Question

An engineer cuts a 1.0-m-long, 0.33-mm-diameter piece of wire, connects it across a 1.5 V battery, and finds that the current in the wire is 8.0 A.
Find the resistitvity rho of the piece of the wire.

Answer 1

Answer:

The resistivity of the wire is:

$\rho=1.60\,\,10^{-8}\,\,\Omega\,m$

Explanation:

Recall the formula that connects resistance R with the material's resistivity $\rho$ :

$R=\frac{\rho\,\,L}{A}$

where A stands for the cross-sectional area of the wire, and L for the wire's length.

In our case, the cross-sectional area of a 0.33 mm wire is the area of a circle of 0.165 mm radius (0.000165 m) which we can calculate as:

$A=\pi\,\,R^2=\pi\,\,0.000165^2\,\,m^2=8.55\,\,10^{-8} \,\,m^2$

Since we don't have the actual resistance, but the information on the current on the wire when applying a potential difference, we use Ohm's Law to get the resistance R of the wire:

$V=I\,\,R\\1.5 = 8 \,\, R\\R = \frac{1.5}{8} \, \Omega\\R= 0.1875\,\,\Omega$

Now, using the resistance formula shown at the beginning, we solve for the resistivity $\rho$ :

$R=\frac{\rho\,\,L}{A}\\0.1875\,\,\Omega=\frac{\rho\,\,(1\,\,m)}{(8.55\,\,10^{-8}\,m^2)}\\\rho=0.1875\,*\,8.55\,\,10^{-8}} \,\,\Omega\,m\\\rho=1.60\,\,10^{-8}\,\,\Omega\,m$