Two long parallel copper wires carry current of 5A each in opposite directions. If the wires are separated by a distance distance of 0.5m, then the force per unit length between the two wires is

Options

A. `10^(-5)N//m`, attractive
B. `10^(-5) N//m` repulsive
C. `2 xx 10^(-5) N//m`, attractive
D. `2 xx 10^(-5) N//m`, repulsive

Answer

To solve the problem of finding the force per unit length between two long parallel copper wires carrying currents in opposite directions, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have two long parallel wires, both carrying a current of 5 A.
- The currents are in opposite directions.
- The distance between the wires is 0.5 m.

2. **Identify the Formula for Force Between Wires**:
- The force per unit length \( F/L \) between two parallel wires carrying currents \( I_1 \) and \( I_2 \) is given by the formula:
\[ F/L = \frac{\mu_0 I_1 I_2}{2 \pi d} \]
where \( \mu_0 \) is the permeability of free space, \( d \) is the distance between the wires, and \( I_1 \) and \( I_2 \) are the currents in the wires.

3. **Substitute the Known Values**:
- Given:
- \( I_1 = 5 \, \text{A} \)
- \( I_2 = 5 \, \text{A} \)
- \( d = 0.5 \, \text{m} \)
- \( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \)

Substitute these values into the formula:
\[ F/L = \frac{(4 \pi \times 10^{-7}) \times (5) \times (5)}{2 \pi \times 0.5} \]

4. **Simplify the Expression**:
- First, simplify the numerator:
\[ 4 \pi \times 10^{-7} \times 5 \times 5 = 100 \pi \times 10^{-7} \]
- Now, simplify the denominator:
\[ 2 \pi \times 0.5 = \pi \]
- Now, substitute back into the equation:
\[ F/L = \frac{100 \pi \times 10^{-7}}{\pi} \]

5. **Cancel Out \(\pi\)**:
- The \(\pi\) in the numerator and denominator cancels out:
\[ F/L = 100 \times 10^{-7} \]

6. **Final Calculation**:
- Simplifying gives:
\[ F/L = 10^{-5} \, \text{N/m} \]

7. **Determine the Nature of the Force**:
- Since the currents are in opposite directions, the force between the wires is repulsive.

### Final Answer:
The force per unit length between the two wires is:
\[ F/L = 10^{-5} \, \text{N/m} \quad \text{(repulsive)} \]