A long straight wire is carrying a current of 12 A . The magnetic field at a distance of 8 cm is
`(mu_(0) =4pi xx 10^(-7) N A ^(2))`

Options

A. `2xx10^(-4) wb//m^(2)`
B. `3xx10^(-5) Wb//m^(2)`
C. `4xx10^(-4)Wb//m^(2)`
D. `4xx10^(-5)Wb//m^(2)`

Answer

To solve the problem of finding the magnetic field around a long straight wire carrying a current, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Formula**:
The magnetic field \( B \) around a long straight conductor carrying a current \( I \) at a distance \( R \) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi R} \]
where \( \mu_0 \) is the permeability of free space.

2. **Substitute the Given Values**:
From the problem, we have:
- Current \( I = 12 \, \text{A} \)
- Distance \( R = 8 \, \text{cm} = 8 \times 10^{-2} \, \text{m} \)
- Permeability \( \mu_0 = 4 \pi \times 10^{-7} \, \text{N/A}^2 \)

Plugging these values into the formula:
\[ B = \frac{4 \pi \times 10^{-7} \, \text{N/A}^2 \times 12 \, \text{A}}{2 \pi \times (8 \times 10^{-2} \, \text{m})} \]

3. **Simplify the Expression**:
Cancel out \( \pi \) from the numerator and denominator:
\[ B = \frac{4 \times 10^{-7} \times 12}{2 \times (8 \times 10^{-2})} \]

4. **Calculate the Values**:
- Calculate the numerator:
\[ 4 \times 12 = 48 \]
- Calculate the denominator:
\[ 2 \times 8 \times 10^{-2} = 16 \times 10^{-2} = 0.16 \]
- Now substitute these values back into the equation:
\[ B = \frac{48 \times 10^{-7}}{0.16} \]

5. **Final Calculation**:
- Divide \( 48 \) by \( 0.16 \):
\[ 48 \div 0.16 = 300 \]
- Therefore:
\[ B = 300 \times 10^{-7} = 3 \times 10^{-5} \, \text{Wb/m}^2 \]

6. **Conclusion**:
The magnetic field at a distance of 8 cm from the wire is:
\[ B = 3 \times 10^{-5} \, \text{Wb/m}^2 \]
This matches with option B.