Question

The diagram below shows three parallel wires X, Y and Z placed in a horizontal plane. Wires X and Z carry current I in opposite directions. Wire Y carries a current of 3I in the direction shown and is equidistant from the other wires.

The magnitude of the force per unit length between two parallel wires placed a distance apart, each carrying a current of I is 2.0 x 10-6 N m-1. What is the direction and magnitude of the net force per unit length acting on the wire Z?

(The force per unit length is proportional to the product of the currents in the two wires and inversely proportional to the separation between them.)

Answer

Based on the information regarding the forces per unit length between two wires,

\( {F \over m} = k {I_1 I_2 \over r} \)

Based on the value of the force given,

The force between Y and Z is attractive because the currents in both wires are flowing in the same direction.

The force between X and Z is repulsive because the currents in both wires are flowing in the opposite directions.

Hence, the resultant force per unit length is 5.0 x 10^-6 N m^-1 upwards towards Y.