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 solve the problem of finding 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**:
- Distance between the wires, \( R = 2 \, \text{m} \)
- Current in wire 1, \( I_1 = 2 \, \text{A} \)
- Current in wire 2, \( I_2 = 5 \, \text{A} \)

2. **Use the Formula for the Force Between Parallel Currents**:
The formula for the force per unit length \( F/L \) between two parallel wires carrying currents \( I_1 \) and \( I_2 \) separated by a distance \( R \) 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**:
\[ \frac{F}{L} = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{2 \times 5}{2} \]

4. **Simplify the Expression**:
- The \( \pi \) terms cancel out:
\[ \frac{F}{L} = \frac{4 \times 10^{-7}}{2} \cdot \frac{10}{2} \]
- Simplifying further:
\[ \frac{F}{L} = 2 \times 10^{-7} \cdot 5 = 10 \times 10^{-7} \]

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

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

### Answer:
The correct option is 3: \( 1 \times 10^{-6} \, \text{N/m} \).

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