A current of `1//(4pi)` ampere is flowing in a long straight conductor. The line integral of magnetic induction around a closed path enclosing the current carrying conductor is

Options

A. `10^-7Wbm^-1`
B. `4pi^-7Wbm^-1`
C. `16pi^2xx10^-7Wbm^-1`
D. zero.

Explanation

To solve the problem, we will use Ampere's Circuital Law, which states that the line integral of the magnetic field \( \mathbf{B} \) around a closed loop is equal to the permeability of free space \( \mu_0 \) times the current \( I \) enclosed by that loop. The formula can be expressed as:

\[ \oint \mathbf{B} \cdot d\mathbf{L} = \mu_0 I \]

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Current \( I = \frac{1}{4\pi} \) A.
- Permeability of free space \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

2. **Apply Ampere's Circuital Law**:
- According to Ampere's Circuital Law:
\[ \oint \mathbf{B} \cdot d\mathbf{L} = \mu_0 I \]

3. **Substitute the Known Values**:
- Substitute the values of \( \mu_0 \) and \( I \) into the equation:
\[ \oint \mathbf{B} \cdot d\mathbf{L} = (4\pi \times 10^{-7}) \left(\frac{1}{4\pi}\right) \]

4. **Simplify the Expression**:
- The \( 4\pi \) terms cancel out:
\[ \oint \mathbf{B} \cdot d\mathbf{L} = 10^{-7} \, \text{T m} \]

5. **Conclusion**:
- The line integral of magnetic induction around the closed path enclosing the current-carrying conductor is:
\[ \oint \mathbf{B} \cdot d\mathbf{L} = 10^{-7} \, \text{Wb/m}^2 \]

### Final Answer:
The line integral of magnetic induction around the closed path is \( 10^{-7} \, \text{Wb/m}^2 \).

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