Question

A copper wire that is \(1 \mathrm{~m}\) long and has a radius of \(0.5 \mathrm{~mm}\) is stretched to a length of \(2 \mathrm{~m}\). 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

Based on the given solution, determine the fractional change in resistance for both copper and aluminum wires when they are stretched.

Step-by-step solution

Determine the resistivity of copper and aluminum

To find the fractional change in resistance, we must first know the resistivities of both copper and aluminum. These are constants for these materials and can be found in a physics reference or textbook. The resistivity of copper is approximately \(1.7 \times 10^{-8} \ \Omega . m\), and the resistivity of aluminum is approximately \(2.8 \times 10^{-8} \ \Omega . m\).

Calculate the initial resistance of the wires

The resistance R of a wire can be calculated using the formula: $$R = \frac{\rho L}{A}$$, where \(\rho\) is the resistivity, \(L\) is the length of the wire, and \(A\) is the cross-sectional area. For a cylindrical wire, the cross-sectional area can be described as \(A = \pi r^2\). Thus, the resistance formula for the copper and aluminum wires initially becomes:$$R_{Cu} = \frac{(1.7 \times 10^{-8} \ \Omega . m)(1 \ \mathrm{m})}{\pi (0.5 \times 10^{-3} \ \mathrm{m})^2}$$$$R_{Al} = \frac{(2.8 \times 10^{-8} \ \Omega . m)(1 \ \mathrm{m})}{\pi (0.5 \times 10^{-3} \ \mathrm{m})^2}$$. Evaluate these expressions to find the initial resistance of both wires.

Calculate the resistance of the wires after stretching

After the wires are stretched, the length of the copper and aluminum wires is now \(2 \ \mathrm{m}\). It is important to note that the volume of the wire remains constant, so when the wire is stretched, the radius of the wire will also change. The initial volume can be calculated using the formula \(V = A L\). After stretching, the volume needs to be kept constant, so \(V = A' L' = A(2L)\). Therefore, we get \(A' = \frac{A}{2}\), and the new radius can be calculated as \(r' = \sqrt{\frac{A'}{\pi}}\). Now, we can find the resistance of the wires after stretching using the resistance formula:$$R'_{Cu} = \frac{(1.7 \times 10^{-8} \ \Omega . m)(2 \ \mathrm{m})}{\pi (\sqrt{\frac{A}{2\pi}})^2}$$$$R'_{Al} = \frac{(2.8 \times 10^{-8} \ \Omega . m)(2 \ \mathrm{m})}{\pi (\sqrt{\frac{A}{2\pi}})^2}$$. Evaluate these expressions to find the resistance of both wires after stretching.

Calculate the fractional change in resistance

Now that we have the resistance before and after stretching, we can find the fractional change in resistance using the formula \(\Delta R / R = \frac{R' - R}{R}\). Substitute the values for the copper wire to get:$$\frac{\Delta R}{R}_{Cu} = \frac{R'_{Cu} - R_{Cu}}{R_{Cu}}$$. Similarly, substitute the values for the aluminum wire to get:$$\frac{\Delta R}{R}_{Al} = \frac{R'_{Al} - R_{Al}}{R_{Al}}$$. Calculate these expressions to find the fractional change in the resistance of the copper and aluminum wires.

Key concepts

Resistivity of Materials

When it comes to understanding electrical resistance, a fundamental concept is the resistivity of materials. Resistivity, denoted by the Greek letter rho \( \rho \), is a measure of how strongly a material opposes the flow of electric current. It's a characteristic property of the material itself, which means that it doesn't change with the shape or size of the material in question.
For students, it's crucial to remember that different materials have different resistivities. For example, metals such as copper and aluminum have relatively low resistivities, making them excellent conductors of electricity. In contrast, insulators like rubber have high resistivities and do a great job of preventing current flow.
When solving physics problems, the values of resistivity for copper (approximately \(1.7 \times 10^{-8} \ \Omega \cdot m\)) and for aluminum (approximately \(2.8 \times 10^{-8} \ \Omega \cdot m\)) can be found in reference tables. These values should be used in calculations involving the resistance of wires made from these materials.

Resistance Calculation

To calculate the resistance of a wire, you can use the formula \(R = \frac{\rho L}{A}\), where \(R\) is resistance, \(\rho\) is the material's resistivity, \(L\) is the length of the wire, and \(A\) is the cross-sectional area.
For students dealing with cylindrical wires, as is common in many physics exercises, remember that cross-sectional area is a key factor and should be computed considering the geometry of the wire. For a cylinder, \(A = \pi r^2\), where \(r\) is the radius of the wire. Always ensure the units are consistent to avoid calculation errors; for instance, when working with meters and millimeters, convert them all to the same unit before proceeding.

Cylindrical Wire Resistance

A cylindrical wire, such as the one often found in electrical applications, has a resistance that can be calculated as mentioned earlier. What many students should realize, however, is that when the length of a wire changes, its radius also changes if the volume remains constant, impacting the resistance.
If we take a wire that's initially \(1 \mathrm{m}\) long with a radius of \(0.5 \mathrm{mm}\), and stretch it to \(2 \mathrm{m}\), the cross-sectional area will decrease since \(A' = \frac{A}{2}\) to maintain the same volume. Consequently, this results in a change in the wire's resistance that can dramatically affect the wire's ability to conduct electricity. Such a process is crucial in the calculation of fractional change in resistance, as the area impacts resistance directly.

Fractional Change in Resistance

The fractional change in resistance \(\Delta R / R\) quantifies how much the resistance of a wire changes relative to its original resistance. It's calculated using the formula \(\frac{R' - R}{R}\), where \(R'\) is the final resistance after some change, and \(R\) is the initial resistance. This measure is unitless since it is a ratio.
In practical exercises or real-world scenarios, if a cylindrical wire is stretched and its length is doubled, its radius will decrease as the volume remains constant. The new resistance \(R'\) is higher because the length of the wire has increased while the cross-sectional area has decreased. The ratio \(\frac{\Delta R}{R}\) for materials with different resistivities, such as copper and aluminum, will differ even if the wires have the same initial dimensions, underlining the significance of resistivity in determining resistance changes.