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 into \( \text{m}^2 \): \( 1 \, \text{mm}^2 = 10^{-6} \, \text{m}^2 \). Use \( \rho = \frac{V}{I} \cdot \frac{A}{L} \) for quick calculation.

Answer 1

Concept: Resistivity of a material is given by: \[ \rho = \frac{RA}{L} \] Where:

\( R \) = resistance of wire
\( A \) = cross-sectional area
\( L \) = length of wire
Also, from Ohm’s law: \[ R = \frac{V}{I} \]
Step 1: Calculate resistance. \[ R = \frac{V}{I} = \frac{2}{4.0} = 0.5 \, \Omega \]
Step 2: Convert area into SI units. \[ 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \]
Step 3: Substitute into resistivity formula. \[ \rho = \frac{RA}{L} = \frac{0.5 \times 1 \times 10^{-6}}{1} \] \[ \rho = 0.5 \times 10^{-6} = 5 \times 10^{-7} \, \Omega \cdot \text{m} \] Final Answer: \[ \rho = 5 \times 10^{-7} \, \Omega \cdot \text{m} \]

Answer 2

Instead of first finding the resistance and then converting it into resistivity, we can reach the resistivity directly using the microscopic form of Ohm's law, which states that resistivity is the ratio of the electric field inside the conductor to the current density flowing through it, \( \rho = \dfrac{E}{J} \).

1. Step 1: Electric field inside the wire. The potential difference of \( 2 \, \text{V} \) is applied across the full length of the wire, \( L = 1 \, \text{m} \), so the electric field along the wire is \[ E = \frac{V}{L} = \frac{2}{1} = 2 \, \text{V/m}. \]
2. Step 2: Current density through the wire. The cross-sectional area must first be converted to SI units: \[ A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2. \] The current density is the current per unit cross-sectional area: \[ J = \frac{I}{A} = \frac{4.0}{1 \times 10^{-6}} = 4 \times 10^{6} \, \text{A/m}^2. \]
3. Step 3: Resistivity from the field-density ratio. \[ \rho = \frac{E}{J} = \frac{2}{4 \times 10^{6}} = 0.5 \times 10^{-6} = 5 \times 10^{-7} \, \Omega \cdot \text{m}. \]

So the resistivity of the wire's material is \( \rho = 5 \times 10^{-7} \, \Omega \cdot \text{m} \).