Answer
See answers.
Work Step by Step
The center of the third wire is 5.6 mm from the left wire, center to center. The force on the left wire will repel it to the left, since the currents run in opposite directions.
Use equation 20–7 to calculate the force per unit length.
$$\frac{F_{far}}{L_{far}} =\frac{\mu_o}{2\pi}\frac{I_1I_2}{d_{far}} $$
$$ =\frac{4\pi\times10^{-7}}{2\pi}\frac{ (25.0A)(24.5A)}{0.0056m} =0.022N/m$$
The center of the third wire is 2.8 mm from the right wire, center to center. The force on the right wire will attract it to the right, since the currents run in the same direction.
Use equation 20–7 to calculate the force per unit length.
$$\frac{F_{near}}{L_{near}} =\frac{\mu_o}{2\pi}\frac{I_1I_2}{d_{near}} $$
$$ =\frac{4\pi\times10^{-7}}{2\pi}\frac{ (25.0A)(24.5A)}{0.0028m} =0.044N/m$$
