Question

22. A $3.0$-mm-diameter wire carries a $12A$ current when the electric field is $0.085\mathrm{V}/\mathrm{m}$. What is the wire's resistivity?

Short answer

The wire's resistivity is $5\times {10}^{-8}\Omega \mathrm{m}$.

Step-by-step solution

Given information

A $3.0$-mm-diameter wire carries a $12A$current when the electric field is $0.085\mathrm{V}/\mathrm{m}$.

Explanation

The current density is,$J=\sigma E$,
Rewrite as,

$J=\frac{E}{\rho }$

From this, the resistivity is,$\rho =\frac{E}{J}$

The current density by applying its definition:

$J=\frac{I}{A}=\frac{4I}{\pi d^2}$

Substituting this expression in the one for the resistivity,
$\rho =\frac{\pi E{d}^2}{4I}$
$\rho =\frac{\pi \times 0.085V/m\times 0.{003}^{2}m}{4\times 12A}$

$=5\times {10}^{-8}\Omega \mathrm{m}$

Final answer

The wire's resistivity is $5\times {10}^{-8}\Omega \mathrm{m}$.