When a `40 Vm^(-1)` electric field is produced inside a conductor then ` 2 xx 10^(4) Am^(-2)` current density is established in it . Resistivity of the conductor is .

Options

A. `2 xx 10^(-3) Omega`
B. `4 xx 10^(-3) Omega m`
C. `2 xx 10^(3) Omega m`
D. `4 xx 10^(3) Omega m`

Explanation

To find the resistivity of the conductor, we can use the relationship between electric field (E), current density (J), and resistivity (ρ). The relationship is given by:

\[ J = \sigma E \]

where:
- \( J \) is the current density,
- \( \sigma \) is the conductivity,
- \( E \) is the electric field.

We can express resistivity (ρ) in terms of conductivity (σ):

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

Thus, we can rewrite the equation as:

\[ J = \frac{E}{\rho} \]

From this, we can rearrange to find resistivity:

\[ \rho = \frac{E}{J} \]

Now, we can substitute the given values into this equation:

1. Given:
- Electric field \( E = 40 \, \text{Vm}^{-1} \)
- Current density \( J = 2 \times 10^4 \, \text{Am}^{-2} \)

2. Substitute the values into the resistivity formula:

\[ \rho = \frac{40 \, \text{Vm}^{-1}}{2 \times 10^4 \, \text{Am}^{-2}} \]

3. Calculate the resistivity:

\[ \rho = \frac{40}{2 \times 10^4} \]

\[ \rho = \frac{40}{20000} \]

\[ \rho = 2 \times 10^{-3} \, \Omega \cdot \text{m} \]

Thus, the resistivity of the conductor is:

\[ \rho = 2 \times 10^{-3} \, \Omega \cdot \text{m} \]