An engineer cuts a $1.0m$long, $0.33mm$diameter piece of wire, connects it across a $1.5\mathrm{V}$battery, and finds that the current in the wire is$8.0\mathrm{A}$. Of what material is the wire made?

Wire length$=1.0m$

Wire diameter$=0.33mm$

Battery$=1.5V$

Current$=8.0A$

In this part of the experiment, we will determine the resistivity (or conductivity) of various materials and compare them to the values given in the book. Resistance takes the form of:

$R=\rho \frac{L}{A}\Rightarrow \rho =\frac{AR}{L}$

In addition, given that the wire is conductive and that the current flowing in it is constant, we can find the resistance by applying Ohm's law.

$R=\frac{U}{I}$

Substituting the second expression into the first one,

$\rho =\frac{AU}{LI}$

The cross-section area is a circle,

We can substitute $A=\frac{\pi D^2}{4}$to find

$\rho =\frac{\pi U{D}^2}{4LI}$

Substitute the values,

role="math" localid="1648976871881" $\rho =\left(\frac{\pi \times 1.5V\times 0.00033m{m}^2}{4\times 8A\times 1m}\right)$

role="math" localid="1648977110953" $=1.6\times {10}^{-8}\Omega m$

Hence, by checking this value in our table provided in the book, we find that the material is silver.