To determine the length (\( L \)) when resistivity (\( \rho \)), resistance (\( R \)), and cross-sectional area (\( A \)) are known, use the formula:
\[ L = \dfrac{R \cdot A}{\rho} \]
where:
- \( L \) is the length (in meters, \( \text{m} \)),
- \( \rho \) is the resistivity (in ohm meters, \( \Omega \cdot \text{m} \)),
- \( R \) is the resistance (in ohms, \( \Omega \)),
- \( A \) is the cross-sectional area (in square meters, \( \text{m}^2 \)).
Problem 1: Length of a Nichrome Wire
Scenario: A nichrome wire has a resistivity of \( 1.1 \times 10^{-6} \, \Omega \cdot \text{m} \), a resistance of \( 2 \, \Omega \), and a cross-sectional area of \( 1 \times 10^{-7} \, \text{m}^2 \). What is its length?
Calculation:
1. Given:
\[ \rho = 1.1 \times 10^{-6} \, \Omega \cdot \text{m} \]
\[ R = 2 \, \Omega \]
\[ A = 1 \times 10^{-7} \, \text{m}^2 \]
2. Substitute into the Length Formula:
\[ L = \dfrac{R \cdot A}{\rho} \]
\[ L = \dfrac{2 \cdot 1 \times 10^{-7}}{1.1 \times 10^{-6}} \]
3. Calculate:
\[ L = \dfrac{2 \times 10^{-7}}{1.1 \times 10^{-6}} \approx 0.182 \, \text{m} \]
Answer: The length of the nichrome wire is approximately \( 0.182 \, \text{m} \).
Problem 2: Length of a Platinum Wire
Scenario: A platinum wire has a resistivity of \( 10.6 \times 10^{-8} \, \Omega \cdot \text{m} \), a resistance of \( 5 \, \Omega \), and a cross-sectional area of \( 5 \times 10^{-8} \, \text{m}^2 \). Determine its length.
Calculation:
1. Given:
\[ \rho = 10.6 \times 10^{-8} \, \Omega \cdot \text{m} \]
\[ R = 5 \, \Omega \]
\[ A = 5 \times 10^{-8} \, \text{m}^2 \]
2. Substitute into the Length Formula:
\[ L = \dfrac{R \cdot A}{\rho} \]
\[ L = \dfrac{5 \cdot 5 \times 10^{-8}}{10.6 \times 10^{-8}} \]
3. Calculate:
\[ L = \dfrac{25 \times 10^{-8}}{10.6 \times 10^{-8}} \approx 2.36 \, \text{m} \]
Answer: The length of the platinum wire is approximately \( 2.36 \, \text{m} \).
