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Magnetic force on wires -direction of B

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations

    3. 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.
  2. jcsd
  3. Nov 16, 2008 #2


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    Where does the cos(60) come from?
  4. Nov 16, 2008 #3
    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.
  5. Nov 17, 2008 #4


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