Concept:
The force per unit length between two long parallel current-carrying conductors separated by distance \(r\) is:
\[
\frac{F}{l} = \frac{\mu_0}{2\pi}\frac{I_1 I_2}{r}
\]
where:
\(I_1, I_2\) are the currents,
\(r\) is the separation between the wires,
Direction of force depends on direction of currents (same direction: attraction, opposite direction: repulsion).
Step 1: Force on the middle wire due to the left wire.
Left wire current \(= 3\text{ A}\) (upward)
Middle wire current \(= 1\text{ A}\) (downward)
Since currents are in opposite directions, the force is repulsive.
Distance between left and middle wire:
\[
r_1 = 3\text{ cm} = 0.03\text{ m}
\]
\[
\frac{F_1}{l} = \frac{\mu_0}{2\pi}\frac{3 \times 1}{0.03}
\]
Step 2: Force on the middle wire due to the right wire.
Right wire current \(= 2\text{ A}\) (downward)
Middle wire current \(= 1\text{ A}\) (downward)
Since currents are in the same direction, the force is attractive.
Distance between middle and right wire:
\[
r_2 = 2\text{ cm} = 0.02\text{ m}
\]
\[
\frac{F_2}{l} = \frac{\mu_0}{2\pi}\frac{2 \times 1}{0.02}
\]
Step 3: Calculate net force per unit length.
\[
\frac{F_1}{l} = \frac{2\times 10^{-7}\times 3}{0.03} = 2\times 10^{-5}\text{ N/m}
\]
\[
\frac{F_2}{l} = \frac{2\times 10^{-7}\times 2}{0.02} = 2\times 10^{-5}\text{ N/m}
\]
Both forces act in the same direction, hence they add:
\[
\frac{F_{\text{net}}}{l} = 4\times 10^{-5}\text{ N/m}
\]
Step 4: Force on \(15\) cm of the middle wire.
\[
l = 15\text{ cm} = 0.15\text{ m}
\]
\[
F = 4\times 10^{-5} \times 0.15 = 6\times 10^{-6}\text{ N}
\]
\[
\Rightarrow F = 6~\mu\text{N}
\]
Rounding as per options:
\[
\boxed{F \approx 7~\mu\text{N}}
\]
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