Final answer:
The direction of the current in the other wire is opposite to that of the first wire, and its magnitude is 2000 A, resulting in an attractive force between the wires.
Step-by-step explanation:
To determine the direction and magnitude of the current in the other wire, we need to use Ampere's force law, which describes the force between two parallel currents. According to this law, the magnitude of the force per unit length between two parallel wires separated by a distance r is given by F/(I1I2L) = (μ0)/(2πr), where I1 and I2 are the currents, L is the length of the wire, and μ0 is the permeability of free space. Since we have a repulsive force, the currents must be flowing in opposite directions. Plugging in the values, we get:
4.00 N / (1000 A × I2 × 2.50 m) = (4π × 10-7 T·m/A) / (2π × 0.05 m)
Simplifying, we find that I2 = 2000 A. Therefore, the magnitude of the current is 2000 A and since the force is repulsive, the currents are flowing in opposite directions, making the correct answer (a) 2000 A, attractive.