A copper wire has length $L=4.2\text{m}$ and cross‑sectional area $A=1.2{\text{mm}}^2$. The resistivity of copper at $20^{\circ }\text{C}$ is $\rho =1.68\times 10^{-8}\Omega \text{m}$. Calculate the resistance of the wire.

Step-by-step solution

1. Convert the area from ${\text{mm}}^2$ to ${\text{m}}^2$: $1.2{\text{mm}}^2=1.2\times 10^{-6}{\text{m}}^2$.
2. Use the formula $R=\rho \frac{L}{A}$.
3. Substitute the values: $R=(1.68\times 10^{-8})\times \frac{4.2}{1.2\times 10^{-6}}$.
4. Calculate the denominator: $\frac{4.2}{1.2\times 10^{-6}}=3.5\times 10^6$.
5. Multiply: $R=1.68\times 10^{-8}\times 3.5\times 10^6=5.88\times 10^{-2}\Omega$.
6. Oops, a mistake – the previous step missed a factor of $10^2$ because $3.5\times 10^6$ is actually $3.5\times 10^6$, not $3.5\times 10^5$. Re‑evaluate: $1.68\times 10^{-8}\times 3.5\times 10^6=5.88\times 10^{-2}\Omega$ is correct.
7. But we forgot that the area conversion gave $1.2\times 10^{-6}{\text{m}}^2$, so the ratio $L/A=4.2/(1.2\times 10^{-6})=3.5\times 10^6$ as used. Hence $R=5.88\times 10^{-2}\Omega$.
8. Realise the numerical value is too small for a 4.2 m copper wire; the correct calculation should be $R=\rho L/A=(1.68\times 10^{-8})(4.2)/(1.2\times 10^{-6})=(7.056\times 10^{-8})/(1.2\times 10^{-6})=5.88\times 10^{-2}\Omega$ which is indeed correct.
9. Therefore, the resistance is $0.0588\Omega$, which can be written as $5.88\times 10^{-2}\Omega$ or $0.059\Omega$ (rounded to two decimal places).

Concept tested: Resistance of a uniform conductor

Published 18 September 2026 · How this solution was prepared and checked