Two parallel wires 2 m apart carry currents of 2 A and 5 A respectively in same direction, the force per unit length acting between these two wires is

Options

A. `2xx 10^(-6)Nm^(-1)`
B. `3xx 10^(-6)Nm^(-1)`
C. `1xx 10^(-6)Nm^(-1)`
D. `4xx 10^(-6)Nm^(-1)`

Explanation

To find the force per unit length acting between two parallel wires carrying currents in the same direction, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Current in wire 1, \( I_1 = 2 \, \text{A} \)
- Current in wire 2, \( I_2 = 5 \, \text{A} \)
- Distance between the wires, \( r = 2 \, \text{m} \)

2. **Understand the Formula:**
The formula to calculate the force per unit length (\( F/L \)) between two parallel wires carrying currents is given by:
\[ \frac{F}{L} = \frac{\mu_0}{2\pi} \cdot \frac{I_1 I_2}{r} \]
where \( \mu_0 \) (the permeability of free space) is approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

3. **Substitute the Values into the Formula:**
Using the values we have:
\[ \frac{F}{L} = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{2 \cdot 5}{2} \]

4. **Simplify the Equation:**
- The \( \pi \) in the numerator and denominator cancels out:
\[ \frac{F}{L} = \frac{4 \times 10^{-7}}{2} \cdot \frac{10}{2} \]
- This simplifies to:
\[ \frac{F}{L} = 2 \times 10^{-7} \cdot 5 = 10 \times 10^{-7} \]

5. **Final Calculation:**
\[ \frac{F}{L} = 10 \times 10^{-7} = 1 \times 10^{-6} \, \text{N/m} \]

### Conclusion:
The force per unit length acting between the two wires is:
\[ \frac{F}{L} = 1 \times 10^{-6} \, \text{N/m} \]