Answer:
[tex]2.5\cdot 10^{-5}N[/tex]
Explanation:
The force per unit length between two current-carrying wires is given by:
[tex]\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}[/tex]
where
[tex]\mu_0[/tex] is the vacuum permeability
[tex]I_1, I_2[/tex] are the two currents
r is the separation between the wires
In this problem, we have
[tex]I_1 =3.55 A[/tex] is the current in the first wire
[tex]I_2 = 2.75 A[/tex] is the current in the second wire
[tex]r=78 mm=0.078 m[/tex] is the separation between the wires
Substituting into the equation, we find
[tex]\frac{F}{L}=\frac{(4\pi \cdot 10^{-7} H/m)(3.55 A)(2.75 A)}{2\pi (0.078 m)}=2.5\cdot 10^{-5}N[/tex]