Question

You apply a potential difference of 4.50 V between 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.

A.What is the resistivity of the wire?

Answer 1

Concepts and reason The concept of ohm's law and resistivity is required to solve the problem. Initially, calculate the resistance of the wire by using ohm's law. Later, calculate the resistivity of the wire by using the relation between the resistance of the wire, length of the area, area of cross-section of the wire, and resistivity of the wire.

Fundamentals

The expression for Ohm's law is, \(V=I R\)

Here, \(V\) is the potential difference across the conductor, \(I\) is the current through the conductor, and \(\mathrm{R}\) is the resistance of the conductor. The resistance of a wire is, \(R=\rho \frac{l}{A}\)

Here, \(\rho\) is the resistivity of the wire, I is the length of the wire, and \(\mathrm{A}\) is the area of cross-section.

Calculate the resistance of the wire. According to Ohm's law, the potential difference between the ends of the wire is, \(V=I R\)

Here, I is the current through the wire and \(\mathrm{R}\) is the resistance of the wire. Rearrange the equation for \(\mathrm{R}\). \(R=\frac{V}{I}\)

Substitute \(4.50 \mathrm{~V}\) for \(\mathrm{V}\) and \(17.6 \mathrm{~A}\) for \(\mathrm{I}\) in equation \(R=\frac{V}{I}\).

\(R=\frac{4.50}{17.6}\)

\(=0.256 \Omega\)

Ohm's law states that the current through a wire is directly proportional to the potential difference between the ends of the wire.

The resistance of a wire is given as, \(R=\rho \frac{l}{A}\)

The cross-section area of the wire is, \(A=\pi r^{2}\)

Here, \(r\) is the radius of the wire. Substitute \(\pi r^{2}\) for \(\mathrm{A}\) in equation \(R=\rho \frac{l}{A}\) and rearrange the equation for \(\rho\).

$$ \begin{array}{l} R=\rho \frac{l}{\pi r^{2}} \\ \rho=\frac{R \pi r^{2}}{l} \end{array} $$

Substitute \(2.50 \mathrm{~m}\) for length, \(0.654 \mathrm{~m}\) for \(\mathrm{r}\), and \(0.256 \Omega\) for \(\mathrm{R}\) in equation \(\rho=\frac{R \pi r^{2}}{l}\).

$$ \rho=\frac{(0.256 \Omega) \pi\left(0.654 \mathrm{~mm}\left(\frac{10^{-3} \mathrm{~m}}{1 \mathrm{~mm}}\right)\right)^{2}}{(2.50 \mathrm{~m})} $$

$$ =1.38 \times 10^{-7} \Omega \cdot \mathrm{m} $$

The measure of the resistance of the material having a unit length and the unit cross-sectional area is known as resistivity. It is the property of the material and it depends on the nature of the material and the temperature of the conductor.

The resistivity of the wire is \(1.38 \times 10^{-7} \Omega \cdot \mathrm{m}\).