A long straight wire carries a current of `pi amp`. The magnetic field due to it will be `5xx10^(-5)"weber"//m^(2)` at what distance from the wire [`mu_(0)=` permeability of air]

Options

A. `10^(4)mu_(0)` metre
B. `(10^(4))/(mu_(0))` metre
C. `10^(6)mu_(0)` metre
D. `(10^(6))/(mu_(0))` metre

Answer

To solve the problem of finding the distance from a long straight wire carrying a current of π amperes, where the magnetic field is given as \(5 \times 10^{-5} \, \text{Wb/m}^2\), we can use the formula for the magnetic field around a long straight conductor.

### Step-by-Step Solution:

1. **Identify the formula for the magnetic field due to a long straight wire**:
The magnetic field \(B\) at a distance \(d\) from a long straight wire carrying current \(I\) is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi d} \]
where \(\mu_0\) is the permeability of free space (approximately \(4\pi \times 10^{-7} \, \text{T m/A}\)).

2. **Substitute the known values into the formula**:
We know:
- \(B = 5 \times 10^{-5} \, \text{Wb/m}^2\)
- \(I = \pi \, \text{A}\)
- \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\)

Plugging these values into the formula gives:
\[ 5 \times 10^{-5} = \frac{(4\pi \times 10^{-7}) \cdot \pi}{2 \pi d} \]

3. **Simplify the equation**:
The \(\pi\) terms cancel out:
\[ 5 \times 10^{-5} = \frac{4 \times 10^{-7}}{2d} \]
This simplifies to:
\[ 5 \times 10^{-5} = \frac{2 \times 10^{-7}}{d} \]

4. **Rearrange the equation to solve for \(d\)**:
Multiplying both sides by \(d\) and then dividing by \(5 \times 10^{-5}\) gives:
\[ d = \frac{2 \times 10^{-7}}{5 \times 10^{-5}} \]

5. **Calculate the value of \(d\)**:
\[ d = \frac{2}{5} \times 10^{-2} = 0.4 \times 10^{-2} = 4 \times 10^{-3} \, \text{m} \]

### Final Answer:
The distance \(d\) from the wire is \(4 \times 10^{-3} \, \text{m}\) or \(4 \, \text{mm}\).