A long straight wire carries an electric current of `2A`. The magnetic induction at a perpendicular distance of `5m` from the wire is `( mu_(0)4 pi xx 10^(7)Hm^(-1))`

Options

A. `4xx10^(-8)T`
B. `8xx10^(-8)T`
C. `12xx10^(-8)T`
D. `16xx10^(-8)T`

Explanation

To find the magnetic induction (B) at a perpendicular distance from a long straight wire carrying an electric current, we can use the formula:

\[ B = \frac{\mu_0 I}{2 \pi D} \]

Where:
- \(B\) is the magnetic induction,
- \(\mu_0\) is the permeability of free space, given as \(4\pi \times 10^{-7} \, \text{H/m}\),
- \(I\) is the current in amperes,
- \(D\) is the perpendicular distance from the wire in meters.

### Step-by-Step Solution:

1. **Identify the given values:**
- Current, \(I = 2 \, \text{A}\)
- Distance, \(D = 5 \, \text{m}\)
- Permeability of free space, \(\mu_0 = 4\pi \times 10^{-7} \, \text{H/m}\)

2. **Substitute the values into the formula:**
\[ B = \frac{4\pi \times 10^{-7} \times 2}{2\pi \times 5} \]

3. **Simplify the expression:**
- The \(2\pi\) in the numerator and denominator cancels out:
\[ B = \frac{4 \times 10^{-7} \times 2}{10} = \frac{8 \times 10^{-7}}{10} \]

4. **Calculate the final value:**
\[ B = 8 \times 10^{-8} \, \text{T} \]

Thus, the magnetic induction at a perpendicular distance of 5 m from the wire is \(8 \times 10^{-8} \, \text{T}\).