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Magnetism (Current-Carrying Wires)

  1. Feb 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Two straight wires, each with a resistance of 0.170 ohm and a length of 3.90m, are lying parallel to each other on a smooth horizontal table. Their ends are connected by identical, non-conducting, light springs, each spring having an unstretched length of 1.08 cm. A wire of negligible resistance connects the wires at one end. When a switch is closed to connect a battery with a voltage of 49.0 V between the other ends of the wires, the wires move apart and come to rest with a separation of 1.57 cm

    Question: Find the force constants of the springs.




    2. Relevant equations

    F=-kx
    (F/L)=(u0I1I2)/2Pi r




    3. The attempt at a solution

    I tried to solve using the equations above for K, and does not work.
     
  2. jcsd
  3. Feb 23, 2010 #2
    Did you modify the k in the spring equation to take into account the presence of both springs in a parallel arrangement?
     
  4. Feb 24, 2010 #3
    Yes I did. This is what I got:
    F=-2kx
    (F/L)=(uI^2)/(2pi a)

    k=(uI^2L)/(4piax)

    I tried solving this, but got wrong answers. What I did:
    u=1.26e-6 (Constant given to use)
    I=144.1176 (Combining resistors in series and finding current)
    L=3.90 (Length of the wire)
    a=0.0157 (Seperation of the wires)
    x=0.0049 (stretch of the spring)

    Thats what I use, and I got 106 N/m. Can you verify that please?
     
  5. Feb 24, 2010 #4
    due to passing of current the magnetic field is created in the rods hence we know the formula F=vlb from that u can find the force and from F=kx u can find the spring constant
     
  6. Feb 24, 2010 #5
    So what I have is correct? Can you check?
     
  7. Feb 24, 2010 #6

    Redbelly98

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    I agree, good job.

    Charanjit already knew that!
     
  8. Feb 24, 2010 #7
    Thank you. :)
     
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