Two long and parallel straight wires A and B are carrying currents of 4A and 7A in the same directions are separated by a distance of 5cm. The force acting on a 8cm section of wire A is

Options

A. `3 xx 10^(-6)N`
B. `6 xx 10^(-6)N`
C. `9 xx 10^(-6)N`
D. `12 xx 10^(-6)N`

Explanation

To find the force acting on an 8 cm section of wire A, we can use the formula for the magnetic force per unit length between two parallel current-carrying wires. The formula is given by:

\[ F/L = \frac{{\mu_0 \cdot I_A \cdot I_B}}{{2\pi d}} \]

where:
- \( F \) is the force between the wires,
- \( L \) is the length of the wire,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( I_A \) and \( I_B \) are the currents in wires A and B respectively,
- \( d \) is the distance between the two wires.

### Step-by-Step Solution:

1. **Identify the given values:**
- Current in wire A, \( I_A = 4 \, \text{A} \)
- Current in wire B, \( I_B = 7 \, \text{A} \)
- Distance between the wires, \( d = 5 \, \text{cm} = 5 \times 10^{-2} \, \text{m} \)
- Length of wire A for which we need to calculate the force, \( L = 8 \, \text{cm} = 8 \times 10^{-2} \, \text{m} \)

2. **Substitute the values into the force per unit length formula:**
\[ F/L = \frac{{\mu_0 \cdot I_A \cdot I_B}}{{2\pi d}} = \frac{{4\pi \times 10^{-7} \cdot 4 \cdot 7}}{{2\pi \cdot 5 \times 10^{-2}}} \]

3. **Simplify the equation:**
- The \( \pi \) cancels out:
\[ F/L = \frac{{4 \cdot 4 \cdot 7 \cdot 10^{-7}}}{{2 \cdot 5 \times 10^{-2}}} \]

4. **Calculate the force per unit length:**
\[ F/L = \frac{{112 \times 10^{-7}}}{{10}} = 11.2 \times 10^{-7} \, \text{N/m} \]

5. **Calculate the total force on the 8 cm section:**
\[ F = (F/L) \cdot L = (11.2 \times 10^{-7}) \cdot (8 \times 10^{-2}) = 8.96 \times 10^{-8} \, \text{N} \]

6. **Final result:**
\[ F = 8.96 \times 10^{-6} \, \text{N} \]

### Conclusion:
The force acting on the 8 cm section of wire A is approximately \( 8.96 \times 10^{-6} \, \text{N} \).