Magnetic Force Between Current-Carrying Wires

[David21]

Homework Statement

:
[/B]
Three parallel wires of length l each carry current i in the same direction. They’re positioned at the vertices of an equilateral triangle of side a, and oriented perpendicular to the triangle.
Find an expression for the magnitude of the force on each wire.
Express your answer in terms of the variables i, l, a, and appropriate constants.

Homework Equations

:
[/B]
F = [(permittivity constant)(current 1)(current 2)(length)] / (2pi)(distance apart)
This is the equation for the magnetic force between two wires

3. The Attempt at a Solution

u is permittivity constant
l is length
d is distance between wires

I started by trying to find the force on the middle wire:
F on wire 2 = force on 2 from 1 + force on 2 from 3
= [ (u*i1*i2*l) / (2pi*d) ] + [ (u*i3*i2*l) / (2pi*d)]

Since all three currents are the same:
= [ (u*i^2*l) / (2pi*d) ] + [ (u*i^2*l) / (2pi*d) ]
= 2(u*i^2*l) / (2pi*d) = (u*i^2*l) / (pi*d)

On my equilateral triangle, I drew the three wires pointing directly upwards from the vertices of the triangle. Since the base of the triangle is a, the distance that wire 2 is from wires 1 and 3 is a/2:
= (u*i^2*l) / (pi*(a/2)) = 2(u*i^2 *l) / (pi*a)

This is the answer I have but I have made a mistake, I just don't know where.

 

[collinsmark]

Hello David21,

Welcome to Physics Forums!

There might be a couple of issues with your approach.

Homework Statement

:
[/B]
Three parallel wires of length l each carry current i in the same direction. They’re positioned at the vertices of an equilateral triangle of side a, and oriented perpendicular to the triangle.
Find an expression for the magnitude of the force on each wire.
Express your answer in terms of the variables i, l, a, and appropriate constants.

Homework Equations

:
[/B]
F = [(permittivity constant)(current 1)(current 2)(length)] / (2pi)(distance apart)
This is the equation for the magnetic force between two wires

3. The Attempt at a Solution

u is permittivity constant
l is length
d is distance between wires

I started by trying to find the force on the middle wire:
F on wire 2 = force on 2 from 1 + force on 2 from 3
= [ (u*i1*i2*l) / (2pi*d) ] + [ (u*i3*i2*l) / (2pi*d)]

You can't add the magnitudes of the two forces together quite so simply. Remember, forces are vectors. The important thing to take home here is that the two forces acting on wire 2 are not exactly in the same direction.

You'll need to add them together as vectors, meaning you'll need to take the angles into account.

On my equilateral triangle, I drew the three wires pointing directly upwards from the vertices of the triangle. Since the base of the triangle is a, the distance that wire 2 is from wires 1 and 3 is a/2:

I'm a bit lost there. If it's a equilateral triangle, meaning all sides have the same length, and that length is $a$, then isn't a given wire's distance to either of the other wires also $a$? (Maybe I'm missing something )

 

[Estifericate]

The answer to the problem can be found by visualising an equilateral triangle in horizontal direction . By applying the formulas you stated you can obtain the solution.
First consider any of the current carrying conductor.
The force on it will be applied by the rest two and will be directed towards them .
So that the net force on the conductor at a vertex will be along the two sides.
Note that the distance between them will be 'a'.
Now adding them by vector addition, you can get the answer. ( F=i×i×I×cos30÷pi×a)
If the answer is correct let me know I will post a full solution.