1) Magnitude of the force:
The magnetic field generated by a current-carrying wire is
[tex]B= \frac{\mu_0I}{2 \pi r} [/tex]
where
[tex]\mu_0[/tex] is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated
Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance
[tex]r=40.0 cm=0.40 m[/tex],
which is the distance at which the other wire is located:
[tex]B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T [/tex]
Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
[tex]F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N[/tex]
2) direction of the force:
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)
Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive
A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.
