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 relates the magnetic field around a closed loop to the current passing through the loop. The law states that the line integral of the magnetic field \( B \) around a closed path is equal to \( \mu_0 \) times the current \( I \) enclosed by that path.

### 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:
\[ \oint B \cdot dl = \mu_0 I_{\text{enc}} \]
- Here, \( I_{\text{enc}} \) is the current enclosed by the closed path.

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

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

5. **Final Result:**
- The line integral of magnetic induction around the closed path is:
\[ \oint B \cdot dl = 10^{-7} \, \text{T m} \]