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: \[ \rho = 5 \times 10^{-7} \, \Omega \cdot \text{m} \]