Calculate the conductivity of a wire of length 2 m, area of cross-sectionl `2 cm^(2)` and resistance `10^(-4) Omega`.

Answer

To calculate the conductivity of a wire, we can use the formula that relates resistance (R), resistivity (ρ), length (L), and area (A) of the wire. The relationship is given by:

\[ R = \frac{\rho L}{A} \]

Where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area.

Since we want to find the conductivity (σ), we know that:

\[ \sigma = \frac{1}{\rho} \]

From the resistance formula, we can express resistivity in terms of conductivity:

\[ R = \frac{1}{\sigma} \cdot \frac{L}{A} \]

Rearranging this gives us:

\[ \sigma = \frac{L}{R \cdot A} \]

Now, let's substitute the given values into this formula.

1. **Convert the area from cm² to m²**:
- Given area \( A = 2 \, \text{cm}^2 \)
- Convert to m²:
\[ A = 2 \, \text{cm}^2 = 2 \times 10^{-4} \, \text{m}^2 \]

2. **Substitute the values into the conductivity formula**:
- Length \( L = 2 \, \text{m} \)
- Resistance \( R = 10^{-4} \, \Omega \)
- Area \( A = 2 \times 10^{-4} \, \text{m}^2 \)

Now substituting these values into the conductivity formula:

\[ \sigma = \frac{L}{R \cdot A} = \frac{2 \, \text{m}}{(10^{-4} \, \Omega) \cdot (2 \times 10^{-4} \, \text{m}^2)} \]

3. **Calculate the denominator**:
- \( R \cdot A = 10^{-4} \, \Omega \cdot 2 \times 10^{-4} \, \text{m}^2 = 2 \times 10^{-8} \, \Omega \cdot \text{m}^2 \)

4. **Now calculate the conductivity**:
\[ \sigma = \frac{2 \, \text{m}}{2 \times 10^{-8} \, \Omega \cdot \text{m}^2} = \frac{2}{2 \times 10^{-8}} = 10^{8} \, \text{S/m} \]

5. **Convert the conductivity to cm⁻¹**:
- Since \( 1 \, \text{S/m} = 10^{-4} \, \text{S/cm} \):
\[ \sigma = 10^{8} \, \text{S/m} = 10^{8} \times 10^{-4} \, \text{S/cm} = 10^{4} \, \text{S/cm} \]

Thus, the conductivity of the wire is:

\[ \sigma = 10^{4} \, \text{S/cm} \]