91Ó°ÊÓ

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:

$R=\mathrm{ÒÏ}\frac{L}{A}⇒\mathrm{ÒÏ}=\frac{AR}{L}$

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,

$\mathrm{ÒÏ}=\frac{AU}{LI}$

The cross-section area is a circle,

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

$\mathrm{ÒÏ}=\frac{\mathrm{Ï€}U{D}^2}{4LI}$

Substitute the values,

role="math" localid="1648976871881" $\mathrm{ÒÏ}=\left(\frac{\mathrm{Ï€}×1.5V×0.00033m{m}^2}{4×8A×1m}\right)$

role="math" localid="1648977110953" $=1.6×{10}^{-8}\mathrm{Î\copyright }m$

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