Question

Consider an aluminium wire with a diameter of 1.07 mm. What would be the diameter of a copper wire of the same length with the same resistance? Would it be more or less massive than the aluminium wire and by what factor? The density of copper is roughly 8,960 kgm^(−3).

Answer 1

Here rho sign stands for resistivity.

Answer 2

Given:

• Diameter of aluminium wire, $d_{Al}=1.07\text{mm}$

• Resistivity of aluminium, ${\rho }_{Al}=2.65\times 10^{-8}\mathrm{\Omega }\cdot \text{m}$

• Resistivity of copper, ${\rho }_{Cu}=1.68\times 10^{-8}\mathrm{\Omega }\cdot \text{m}$

• Density of copper, ${\rho }_{\text{mass, Cu}}=8960{\text{kg/m}}^3$

• Density of aluminium, ${\rho }_{\text{mass, Al}}=2700{\text{kg/m}}^3$

Step 1: Equal Resistance Condition

Resistance $R$ for a wire is given by:

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

For equal resistance and length:

$$\frac{{\rho }_{Al}}{A_{Al}}=\frac{{\rho }_{Cu}}{A_{Cu}}$$$$A_{Cu}=A_{Al}\cdot \frac{{\rho }_{Cu}}{{\rho }_{Al}}$$

Step 2: Calculate Cross-Sectional Area

$$A_{Al}=\frac{\pi d_{Al}^2}{4}=\frac{\pi (1.07\times 10^{-3}{)}^2}{4}=8.99\times 10^{-7}{\text{m}}^2$$$$A_{Cu}=8.99\times 10^{-7}\cdot \frac{1.68\times 10^{-8}}{2.65\times 10^{-8}}=5.70\times 10^{-7}{\text{m}}^2$$

Step 3: Find Copper Wire Diameter

$$d_{Cu}=\sqrt{\frac{4A_{Cu}}{\pi }}=\sqrt{\frac{4\times 5.70\times 10^{-7}}{\pi }}=0.853\text{mm}$$

Step 4: Compare Masses

Mass is given by $m={\rho }_{\text{mass}}\cdot A\cdot L$. For the same length $L$:

$$\frac{m_{Cu}}{m_{Al}}=\frac{{\rho }_{\text{mass, Cu}}\cdot A_{Cu}}{{\rho }_{\text{mass, Al}}\cdot A_{Al}}=\frac{8960\times 5.70\times 10^{-7}}{2700\times 8.99\times 10^{-7}}=2.10$$

Answer:

• Diameter of copper wire: $0.853\text{mm}$

• Mass comparison: The copper wire is 2.10 times more massive than the aluminium wire