The cross-sectional area (\( A \)) of a material can be determined using the formula:
\[ A = \dfrac{\rho \cdot L}{R} \]
where:
- \( A \) is the cross-sectional area (in square meters, \( \text{m}^2 \)),
- \( \rho \) is the resistivity (in ohm meters, \( \Omega \cdot \text{m} \)),
- \( L \) is the length of the material (in meters, m),
- \( R \) is the resistance (in ohms, \( \Omega \)).
We’ll illustrate how to find the cross-sectional area with five practical examples.
Example 1: Cross-Sectional Area of a Copper Wire
Scenario: A copper wire has a length of \( 2 \, \text{m} \), a resistance of \( 0.0336 \, \Omega \), and a resistivity of \( 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \). What is the cross-sectional area?
Step-by-Step Calculation:
1. Given:
\[ \rho = 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \]
\[ L = 2 \, \text{m} \]
\[ R = 0.0336 \, \Omega \]
2. Substitute Values into the Cross-Sectional Area Formula:
\[ A = \dfrac{\rho \cdot L}{R} \]
\[ A = \dfrac{1.68 \times 10^{-8} \cdot 2}{0.0336} \]
3. Perform the Calculation:
\[ A \approx 1 \times 10^{-6} \, \text{m}^2 \]
Final Value
The cross-sectional area of the copper wire is:
\[ A \approx 1 \times 10^{-6} \, \text{m}^2 \]
Example 2: Cross-Sectional Area of an Aluminum Rod
Scenario: An aluminum rod has a length of \( 5 \, \text{m} \), a resistance of \( 0.06625 \, \Omega \), and a resistivity of \( 2.65 \times 10^{-8} \, \Omega \cdot \text{m} \). Calculate the cross-sectional area.
Step-by-Step Calculation:
1. Given:
\[ \rho = 2.65 \times 10^{-8} \, \Omega \cdot \text{m} \]
\[ L = 5 \, \text{m} \]
\[ R = 0.06625 \, \Omega \]
2. Substitute Values into the Cross-Sectional Area Formula:
\[ A = \dfrac{\rho \cdot L}{R} \]
\[ A = \dfrac{2.65 \times 10^{-8} \cdot 5}{0.06625} \]
3. Perform the Calculation:
\[ A \approx 2 \times 10^{-6} \, \text{m}^2 \]
Final Value
The cross-sectional area of the aluminum rod is:
\[ A \approx 2 \times 10^{-6} \, \text{m}^2 \]
Example 3: Cross-Sectional Area of a Silver Cable
Scenario: A silver cable has a length of \( 10 \, \text{m} \), a resistance of \( 0.318 \, \Omega \), and a resistivity of \( 1.59 \times 10^{-8} \, \Omega \cdot \text{m} \). What is the cross-sectional area?
Step-by-Step Calculation:
1. Given:
\[ \rho = 1.59 \times 10^{-8} \, \Omega \cdot \text{m} \]
\[ L = 10 \, \text{m} \]
\[ R = 0.318 \, \Omega \]
2. Substitute Values into the Cross-Sectional Area Formula:
\[ A = \dfrac{\rho \cdot L}{R} \]
\[ A = \dfrac{1.59 \times 10^{-8} \cdot 10}{0.318} \]
3. Perform the Calculation:
\[ A \approx 0.5 \times 10^{-6} \, \text{m}^2 \]
Final Value
The cross-sectional area of the silver cable is:
\[ A \approx 0.5 \times 10^{-6} \, \text{m}^2 \]
Example 4: Cross-Sectional Area of an Iron Bar
Scenario: An iron bar has a length of \( 3 \, \text{m} \), a resistance of \( 0.0971 \, \Omega \), and a resistivity of \( 9.71 \times 10^{-8} \, \Omega \cdot \text{m} \). Calculate the cross-sectional area.
Step-by-Step Calculation:
1. Given:
\[ \rho = 9.71 \times 10^{-8} \, \Omega \cdot \text{m} \]
\[ L = 3 \, \text{m} \]
\[ R = 0.0971 \, \Omega \]
2. Substitute Values into the Cross-Sectional Area Formula:
\[ A = \dfrac{\rho \cdot L}{R} \]
\[ A = \dfrac{9.71 \times 10^{-8} \cdot 3}{0.0971} \]
3. Perform the Calculation:
\[ A \approx 3 \times 10^{-6} \, \text{m}^2 \]
Final Value
The cross-sectional area of the iron bar is:
\[ A \approx 3 \times 10^{-6} \, \text{m}^2 \]
Example 5: Cross-Sectional Area of a Gold Wire
Scenario: A gold wire has a length of \( 0.5 \, \text{m} \), a resistance of \( 0.122 \, \Omega \), and a resistivity of \( 2.44 \times 10^{-8} \, \Omega \cdot \text{m} \). What is the cross-sectional area?
Step-by-Step Calculation:
1. Given:
\[ \rho = 2.44 \times 10^{-8} \, \Omega \cdot \text{m} \]
\[ L = 0.5 \, \text{m} \]
\[ R = 0.122 \, \Omega \]
2. Substitute Values into the Cross-Sectional Area Formula:
\[ A = \dfrac{\rho \cdot L}{R} \]
\[ A = \dfrac{2.44 \times 10^{-8} \cdot 0.5}{0.122} \]
3. Perform the Calculation:
\[ A \approx 0.1 \times 10^{-6} \, \text{m}^2 \]
Final Value
The cross-sectional area of the gold wire is:
\[ A \approx 0.1 \times 10^{-6} \, \text{m}^2 \]
Summary
To find the cross-sectional area (\( A \)) given the resistivity (\( \rho \)), length (\( L \)), and resistance (\( R \)), use the formula:
\[ A = \dfrac{\rho \cdot L}{R} \]
In the examples provided:
1. Copper wire: \( A \approx 1 \times 10^{-6} \, \text{m}^2 \)
2. Aluminum rod: \( A \approx 2 \times 10^{-6} \, \text{m}^2 \)
3. Silver cable: \( A \approx 0.5 \times 10^{-6} \, \text{m}^2 \)
4. Iron bar: \( A \approx 3 \times 10^{-6} \, \text{m}^2 \)
5. Gold wire: \( A \approx 0.1 \times 10^{-6} \, \text{m}^2 \)
