Question

Two parallel wires 4 cm apart carry currents of 2 A and 4 A respectively in the same direction. The force per unit length in N/m of one wire on the other is:

A. 1 × 10−3 repulsive
B. 1 × 10−3 attractive
C. 4 × 10−5 repulsive
D. 4 × 10−5 attractive
E. none of these

Answer 1

The force per unit length between two parallel wires carrying currents in the same direction is attractive and can be calculated using Ampère's force law. For the wires carrying 2 A and 4 A currents and being 4 cm apart, the force per unit length is 4 × 10-5 N/m.

Step-by-step explanation:

The student's question addresses the magnetic force exerted between two parallel wires carrying electric currents. According to Ampère's force law, the force per unit length between two parallel currents I1 and I2, separated by a distance r, in a medium of permeability µ, is given by the formula F/L = (µ/2π) * (I1*I2/r), where F is the force and L is the length of the wire over which the force is measured. For currents in the same direction, the force is attractive, and for currents in opposite directions, the force is repulsive.

In this case, the two wires, one carrying 2 A and the other 4 A, are 4 cm apart (0.04 m) and the currents are in the same direction. Substituting these values into the formula and using the magnetic constant (µ0 = 4π x 10-7 T·m/A), the force per unit length is calculated as follows:

F/L = (4π x 10-7 T·m/A / 2π) × (2 A × 4 A / 0.04 m) = (2 x 10-7 N/A2) × (8 A2/0.04 m) = 4 x 10-5 N/m

Because the currents are in the same direction, the force is attractive. Therefore, the correct answer is D. 4 × 10-5 attractive.