Two long parallel wires are at a distance of 1 m. Both of them carry 1A of current. The force of attraction per unit length between the two wires is

Options

A. `2xx10^(-7)N//m`
B. `2xx10^(8)N//m`
C. `5xx10^(8)N//m`
D. `10^(7)N//m`

Answer

To solve the problem of finding the force of attraction per unit length between two long parallel wires carrying current, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Given Data**:
- Distance between the wires (R) = 1 m
- Current in wire 1 (I1) = 1 A
- Current in wire 2 (I2) = 1 A

2. **Use the Formula for Force per Unit Length**:
The formula for the force of attraction 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:
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

3. **Substitute the Values into the Formula**:
- Substitute \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)
- Substitute \( I_1 = 1 \, \text{A} \)
- Substitute \( I_2 = 1 \, \text{A} \)
- Substitute \( R = 1 \, \text{m} \)

Thus, the equation becomes:
\[ \frac{F}{L} = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{1 \cdot 1}{1} \]

4. **Simplify the Equation**:
- The \( \pi \) terms cancel out:
\[ \frac{F}{L} = \frac{4 \times 10^{-7}}{2} = 2 \times 10^{-7} \, \text{N/m} \]

5. **Conclusion**:
The force of attraction per unit length between the two wires is:
\[ \frac{F}{L} = 2 \times 10^{-7} \, \text{N/m} \]

### Final Answer:
The force of attraction per unit length between the two wires is \( 2 \times 10^{-7} \, \text{N/m} \).

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