}\sigma =6.3\times {10}^7\raisebox{1ex}{$\left(\mathrm{A}/{\mathrm{m}}^2\right)$}\!\left/ \!\raisebox{-1ex}{$\left(\mathrm{V}/\mathrm{m}\right)$}\right. \end{gathered}$$

Concept

The force as per the test charge is known as the magnitude of electric field.

Calculate the magnitude of the electric field 

First we have to find the resistivity,

$\begin{array}{rcl}\sigma & =& \frac{1}{\rho }\left(\mathrm{where},\mathrm{\rho }\mathrm{is}\mathrm{resistivity}\right)\\ \mathrm{\rho }& =& \frac{1}{\mathrm{\sigma }}\\ & =& \frac{1}{6.3\times {10}^7}\\ & =& 1.58\times {10}^{-8}\mathrm{\Omega m}\end{array}$

Electric Field,

localid="1662180919419" $\begin{array}{cc}\mathrm{E}=\mathrm{\rho }\frac{1}{\mathrm{A}}(\mathrm{where}E\mathrm{is}\mathrm{electric}\mathrm{field})& \\ =1.58\times {10}^{-8}\times \frac{0.3\mathrm{A}}{1\times {10}^{-6}{\mathrm{m}}^2}& \\ =4.76\times {10}^{-3}\mathrm{V}/\mathrm{m}& \end{array}$

Hence, the magnitude of the electric field required to drive this current through the wire is $4.76\times {10}^{-3}\mathrm{V}/\mathrm{m}$