Question

21. A $0.0075\mathrm{V}/\mathrm{m}$ electric field creates a $3.9\mathrm{mA}$ current in a $1.0$-mm-diameter wire. What material is the wire made of?

Short answer

The material is the wire made of Nichrome.

Step-by-step solution

Given information

The electric field in the wire is $0.0075\mathrm{V}/\mathrm{m}$.The current through the wire is $3.9\mathrm{mA}$. The diameter of the wire is $1.0$-mm-diameter wire.

So, the radius is$r=0.5\mathrm{mm}=5\times {10}^{-4}\mathrm{m}$.

Explanation

Determine the material from its conductivity $\sigma$ by,
$J=\sigma E$(1)
where$J=\frac{I}{\pi r^2}$is the current density . That means the current per square meter passing through a cross section of the wire -- and$E$ is the electric field strength in the material.
Plugging the formula for the current density into equation (1) and solving for $\sigma$,
$\sigma =\frac{I}{E\pi r^2}$(2)

Explanation

The Calculation: Now we can simply plug these values into equation (2):$\sigma =\frac{I}{E\pi r^2}$
$=\frac{\left(3.9\times {10}^{-3}\mathrm{A}\right)}{(0.0075\mathrm{V}/\mathrm{m})\pi {\left(5\times {10}^{-4}\mathrm{m}\right)}^2}$

$=6.6\times {10}^5{\mathrm{\Omega }}^{-1}\times {\mathrm{m}}^{-1}$

The only material with a conductivity near$6.6\times {10}^5{\Omega }^{-1}\times {\mathrm{m}}^{-1}$ is Nichrome.

Final answer

The material the wire is made from Nichrome.