Question

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

A) 1ᴇ−3, repulsive
B) 1ᴇ−3, attractive
C) 4ᴇ−5, repulsive
D) 4ᴇ−5, attractive
E) none of these

Answer 1

The force per unit length between two parallel wires carrying currents in opposite directions can be calculated using Ampère's force law. With given currents of 2A and 4A and a separation of 4cm, the force per unit length is 4×-5 N/m and is repulsive, making option C the correct answer.

Step-by-step explanation:

The student has asked about the magnetic force per unit length between two parallel current-carrying wires. According to Ampère's force law, the force per unit length between two parallel currents I1 and I2, separated by a distance r, is given by the formula FL = (µ0/2π) * (I1 * I2)/r, where µ0 is the permeability of free space (µ0 = 4π x 10-7 T·m/A).

Using the given values: I1 = 2 A, I2 = 4 A, and r = 4 cm = 0.04 m, the force per unit length is calculated as:

FL = (4π x 10-7 T·m/A) / (2π) * (2 A * 4 A) / 0.04 m

This simplifies to FL = 4 x 10-7 T·m/A * 8 A2 / 0.04 m = 8 x 10-5 N/m.

Since the currents are in opposite directions, the force between the wires will be repulsive. Therefore, the correct answer is (C) 4×-5, repulsive.