Two long straight parallel conductors separated by a distance of `0.5m` carry currents of `5A` and `8A` in the same direction. The force per unit length experienced by each other is

Options

A. `1.6xx10^(-5)N` (attractive)
B. `1.6xx10^(-5)N` (repulsive)
C. `16xx10^(-5)N` (attractive)
D. `16xx10^(-5)N` (repulsive)

Answer

To solve the problem, we need to calculate the force per unit length experienced by two long straight parallel conductors carrying currents in the same direction. We will use the formula for the magnetic force between two parallel conductors.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Current in the first conductor, \( I_1 = 5 \, \text{A} \)
- Current in the second conductor, \( I_2 = 8 \, \text{A} \)
- Distance between the conductors, \( d = 0.5 \, \text{m} \)

2. **Use the Formula for Force per Unit Length:**
The formula for the force per unit length \( F/L \) between two parallel conductors is given by:
\[ \frac{F}{L} = \frac{\mu_0}{4\pi} \cdot \frac{I_1 I_2}{d} \]
where \( \mu_0 \) (the permeability of free space) is approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

3. **Substituting the Values:**
First, we can simplify \( \frac{\mu_0}{4\pi} \):
\[ \frac{\mu_0}{4\pi} = 10^{-7} \, \text{T m/A} \]
Now substituting the values into the formula:
\[ \frac{F}{L} = 10^{-7} \cdot \frac{5 \cdot 8}{0.5} \]

4. **Calculating the Product of Currents:**
\[ 5 \cdot 8 = 40 \]

5. **Calculating the Division by Distance:**
\[ \frac{40}{0.5} = 80 \]

6. **Putting it All Together:**
Now substituting back into the equation:
\[ \frac{F}{L} = 10^{-7} \cdot 80 = 8 \times 10^{-6} \, \text{N/m} \]

7. **Expressing in Standard Form:**
We can express \( 8 \times 10^{-6} \) as:
\[ 1.6 \times 10^{-5} \, \text{N/m} \]

8. **Direction of the Force:**
Since both currents are in the same direction, the force between the conductors is attractive.

### Final Answer:
The force per unit length experienced by each conductor is:
\[ 1.6 \times 10^{-5} \, \text{N/m} \, \text{(attractive)} \]