Two long current-carrying wires run parallel to each other and are separated by a distance of 4.00 cm. If the current in one wire is 1.50 A and the current in the other wire is 3.15 A running in the opposite direction, determine the magnitude and direction of the force per unit length the wires exert on each other.

The magnitude of force per unit length the wires exert on each other is 2.36 × 10^-5 N/m.

Since the wires carry currents running in the opposite direction, the force is therefore repulsive.

Explanation: Please see the attachments below

The magnitude of the force F =

The direction of the force is in opposite direction.

Explanation:

The expression for force per unit length between two parallel wires carrying current and can be written as:

permeability constant =

a = distance between the wires = 4.00 cm = 0.04 cm

replacing our values into the above equation; we have:

Hence, If the current flows in the same direction, then , the force is said to be attractive (+ve) . However, if the direction of the current flow is opposite; then the force is said to be repulsive (-ve).

From the question given, the current flows in opposite direction in the wires, thus the force is said to be repulsive.

Thus ; F =

where are the vectors ?

the inner core.

good luck: )

klicker