The force per unit length between the two wires is [tex]6.0\cdot 10^{-5} N/m[/tex]
Explanation:
The magnitude of the force per unit length exerted 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 = 4\pi \cdot 10^{-7} Tm/A[/tex] is the vacuum permeability
[tex]I_1, I_2[/tex] are the currents in the two wires
r is the separation between the two wires
For the wires in this problem, we have
[tex]I_1 = 1.75 A[/tex]
[tex]I_2 = 3.45 A[/tex]
r = 2.00 cm = 0.02 m
Substituting into the equation, we find
[tex]\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(1.75)(3.45)}{2\pi (0.02)}=6.0\cdot 10^{-5} N/m[/tex]
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