Question

You apply a potential difference of 4.50 Vbetween the ends of a wire that is 2.50 m in length and 0.654 mm in radius. The resulting current through the wire is 17.6 A. What is the resistivity of the wire?

Short answer

The resistivity of the wire is$13.8\times {10}^{-8}\Omega m$ .

Step-by-step solution

Determine the area of the wire

The given data is:

$V=4.50V,L=2.50,m=0.654mm,andl=17.6A$

Now, the area of the wire can be calculated as:

$$\begin{gathered} A={\mathrm{\pi r}}^2 \\ \mathrm{A}=\mathrm{\pi }\times {\left(0.654\times {10}^{-3}\right)}^2 \\ \mathrm{A}=1.344\times {10}^{-6}{\mathrm{m}}^2 \end{gathered}$$

Determine the resistance of the wire

Now,

$$\begin{gathered} R=\frac{V}{I} \\ R=\frac{4.50}{17.6} \\ R=0.256\Omega \end{gathered}$$

Thus, the resistance of the wire is $0.256\Omega$

Determine the resistivity of the wire

The formula to find resistivity is:

$\rho =\frac{RA}{L}$

Substitute the given values

$$\begin{gathered} \rho =\frac{\left(0.256\Omega \right)\left(1.344\times {10}^{-6}{m}^2\right)}{2.50m} \\ \rho =13.8\times {10}^{-8}\Omega m \end{gathered}$$

Therefore, the resistivity of the wire is $13.8\times {10}^{-8}\Omega m$.