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