The figure shows a cross section of three parallel wires each carrying a current of 15 A. The currents in wires A and C are out of the paper, while that in wire B is into the paper. If the distance R = 5.0 mm, what is the magnitude of the force on a 4.0-m length of wire C?

0.018 N

Explanation:

From the given information;

We understood that A and C possess the same current direction, thus the force between them is said to be an attractive force.

Also, between B and C, there exist different current directions, thus, the force between them is said to be a repulsive force.

So, the force per unit length magnitude are as follows:

Similarly;

Thus, the net force acting on 4.0 - m length of the wire C is:

Thus, the magnitude of the force on a 4.0-m length of wire C = 0.018 N

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