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

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

Answer 2

Let's derive a single combined formula for resistivity directly from Ohm's law and the definition of resistivity, rather than calculating the resistance as a separate intermediate step.

Ohm's law for the wire states \( V = IR \), so \( R = \dfrac{V}{I} \).
The definition of resistivity relates resistance to the wire's geometry: \( R = \dfrac{\rho L}{A} \), so \( \rho = \dfrac{RA}{L} \).

Substituting the expression for \( R \) into the resistivity relation gives a single formula in terms of the directly measured quantities: \[ \rho = \frac{V A}{I L}. \]

Now plug in the given values in one pass. The area must be in SI units first: \[ A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2. \] \[ \rho = \frac{2 \times (1 \times 10^{-6})}{4.0 \times 1} = \frac{2 \times 10^{-6}}{4.0} = 0.5 \times 10^{-6} \, \Omega \cdot \text{m}. \]

Writing this in standard scientific notation: \[ \rho = 5 \times 10^{-7} \, \Omega \cdot \text{m}. \]

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