If `vec(j)` and `vec(E )` are current density and electric field respectively inside a current carrying conductor , then correct relation is

Options

A. `vec(j ) = sigma vec(E )`
B. `vec(j ) xx vec( E ) = 0 `
C. `vec( j) . vec( E ) gt 0`
D. All of these

Answer

To solve the question regarding the relationship between current density (J) and electric field (E) inside a current-carrying conductor, we can follow these steps:

### Step-by-Step Solution:

1. **Define Current Density (J)**:
Current density is defined as the current flowing per unit area of the conductor. Mathematically, it is given by:
\[ J = \frac{I}{A} \]
where \(I\) is the current and \(A\) is the cross-sectional area of the conductor.

**Hint**: Remember that current density is a vector quantity that indicates the flow of electric charge.

2. **Define Electric Field (E)**:
The electric field inside the conductor can be defined in terms of the potential difference (V) across the length (L) of the conductor:
\[ E = \frac{V}{L} \]

**Hint**: Electric field direction is the direction of force on a positive charge, which is the same direction as the current in a conductor.

3. **Relate Current (I) and Electric Field (E)**:
According to Ohm's law, the potential difference (V) can also be expressed in terms of current (I) and resistance (R):
\[ V = I \cdot R \]
The resistance \(R\) can be expressed using resistivity (\(\rho\)):
\[ R = \frac{\rho L}{A} \]

**Hint**: Ohm's law connects voltage, current, and resistance in a linear manner.

4. **Substitute for V and I**:
We can substitute the expressions for \(V\) and \(I\) into the equation:
\[ E \cdot L = (J \cdot A) \cdot \left(\frac{\rho L}{A}\right) \]
Simplifying this gives:
\[ E = J \cdot \rho \]

**Hint**: Be careful with units and ensure that all terms are consistent.

5. **Express in Terms of Conductivity**:
Conductivity (\(\sigma\)) is the reciprocal of resistivity:
\[ \sigma = \frac{1}{\rho} \]
Thus, we can rewrite the relationship:
\[ J = \sigma E \]

**Hint**: This relationship indicates that current density is proportional to the electric field, with conductivity as the proportionality constant.

6. **Evaluate Additional Options**:
- **Option B**: \(J \times E = 0\) is true because the angle between J and E is zero (they are in the same direction).
- **Option C**: \(J \cdot E > 0\) is also true since the dot product of two vectors in the same direction is positive.

**Hint**: Understand the geometric interpretations of dot and cross products for vectors.

7. **Conclusion**:
All three options provided in the question are correct, leading to the conclusion that the correct answer is:
\[ \text{Option D: All of these} \]

**Hint**: Always check if multiple options can be correct in physics problems, especially when dealing with vector quantities.