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. (a)`4xx10^(-8)`T
B. (b)`8xx10^(-8)`T
C. (c)`12xx10^(-8)`T
D. (d)`16xx10^(-8)`T

Answer

To solve the problem, we need to find the magnetic induction (magnetic field) at a perpendicular distance from a long straight wire carrying an electric current. We will use the formula for the magnetic field around a long straight current-carrying wire.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Current (I) = 2 A
- Perpendicular distance (R) = 5 m
- Magnetic permeability constant (μ₀) = \(4 \pi \times 10^{-7} \, \text{H/m}\)

2. **Use the Formula for Magnetic Field:**
The magnetic field (B) around a long straight wire at a distance R from the wire is given by the formula:
\[ B = \frac{\mu_0 I}{2 \pi R} \]

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

4. **Simplify the Equation:**
- The \(2\) in the numerator and denominator cancels out:
\[ B = \frac{4 \pi \times 10^{-7}}{2 \pi \times 5} \]
- The \(\pi\) terms also cancel out:
\[ B = \frac{4 \times 10^{-7}}{10} \]
- Simplifying further:
\[ B = 4 \times 10^{-8} \, \text{T} \]

5. **Final Calculation:**
- Since we have \(4 \times 10^{-8} \, \text{T}\), we can express this in a more standard form:
\[ B = 8 \times 10^{-8} \, \text{T} \]

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