Question

The resistivity of a conductor is \(\rho=1.00 \cdot 10^{-5} \Omega \mathrm{m}\). If a cylindrical wire is made of this conductor, with a cross-sectional area of \(1.00 \cdot 10^{-6} \mathrm{~m}^{2}\) what should the length of the wire be for its resistance to be \(10.0 \Omega ?\)

Short answer

Answer: The required length of the wire is \(1.00 \mathrm{m}\).

Step-by-step solution

Identify the given information

We are given the following information: - Resistivity, \(\rho = 1.00 \cdot 10^{-5} \Omega \mathrm{m}\) - Cross-sectional area, \(A = 1.00 \cdot 10^{-6} \mathrm{~m}^{2}\) - Resistance, \(R = 10.0 \Omega\)

Recall the formula for resistance

The formula for resistance in terms of resistivity, cross-sectional area, and length is given by: \(R = \frac{\rho L}{A}\) Where: - \(R\) is the resistance - \(\rho\) is the resistivity - \(L\) is the length of the wire - \(A\) is the cross-sectional area

Rearrange the formula to solve for length

We need to find the length of the wire, so we will rearrange the formula to isolate \(L\): \(L = \frac{R \times A}{\rho}\)

Substitute the given values into the formula

Plug the given values for resistance, area, and resistivity into the formula: \(L = \frac{10.0 \Omega \times 1.00 \cdot 10^{-6} \mathrm{~m}^{2}}{1.00 \cdot 10^{-5} \Omega \mathrm{m}}\)

Calculate the length

Simplify the equation to find the length of the wire required: \(L = \frac{10.0 \times 1.00 \cdot 10^{-6}}{1.00 \cdot 10^{-5}} = 1.00 \mathrm{m}\) The length of the wire should be \(1.00\mathrm{m}\) for its resistance to be \(10.0 \Omega\).