The ratio of magnitude of current density and magnitude of electric field inside the conductor is

Options

A. Independent of temperature
B. Independent of dimension of conductor
C. Independent of material of conductor
D. All of these

Answer

To solve the question regarding the ratio of the magnitude of current density (J) to the magnitude of electric field (E) inside a conductor, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Ohm's Law**:
Ohm's law states that the voltage (V) across a conductor is proportional to the current (I) flowing through it, given by the equation:
\[ V = I \times R \]
where R is the resistance of the conductor.

2. **Relating Current to Current Density**:
The current density (J) is defined as the current (I) per unit area (A) through which the current flows:
\[ J = \frac{I}{A} \]

3. **Expressing Resistance in Terms of Resistivity**:
The resistance (R) of a conductor can be expressed in terms of its resistivity (ρ), length (L), and cross-sectional area (A):
\[ R = \frac{\rho L}{A} \]

4. **Substituting Resistance in Ohm's Law**:
By substituting the expression for R into the equation for Ohm's law, we get:
\[ V = I \times \frac{\rho L}{A} \]

5. **Rearranging to Find Current Density**:
Dividing both sides by A, we can express the current density (J):
\[ J = \frac{I}{A} = \frac{V}{\frac{\rho L}{A}} = \frac{V \cdot A}{\rho L} \]

6. **Simplifying the Expression**:
Since \( \frac{V}{L} \) is the electric field (E), we can rewrite the expression for J:
\[ J = \frac{1}{\rho} \cdot E \]

7. **Finding the Ratio of Current Density to Electric Field**:
To find the ratio of the magnitude of current density (J) to the magnitude of electric field (E), we rearrange the equation:
\[ \frac{J}{E} = \frac{1}{\rho} \]

8. **Relating to Conductivity**:
The reciprocal of resistivity (ρ) is conductivity (σ):
\[ \sigma = \frac{1}{\rho} \]
Therefore, we can express the ratio as:
\[ \frac{J}{E} = \sigma \]

### Final Answer:
The ratio of the magnitude of current density (J) to the magnitude of electric field (E) inside the conductor is:
\[ \frac{J}{E} = \sigma \]