A coaxial cable consists of a thin inner conductor fixed along the axis of a hollow outer conductor. The two conductor carry equal currents in opposite directions. Let `B_1 and B_2` be the magnetic fields in the region between the conductors and outside the conductor, respectively. Then,

Options

A. `B_1!=0, B_2!=0`
B. `B_1=B_2=0`
C. `B_1!=0, B_2=0`
D. `B_1=0, B_2!=0`

Explanation

To solve the problem regarding the magnetic fields \( B_1 \) and \( B_2 \) in a coaxial cable with equal currents in opposite directions, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**:
- We have a coaxial cable with a thin inner conductor and a hollow outer conductor.
- The inner conductor carries a current \( I \) in one direction, while the outer conductor carries an equal current \( I \) in the opposite direction.

2. **Identify Regions**:
- There are two regions to consider:
- Region between the inner and outer conductors (where we will find \( B_1 \)).
- Region outside the outer conductor (where we will find \( B_2 \)).

3. **Apply Ampere's Circuital Law**:
- Ampere's Circuital 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.
- The formula is given by:
\[ \oint B \cdot dl = \mu_0 I_{\text{enc}} \]

4. **Calculate \( B_1 \) (Magnetic Field Between Conductors)**:
- For the region between the inner and outer conductors, the enclosed current \( I_{\text{enc}} \) is simply \( I \) (the current through the inner conductor).
- We can choose a circular path of radius \( r \) (where \( r \) is less than the radius of the outer conductor).
- The integral becomes:
\[ B_1 \cdot (2\pi r) = \mu_0 I \]
- Solving for \( B_1 \):
\[ B_1 = \frac{\mu_0 I}{2\pi r} \]

5. **Calculate \( B_2 \) (Magnetic Field Outside the Conductors)**:
- For the region outside the outer conductor, the enclosed current \( I_{\text{enc}} \) is zero because the currents in the inner and outer conductors cancel each other out.
- For a circular path of radius \( R \) (where \( R \) is greater than the radius of the outer conductor):
\[ B_2 \cdot (2\pi R) = \mu_0 (I - I) = 0 \]
- Thus, we find:
\[ B_2 = 0 \]

6. **Conclusion**:
- We conclude that:
- \( B_1 \neq 0 \) (the magnetic field exists in the region between the conductors).
- \( B_2 = 0 \) (the magnetic field outside the conductors is zero).

### Final Answer:
- \( B_1 \) is not equal to zero, and \( B_2 \) is equal to zero.