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Two long parallel wires carry currents I1 and I2 and are separated by a distance r. What is the force per unit length between them?

AF/l = mu0 I1 I2 / (2 pi r)
BF/l = mu0 I1 I2 / (2 r)
CF/l = 2 pi r / (mu0 I1 I2)
DF/l = mu0 I1 I2 r / (2 pi)
Answer & Solution
Correct answer: A. F/l = mu0 I1 I2 / (2 pi r)
1. Wire 1 creates a field at wire 2 of magnitude mu0 I1 / (2 pi r). 2. That field is uniform along wire 2 and perpendicular to it, so the force on wire 2 is I2 l B1. 3. Substituting the field and dividing by the length gives F/l = mu0 I1 I2 / (2 pi r). 4. By the third law of Newton, each wire feels the same size of force. 5. Putting r on top would make wires far apart push harder than wires close together, which is wrong. _Source: OpenStax College Physics (CC BY 4.0), Ch 22 "Magnetism", section 22.10 Magnetic Force between Two Parallel Conductors_
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