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What current will allow the wire to float ?

  1. Oct 22, 2007 #1
    What current will allow the wire to "float"?

    The linked image is a cross section through three long wires with linear mass density 50 g/m. They each carry equal currents in the directions shown. The lower two wires are 4.0 cm apart and are attached to a table. What current I will allow the upper wire to "float" so as to form an equilateral triangle with the lower wires?

    The force that the upper wire feels is a combined force from the two other wires and the normal force of gravity. I am just not sure how to calculate the normal force of gravity with the linear mass density. Also the current that is needed will have to produce a magnetic field that combined equals the force downward on the wire from gravity, correct?

    The equation for the force between two wires carrying identical currents with an unknown length is
    F/L= ([tex]\mu[/tex]I)/([tex]\mu[/tex]r)

    For gravity

    F= m*9.8m/s

    i have a linear mass density of 50g/m.. I can take and divide both sides by L to get

    F/L = (m*9.8m/s)/L but I'm afraid that doesn't help me.. Hmmm. Ill keep trying. Please any hints are helpful
     
    Last edited: Oct 22, 2007
  2. jcsd
  3. Oct 22, 2007 #2
    Please Help me out. I am missing something but i cant seem to connect the dots
     
  4. Oct 22, 2007 #3

    mgb_phys

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    If you have a linear density of M kg/m then you have a force of 9.8M N/m.
    So F=9.8*0.05 N/m= 0.49N/m
     
  5. Oct 23, 2007 #4
    so then i can set that force equal to the force between the wires and solve for I.
    I get a value of 49kA when i do this.. this seems like a lot but it would also seem that an enormous amount of current would be needed to overcome the force of gravity. let me know what you think.
     
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