Three coplanar parallel wires, each carrying a current of 10 A in...

Let the three wires be 1, 2, and 3. The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $r$ is given by $F = \dfrac{\mu_0 I_1 I_2}{2 \pi r}$.

For the middle wire (wire 2), the forces exerted by wire 1 and wire 3 are equal in magnitude and opposite in direction because the currents and distances are the same. Thus, the net force on the middle wire is zero.

For wire 1, both wire 2 and wire 3 exert attractive forces in the same direction.

$F_1 = F_{12} + F_{13} = \dfrac{\mu_0 I^2}{2 \pi d} + \dfrac{\mu_0 I^2}{2 \pi (2d)} = \dfrac{3 \mu_0 I^2}{4 \pi d}$

Substituting the given values $I = 10 \text{ A}$ and $d = 0.05 \text{ m}$:

$F_1 = \dfrac{3 \times 4 \pi \times 10^{-7} \times (10)^2}{4 \pi \times 0.05} = \dfrac{3 \times 10^{-5}}{0.05} = 6.0 \times 10^{-4} \text{ N m}^{-1}$

By symmetry, the force per unit length on wire 3 is also $6.0 \times 10^{-4} \text{ N m}^{-1}$.