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 need to find the resistivity (ρ) and conductivity (σ) of the material given the current density (J) and electric field (E).

### Step 1: Convert the given values to SI units
- Current density (J) = 10 A/cm² = 10 × 10² A/m² = 1000 A/m²
- Electric field (E) = 5 V/cm = 5 × 10² V/m = 500 V/m

### Step 2: Use the relationship between resistivity, electric field, and current density
The relationship between current density (J), electric field (E), and resistivity (ρ) is given by Ohm's law in its microscopic form:
\[ J = \sigma E \]
Where σ is the conductivity. We can also express resistivity as:
\[ \rho = \frac{E}{J} \]

### Step 3: Calculate the resistivity (ρ)
Using the formula for resistivity:
\[ \rho = \frac{E}{J} \]
Substituting the values:
\[ \rho = \frac{500 \, \text{V/m}}{1000 \, \text{A/m²}} \]
\[ \rho = 0.5 \, \Omega \cdot m \]

### Step 4: Calculate the conductivity (σ)
The conductivity (σ) is the reciprocal of resistivity:
\[ \sigma = \frac{1}{\rho} \]
Substituting the value of ρ:
\[ \sigma = \frac{1}{0.5} = 2 \, \text{S/m} \]

### Step 5: Verify the values
We can also use the relationship:
\[ \sigma = \frac{J}{E} \]
Substituting the values:
\[ \sigma = \frac{1000 \, \text{A/m²}}{500 \, \text{V/m}} = 2 \, \text{S/m} \]

### Final Answers:
- Resistivity (ρ) = 0.5 Ω·m
- Conductivity (σ) = 2 S/m