Three long straight conductors are kept perpendicular to the plane of paper. Currents `2A,3A` are passing through the two conductors into the plane of paper in first two conductors and `5A` current passes through thirst conductor, directed out of the paper. A closed loop encloses the conductors, then the value of `oint bar(B). bar(dl)` over the closed loop is `(` assume current into the paper as negative and out of the paper as positive `)`

Options

A. `2 mu_(0)`
B. zero
C. `- mu_(0)`
D. `+ mu_(0)`

Explanation

To solve the problem, we need to determine the value of the line integral of the magnetic field \( \oint \mathbf{B} \cdot d\mathbf{l} \) over a closed loop that encloses three long straight conductors carrying currents. The currents are given as follows:

- Two conductors have currents of \( 2A \) and \( 3A \) flowing into the plane of the paper (which we will consider as negative).
- One conductor has a current of \( 5A \) flowing out of the plane of the paper (which we will consider as positive).

### Step-by-Step Solution:

1. **Identify the Directions of Currents**:
- The first conductor carries a current of \( 2A \) into the paper (negative).
- The second conductor carries a current of \( 3A \) into the paper (negative).
- The third conductor carries a current of \( 5A \) out of the paper (positive).

2. **Calculate the Net Current Enclosed by the Loop**:
- The total current flowing into the loop (negative) is:
\[ I_{\text{in}} = -2A - 3A = -5A \]
- The total current flowing out of the loop (positive) is:
\[ I_{\text{out}} = +5A \]
- Therefore, the net current \( I_{\text{net}} \) enclosed by the loop is:
\[ I_{\text{net}} = I_{\text{out}} + I_{\text{in}} = 5A - 5A = 0A \]

3. **Apply Ampère's Circuital Law**:
- According to Ampère's Circuital Law, the line integral of the magnetic field around a closed loop is given by:
\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{net}} \]
- Where \( \mu_0 \) is the permeability of free space.

4. **Substitute the Net Current into the Equation**:
- Since we found that \( I_{\text{net}} = 0A \):
\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 \times 0 = 0 \]

5. **Conclusion**:
- Therefore, the value of \( \oint \mathbf{B} \cdot d\mathbf{l} \) over the closed loop is:
\[ \boxed{0} \]