Two current-carrying wires are exactly parallel to one another and both carry 2.5a of current. the two wires are separated by a distance of 15cm. the current in wire 1 moves down and the current in wire 2 also moves down. what is the magnitude of the magnetic force per unit length caused by wire 1 on wire 2. what is the direction of the magnetic force caused by wire 2 on wire 1.

1) Magnitude per unit length:

The magnetic force per unit length between two current-carrying wires is given by:

where

is the vacuum permeability

is the current in each wire

is the distance between the two wires

Substituting the numbers into the equation, we find

2)  direction of the force: attractive

First of all, let's analyze what is the direction of the magnetic field produced by wire 2 at the location of wire 1. Assume that wire 2 is on the left of wire 1. The direction of the current for wire 2 is down, so by using the right-hand rule, we see that the direction of the magnetic field at the location of wire 2 is south.

Now we apply the right-hand rule on wire 2, to find the direction of the force:

- current: down (index finger)

- magnetic field: south (middle finger)

- force: to the left (thumb)

So, the force exerted by wire 2 on wire 1 is towards wire 2 (attractive force)