Question

A copper wire that is \(1.00 \mathrm{~m}\) long and has a radius of \(0.500 \mathrm{~mm}\) is stretched to a length of \(2.00 \mathrm{~m}\). What is the fractional change in resistance, \(\Delta R / R,\) as the wire is stretched? What is \(\Delta R / R\) for a wire of the same initial dimensions made out of aluminum?

Short answer

The fractional change in resistance when a wire is stretched can be calculated using the formula: $$\frac{\Delta R}{R} = \frac{R_2 - R_1}{R_1}$$, where \(R_1\) is the initial resistance and \(R_2\) is the final resistance. The fractional change depends on the resistivities of the materials (\(\rho_c\) for copper and \(\rho_a\) for aluminum) as well as the initial and final lengths and cross-sectional areas of the wires. The difference in fractional change between copper and aluminum wires arises from their different resistivities. When comparing the two materials, the relative difference in their fractional change in resistance will depend on the resistivities \(\rho_c\) and \(\rho_a\), with all other factors remaining the same.

Step-by-step solution

1. Calculate the initial cross-sectional area of the wire

Firstly, we need to determine the initial cross-sectional area of the wire. The wire is cylindrical, so its cross-sectional area \(A_1\) can be calculated as: $$A_1= \pi r^2$$ where \(r=0.5 \times 10^{-3}\) (since given radius is in millimeters). Plugging in the value of \(r\), we get: $$A_1= \pi (0.5 \times 10^{-3})^2$$

2. Calculate the final cross-sectional area of the wire

As the wire is stretched, its volume remains constant, so the initial volume must be equal to the final volume. Therefore: $$L_1 A_1 = L_2 A_2$$ Plugging in the given values \(L_1=1.00\,\text{m}, L_2=2.00\,\text{m}\) and the initial area \(A_1 = \pi(0.5 \times 10^{-3})^2\), we get: $$A_2 = \frac{L_1 A_1}{L_2}$$

3. Calculate the initial and final resistances

Now, we can calculate the initial and final resistances for both copper and aluminum wires. The resistance is given by the equation \(R=\frac{\rho L}{A}\). Let's denote the resistivities of copper and aluminum as \(\rho_c\) and \(\rho_a\), respectively. The initial and final resistances for copper are: $$R_{1c}=\frac{\rho_c L_1}{A_1}$$ $$R_{2c}=\frac{\rho_c L_2}{A_2}$$ Similarly, the initial and final resistances for aluminum are: $$R_{1a}=\frac{\rho_a L_1}{A_1}$$ $$R_{2a}=\frac{\rho_a L_2}{A_2}$$

4. Calculate the fractional change in resistance

The fractional change in resistance for both copper and aluminum wires can be calculated by the following equation: $$\frac{\Delta R}{R} = \frac{R_2 - R_1}{R_1}$$ For copper, we have: $$\frac{\Delta R}{R} = \frac{R_{2c} - R_{1c}}{R_{1c}} = \frac{\frac{\rho_c L_2}{A_2} - \frac{\rho_c L_1}{A_1}}{\frac{\rho_c L_1}{A_1}}$$ For aluminum, we have: $$\frac{\Delta R}{R} = \frac{R_{2a} - R_{1a}}{R_{1a}} = \frac{\frac{\rho_a L_2}{A_2} - \frac{\rho_a L_1}{A_1}}{\frac{\rho_a L_1}{A_1}}$$ We can now see that the only differences between copper and aluminum are their respective resistivities, which means the only relative difference between their fractional change in resistance will depend on \(\rho_c\) and \(\rho_a\). Calculating the values using the respective resistivities and the derived equations will give us the fractional change in resistance for a copper wire and an aluminum wire.

Key concepts

Electrical Resistance

Electrical resistance is a fundamental concept in the field of electronics and physics, which describes how difficult it is for an electric current to pass through a conductor. It is commonly represented by the symbol \( R \) and is measured in ohms (Ω). The resistance of a material is determined by its physical properties, such as length, cross-sectional area, and a property known as the resistivity.

The formula for calculating the resistance of a wire is \( R = \frac{\rho L}{A} \), where \( R \) is the resistance, \( \rho \) (rho) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is its cross-sectional area. As the length of the wire increases or the cross-sectional area decreases, resistance increases. This relationship becomes apparent when a wire is stretched—the length increases, the area reduces, and consequently, the resistance goes up.

Resistivity of Materials

Resistivity, represented by the Greek letter \( \rho \), is a material property that quantifies how strongly a material opposes the flow of electric current. Materials with low resistivity, such as copper and aluminum, are excellent conductors of electricity, whereas those with high resistivity act as insulators.

Each material has a characteristic resistivity that is affected by factors such as temperature. For instance, as temperature increases, the resistivity of a conductor usually increases. In problems involving resistance changes, like the stretching of a wire, the resistivity of the material plays a crucial role because it determines how the material's resistance will change with changes in length and cross-sectional area.

Cross-Sectional Area

The cross-sectional area of a wire is the area of the wire's cut surface perpendicular to its length. For a cylindrical wire, this area can be calculated using the formula \( A = \pi r^2 \) where \( r \) is the radius of the wire.

When the length of a wire is changed, under the conservation of volume, the cross-sectional area also changes to compensate. A stretched wire becomes longer and thinner, leading to a decrease in cross-sectional area. Since resistance is inversely proportional to the cross-sectional area, a decrease in the area results in an increase in resistance, contributing to the fractional change in resistance when a wire is stretched.

Conservation of Volume

The principle of conservation of volume states that for a solid object deforming under forces such as stretching or compressing, the volume of the object remains constant if the material is incompressible and no material is added or removed. In the context of stretching wires, even though the length changes, the volume before and after stretching is the same.

This concept allows us to say that the initial volume \( L_1 A_1 \) must be equal to the final volume \( L_2 A_2 \). Thus, when a wire is stretched and its length is increased by a certain factor, its cross-sectional area must reduce proportionally to maintain the same volume. This relationship between length and cross-sectional area due to conservation of volume is essential when analyzing how stretching a wire affects its electrical resistance.