Chapter 18, Problem 73GP

To determine

The diameter and the length of the copper wire.

Answer to Problem 73GP

The diameter of the wire is $2.58\times {10}^{-4}m$

The length of the wire is $38.8m$ .

Explanation of Solution

Given info:

Mass, $m=18.0g=0.018kg$

Resistance, $R=12.5\Omega$

Resistivity, $\rho =1.68\times {10}^{-8}\Omega \cdot m$

Formula used:

Resistance:

  $R=\frac{\rho L}{A}$

Where,

  $\rho$ is the resistivity.

  $L$ is the length of wire

  $A$ is the cross-sectional area of the wire.

Calculation:

First calculate the length of the wire by using the formula:

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

  $m={\rho }_{m}LA={\rho }_{m}L\frac{\rho L}{R}$

  $\Rightarrow L=\sqrt{\frac{mR}{{\rho }_m\rho }}$

Here, ${\text{ρ}}_{\text{m}}$ is the density of the wire.

Substitute the given values:

  $L=\sqrt{\frac{\left(0.081kg\right)\left(12.5\Omega \right)}{\left(8.9\times {10}^{3}kg/m^3\right)\left(1.68\times {10}^{-8}\Omega \cdot m\right)}}$

  $L=38.79m\approx 38.8m$

Now, calculate the diameter of the wire by using the formula:

  $R=\frac{\rho L}{A}\Rightarrow A=\frac{\rho L}{R}\Rightarrow A=\pi {\left(\frac{1}{2}d\right)}^2\Rightarrow d=\sqrt{\frac{4\rho L}{\pi R}}$

  $d=\sqrt{\frac{4\left(1.68\times {10}^{-8}\Omega \cdot m\right)\left(38.79m\right)}{\pi \left(12.5\Omega \right)}}=2.58\times {10}^{-4}m$

Conclusion: The diameter and the length of the wire are $2.58\times {10}^{-4}m$ and $38.8m$ respectively.