Answer:
[tex]F_{1}=\frac{\mu _{0}I_{1}I_{2}}{2\pi a}l+\frac{\mu _{0}I_{1}I_{3}}{2\pi a}l=\frac{\mu _{0}I_{1}}{2\pi a}l(I_{2}+I_{3})[/tex]
Explanation:
To solve this problem we can use the expression for the force between two wires with currents I1 and I2, at a distance of r.
[tex]F_{1}=F_{2}=\frac{\mu _{0}I_{1}I_{2}}{2\pi r}L[/tex]
where L is the length of the wires. Wires feel the same force and the currents are in the same direction.
For our problem the three wires the force is the same. By taking into account that the distance among them is a we have
[tex]F_{1}=F_{2}=F_{3}[/tex]
and the force on on of the wire is the sum of the effects of the others
[tex]F_{1}=\frac{\mu _{0}I_{1}I_{2}}{2\pi a}l+\frac{\mu _{0}I_{1}I_{3}}{2\pi a}l=\frac{\mu _{0}I_{1}}{2\pi a}l(I_{2}+I_{3})[/tex]
where we have taken r=a (wires are equidistant in a equilateral triangle)and L=l for our problem
I hope this is useful for you
regards
The expression for the magnitude of the force on each wire in terms of the
variables I, l, a is: [tex]\mathbf{F_{net} =\dfrac{\mu_o}{4 \pi}\Big( \dfrac{2 I^2 (L)\sqrt{3}}{a}\Big) }[/tex]
If the current of the wire is perpendicular to the plane, then the force of a wire on another wire can be expressed by using the relation:
[tex]\mathbf{F = \dfrac{\mu_o}{4 \pi}\Big( \dfrac{2 I (I) (L)}{a}\Big) }[/tex]
[tex]\mathbf{F = \dfrac{\mu_o}{4 \pi}\Big( \dfrac{2 I^2 (L)}{a}\Big) }[/tex]
Since the force-carrying the current are equilateral are to each other(at an angle of 60° to each other), then they have the same angles, and the magnitude of the net force can be computed as:
[tex]\mathbf{F_{net} = \sqrt{F^2 +F^2 +2(F)(F) \ cos 60}}[/tex]
[tex]\mathbf{F_{net} = \sqrt{2F^2 +2(F)^2 \ cos 60}}[/tex]
[tex]\mathbf{F_{net} =F \sqrt{2 +2 \ cos 60}}[/tex]
Recall that:
∴
[tex]\mathbf{F_{net} =F \sqrt{2 +2( \dfrac{1}{2})}}[/tex]
[tex]\mathbf{F_{net} =F \sqrt{(2 +1)}}[/tex]
[tex]\mathbf{F_{net} =F \sqrt{3}}[/tex]
where:
∴
[tex]\mathbf{F_{net} =\dfrac{\mu_o}{4 \pi}\Big( \dfrac{2 I^2 (L)\sqrt{3}}{a}\Big) }[/tex]
Therefore, the expression for the magnitude of the force on each wire in
terms of the variables I, l, a is: Â [tex]\mathbf{F_{net} =\dfrac{\mu_o}{4 \pi}\Big( \dfrac{2 I^2 (L)\sqrt{3}}{a}\Big) }[/tex]
Learn  more about the magnitude of the force here:
https://brainly.com/question/9774180?referrer=searchResults
