Magnetic force on wires -direction of B

[bcjochim07]

Homework Statement

Three wires have linear density of 50 g/m. They each carry the same current. The bottom two are 4 cm apart and carry currents into the page. What current I will allow the third wire to float so as to form an equilateral triangle with the other two?

Homework Equations

The Attempt at a Solution

I think I am really close, but I'm just concerned because I'm having some problems with magnetic field directions and components.

If I start with the magnetic field, I think that the y- components of the magnetic fields of the bottom two wires cancel. So the net B-field is:

2* cos(60) * (1.257*10-6)I / 2pi*.04 = (5*10^-6)I

F=IlBsin(theta) for the force on a current carrying wire

F= I^2 * (5*10-6) * l = (9.8)(.05)*l
I= 313A.

This isn't quite correct, however, I find that if I turn to forces first, and seeing that the bottom two have forces where the x-components cancel.

F = I^2 * l * (5*10-6)*sin(60) = .05*9.8l
I = 237 A This is correct. But I am confused by why the two give different answers.

 

[Redbelly98]

2* cos(60) * (1.257*10-6)I / 2pi*.04 = (5*10^-6)I

Where does the cos(60) come from?

 

[bcjochim07]

Isn't the magnetic fields from the bottom wires at an angle of 60 degrees to horizontal? So with cos(60) I am getting the x components from those mag. fields and adding them together.

 

[Redbelly98]

Isn't the magnetic fields from the bottom wires at an angle of 60 degrees to horizontal?

I'm getting that B is 60 degrees from vertical.