1) Magnitude per unit length: [tex]8.3\cdot 10^{-6} N[/tex]
The magnetic force per unit length between two current-carrying wires is given by:
[tex]\frac{F}{\Delta L}=\frac{\mu_0 I_1 I_2}{2 \pi r}[/tex]
where
[tex]\mu_0 = 4\pi \cdot 10^{-7} Tm/A[/tex] is the vacuum permeability
[tex]I_1 =I_2 =2.5 A[/tex] is the current in each wire
[tex]r=15 cm=0.15 m[/tex] is the distance between the two wires
Substituting the numbers into the equation, we find
[tex]\frac{F}{\Delta L}=\frac{(4\pi \cdot 10^{-7} Tm/A)(2.5 A)(2.5 A)}{2 \pi (0.15 m)}=8.3\cdot 10^{-6} N[/tex]
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)
