a) The forces on the wires are repulsive. This is because when the right hand is wrapped around the wire in the direction of the current, the thumb points away from the other wire. This means that the two currents are pushing away from each other, creating a repulsive force.
b) The magnetic field magnitude that each produces at the other's location is B = μ_0I/2πd, where μ_0 is the permeability of free space, I is the current, and d is the distance between the wires. Therefore, the magnetic field magnitude is B = (4π x 10^-7 Tm/A)(3.0 A)/(2π x 0.1 m) = 7.85 x 10^-5 T.
c) The magnitude of the force per unit length they exert on each other is F/L = B^2/μ_0. Therefore, the magnitude of the force per unit length is F/L = (7.85 x 10^-5 T)^2/(4π x 10^-7 Tm/A) = 0.62 N/m.
Two long, straight, parallel wires 10 cm apart carry currents in opposite directions. (a) Use the right-hand source and force rules to determine whether the forces on the wires are (1) attractive or (2) repulsive. Show your reasoning. (b) If the wires carry equal currents of 3.0 A, what is the magnetic field magnitude that each produces at the other's location? (c) use the result of part (b) to determine the magnitude of the force per unit length they exert on each other.
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