Question

The resistivity of gold is 2.44 x 10-8 ohms.m at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a current of 940 mA. What is the electric field in the wire?
Answer

0.090 V/m
0.028 V/m
0.046 V/m
0.0090 V/m
0.036 V/m

Answer 1

Concepts and reason

The concept needed to solve this problem is the expression for electric field inside a current carrying conductor.

Write the expression for electric field inside a current conductor. Substitute all the values in this equation and solve for required value of electric field.

Fundamentals

The expression for electric field (E) inside a current conductor is,

$E = \rho J$

Here, J is the current density and $\rho$ is the resistivity of the conductor.

The current density is defined as the current divided by cross section area of the conductor.

$J = \frac{I}{A}$

The cross section area of a wire of radius r can be calculated using the following formula:

$A = \pi {r^2}$

The relation between the radius and diameter of a circle is,

$r = \frac{d}{2}$

The cross section area of a wire of radius r can be calculated using the following formula:

$A = \pi {r^2}$

The relation between the radius and diameter of a circle is,

$r = \frac{d}{2}$

Substitute $\frac{d}{2}$ for r in the equation $A = \pi {r^2}$ .

$\begin{array}{c}\\A = \pi {\left( {\frac{d}{2}} \right)^2}\\\\ = \frac{{\pi {d^2}}}{4}\\\end{array}$

Substitute $\frac{{\pi {d^2}}}{4}$ for A in the equation $J = \frac{I}{A}$ .

$\begin{array}{c}\\J = \frac{I}{{\left( {\frac{{\pi {d^2}}}{4}} \right)}}\\\\ = \frac{{4I}}{{\pi {d^2}}}\\\end{array}$

Convert the unit of diameter from millimeter to meter:

$\begin{array}{c}\\d = 0.9{\rm{ mm}}\left( {\frac{{{{10}^{ - 3}}{\rm{ m}}}}{{1{\rm{ mm}}}}} \right)\\\\ = 0.9 \times {10^{ - 3}}{\rm{ m}}\\\end{array}$

Convert unit of current from mA to A:

$I = 940{\rm{ mA}}\left( {\frac{{{{10}^{ - 3}}{\rm{ A}}}}{{1{\rm{ mA}}}}} \right)$

Substitute $0.9 \times {10^{ - 3}}{\rm{ m}}$ for d and $940 \times {10^{ - 3}}{\rm{ A}}$ for I.

$\begin{array}{c}\\J = \frac{{4\left( {940 \times {{10}^{ - 3}}{\rm{ A}}} \right)}}{{\pi {{\left( {0.9 \times {{10}^{ - 3}}{\rm{ m}}} \right)}^2}}}\\\\ = {\rm{1}}{\rm{.478}} \times {10^6}{\rm{ A/}}{{\rm{m}}^2}\\\end{array}$

The expression for electric field inside a current conductor is,

$E = \rho J$

Substitute ${\rm{1}}{\rm{.478}} \times {10^6}{\rm{ A/}}{{\rm{m}}^2}$ for J and $2.44 \times {10^{ - 8}}{\rm{ }}\Omega$ for $\rho$ .

$\begin{array}{c}\\E = \left( {{\rm{1}}{\rm{.478}} \times {{10}^6}{\rm{ A/}}{{\rm{m}}^2}} \right)\left( {2.44 \times {{10}^{ - 8}}{\rm{ }}\Omega } \right)\\\\ = {\rm{0}}{\rm{.0360632 V/m}}\\\\ = {\rm{0}}{\rm{.036 V/m}}\\\end{array}$

Ans:

The value of electric field inside the wire is ${\rm{0}}{\rm{.036 V/m}}$ .