Question

A current of \(4.0\,\text{A}\) flows through a wire of length \(1\,\text{m}\) and cross-sectional area \(1.0\,\text{mm}^2\), when a potential difference of \(2\,\text{V}\) is applied across its ends. Calculate the resistivity of the material of the wire.

Hint:

Always convert area from \(\text{mm}^2\) to \(\text{m}^2\) before using \[ \rho=\frac{RA}{L}. \] Remember: \[ 1\,\text{mm}^2=10^{-6}\,\text{m}^2. \]

Answer 1

Concept: The resistivity of a material is an intrinsic property that measures how strongly the material opposes the flow of electric current. The relation connecting resistance and resistivity is \[ R=\rho\frac{L}{A}, \] where
• \(R\) = resistance of the conductor,
• \(\rho\) = resistivity of the material,
• \(L\) = length of the conductor,
• \(A\) = cross-sectional area. The resistance can first be determined using Ohm's law. ::contentReference[oaicite:0]{index=0}

Step 1: Calculate the resistance of the wire using Ohm's law. Given, \[ V=2\,\text{V} \] and \[ I=4\,\text{A}. \] Using Ohm's law, \[ R=\frac{V}{I}. \] Substituting the values, \[ R=\frac{2}{4}. \] \[ R=0.5\,\Omega. \] Therefore, \[ \boxed{R=0.5\,\Omega}. \]

Step 2: Convert the cross-sectional area into SI units. Given, \[ A=1.0\,\text{mm}^2. \] Since \[ 1\,\text{mm}=10^{-3}\,\text{m}, \] therefore \[ 1\,\text{mm}^2=(10^{-3})^2\,\text{m}^2. \] \[ A=10^{-6}\,\text{m}^2. \]

Step 3: Use the resistivity formula. The relation is \[ R=\rho\frac{L}{A}. \] Rearranging, \[ \rho=\frac{RA}{L}. \] Substituting \[ R=0.5\,\Omega, \qquad A=10^{-6}\,\text{m}^2, \qquad L=1\,\text{m}, \] we get \[ \rho = \frac{0.5\times10^{-6}}{1}. \] \[ \rho = 5\times10^{-7}\,\Omega\text{m}. \]

Step 4: Write the final answer. Hence, the resistivity of the material of the wire is \[ \boxed{\rho=5\times10^{-7}\,\Omega\text{m}}. \]