A coaxial cable made up of two conductors. The inner conductor is solid and is of radius `R_1` and the outer conductor is hollow of inner radius `R_2` and outer radius `R_3`. The space between the conductors is filled with air. The inner and outer conductors are carrying currents of equal magnitudes and in opposite directions. Then the variation of magnetic field with distance from the axis is best plotted as

Options

A.
B.
C.
D.

Explanation

To solve the problem of the variation of the magnetic field with distance from the axis in a coaxial cable with two conductors carrying equal currents in opposite directions, we will apply Ampere's circuital law step by step.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have an inner solid conductor with radius \( R_1 \).
- The outer conductor is hollow, with an inner radius \( R_2 \) and outer radius \( R_3 \).
- Both conductors carry equal currents \( I \) in opposite directions.

2. **Applying Ampere's Circuital Law**:
- Ampere's law states that the line integral of the magnetic field \( B \) around a closed loop is equal to \( \mu_0 \) times the current enclosed by that loop:
\[ \oint B \cdot dl = \mu_0 I_{\text{enc}} \]

3. **Case 1: Inside the Inner Conductor (\( r < R_1 \))**:
- For a circular loop of radius \( r \) inside the inner conductor, the enclosed current \( I_{\text{enc}} \) can be calculated using the ratio of areas:
\[ I_{\text{enc}} = I \cdot \frac{A_{\text{loop}}}{A_{\text{total}}} = I \cdot \frac{\pi r^2}{\pi R_1^2} = I \cdot \frac{r^2}{R_1^2} \]
- Applying Ampere's law:
\[ B \cdot 2\pi r = \mu_0 I \cdot \frac{r^2}{R_1^2} \]
- Solving for \( B \):
\[ B = \frac{\mu_0 I r}{2\pi R_1^2} \]
- This shows that \( B \) is directly proportional to \( r \) for \( r < R_1 \).

4. **Case 2: Between the Conductors (\( R_1 < r < R_2 \))**:
- For a loop in the air gap between the inner and outer conductors, the enclosed current is just \( I \) (the current in the inner conductor):
\[ B \cdot 2\pi r = \mu_0 I \]
- Solving for \( B \):
\[ B = \frac{\mu_0 I}{2\pi r} \]
- This indicates that \( B \) is inversely proportional to \( r \) for \( R_1 < r < R_2 \).

5. **Case 3: Outside the Outer Conductor (\( r > R_3 \))**:
- For \( r > R_3 \), the total current enclosed is zero because the currents in the inner and outer conductors cancel each other out:
\[ B \cdot 2\pi r = 0 \implies B = 0 \]

6. **Graphing the Magnetic Field**:
- For \( r < R_1 \): \( B \) increases linearly with \( r \).
- For \( R_1 < r < R_2 \): \( B \) decreases inversely with \( r \).
- For \( r > R_3 \): \( B = 0 \).

### Conclusion:
The variation of the magnetic field with distance from the axis can be summarized as:
- Increases linearly for \( r < R_1 \)
- Decreases inversely for \( R_1 < r < R_2 \)
- Becomes zero for \( r > R_3 \)

Thus, the best plot for the variation of the magnetic field with distance from the axis corresponds to option C.