The current density in a wire is `10A//cm^(2)` and the electric field in the wire is 5 V/cm. If p = resistivity of material, `sigma`= conductivity of the material then (in SI unit)

Options

A. `p=5xx10^(-3)`
B. `p=200`
C. `sigma=5xx10^(-3)`
D. `sigma=200`

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert Current Density to SI Units
The given current density \( j \) is \( 10 \, \text{A/cm}^2 \). We need to convert this to SI units (A/m²).

\[ j = 10 \, \text{A/cm}^2 = 10 \, \text{A} \times \left( \frac{100 \, \text{cm}}{1 \, \text{m}} \right)^2 = 10 \times 10^4 \, \text{A/m}^2 = 10^5 \, \text{A/m}^2 \]

### Step 2: Convert Electric Field to SI Units
The electric field \( E \) is given as \( 5 \, \text{V/cm} \). We need to convert this to SI units (V/m).

\[ E = 5 \, \text{V/cm} = 5 \, \text{V} \times \left( \frac{100 \, \text{cm}}{1 \, \text{m}} \right) = 5 \times 10^2 \, \text{V/m} = 500 \, \text{V/m} \]

### Step 3: Calculate Conductivity \( \sigma \)
Using the relationship between current density \( j \), electric field \( E \), and conductivity \( \sigma \):

\[ j = \sigma E \]

We can rearrange this to find \( \sigma \):

\[ \sigma = \frac{j}{E} \]

Substituting the values we found:

\[ \sigma = \frac{10^5 \, \text{A/m}^2}{500 \, \text{V/m}} = \frac{10^5}{500} = 200 \, \text{S/m} \]

### Step 4: Calculate Resistivity \( \rho \)
The resistivity \( \rho \) is the reciprocal of conductivity:

\[ \rho = \frac{1}{\sigma} = \frac{1}{200} = 0.005 \, \Omega \cdot \text{m} = 5 \times 10^{-3} \, \Omega \cdot \text{m} \]

### Final Results
- Conductivity \( \sigma = 200 \, \text{S/m} \)
- Resistivity \( \rho = 5 \times 10^{-3} \, \Omega \cdot \text{m} \)

### Conclusion
The answers are:
- \( \sigma = 200 \, \text{S/m} \) (Option 4)
- \( \rho = 5 \times 10^{-3} \, \Omega \cdot \text{m} \) (Option 1)

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