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Current Between Long Straight Conducting Wires

  1. Mar 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Two identical long straight conducting wires with a mass per unit length of 25.0 g/m are resting parallel to each other on a table. The wires are separated by 2.5 mm and are carrying currents in opposite directions.

    a) If the coefficient of static friction between the wires and the table is 0.035, what minimum current is necessary to make the wires start to move?
    b) Do the wires move closer together or farther apart?

    2. Relevant equations
    F(b) = ILB
    B = muo(0)I/2pi(r)
    F(sf) = muo(s)F(n)

    3. The attempt at a solution
    Okay, so I realized that for the wires to start moving, the force due to the magnetic field has to equal (match) the force of friction, so I equate the two:

    F(b) = ILB = muo(s)F(n)

    Then, I realize that F(n) = mg

    ILB = muo(s)mg

    If you divide out the length, you can get the mass per unit length, so you get:

    IB = 0.025muo(s)g
    I = 0.025muo(s)g / B

    I know the equation to solve for the magnetic field, but at which point am I solving it for so I know the radius to use? You can't really use the center between the wires because the currents are in opposite direction and they are equal, so in the center of the two wires, the magnetic field is 0. Any ideas?

    Thank you!
  2. jcsd
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