Final answer:
The current in the wires when the force per meter between them is 0.225 N/m and they are separated by 2.00 cm is approximately 225 A, corresponding to option (b). The force between the wires is attractive.
Step-by-step explanation:
The question pertains to the magnetic force between two parallel conductors, which is a concept in Physics. To find the current in the wires when the force per meter between them is 0.225 N/m and they are separated by 2.00 cm, we can use the formula derived from Ampere's Law for the magnetic force between two parallel currents:
F/L = (μ₀/2π) * (I1 * I2 / d)
Where F/L is the force per unit length, μ₀ is the permeability of free space (μ₀ = 4π * 10^-7 T·m/A), I1 and I2 are the currents through the wires, and d is the separation between the wires. Since the wires are part of the same circuit and thus carry the same current, we can square one of the currents and rewrite the formula as:
F/L = (μ₀/2π) * (I^2 / d)
After rearranging the formula to solve for I and plugging in the values, we get:
I = √((F/L) * (2π * d) / (μ₀))
Upon calculation, this yields a current I of approximately 225 A, which corresponds to option (b) 225.0 A. Additionally, the force between the wires is attractive as the currents are in the same direction and the force tends to bring the wires together.