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 currents, we can use the formula for the magnetic force between two parallel currents:

\[ F/L = \frac{\mu_0}{2\pi} \cdot \frac{I_1 I_2}{r} \]

Where:
- \( F/L \) is the force per unit length between the wires,
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \),
- \( I_1 \) and \( I_2 \) are the currents in the wires (in Amperes),
- \( r \) is the distance between the wires (in meters).

### Step-by-Step Solution:

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

2. **Substitute the Values into the Formula**:
\[ F/L = \frac{\mu_0}{2\pi} \cdot \frac{I_1 I_2}{r} \]
Substituting the values:
\[ F/L = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{1 \cdot 1}{1} \]

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

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