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Wires and Force

  1. Jul 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A rectangular loop of wire lies in the same plane as a straight wire, as shown http://i30.tinypic.com/347fupw.jpg". There is a current of 2.5 Amperes in both wires. Determine the magnitude and direction of the net force on the loop.

    2. Relevant equations

    Not sure. I am pretty lost. do i use the equation for the magnetic field of a straight wire = (μ_0/4pi)*(2I/r)?

    3. The attempt at a solution

    i am thinking find the electric field of the wire to find what the field will be on each point on the loop and from there find the force on each point on the loop. then, maybe i can use an integral to add all the forces on the loop to find the net force?

    as you can tell, I am very lost, and a little behind in class. If you don't want to answer the problem (with explanation), maybe you could point me in the right direction. any help is appreciated.

    thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Jul 27, 2009 #2
    In order to obtain both the magnitude and direction of the force on a charge, q1 at position , experiencing a field due to the presence of another charge, q2 at position 'r2, the full vector form of Coulomb's law is required.

    The presence of a loop requires the formula above. Good luck and your welcome.
     
  4. Jul 27, 2009 #3

    rl.bhat

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    Homework Helper

    Force on a current carrying conductor in a magnetic field is given by
    F = I*BXL or F = I*B*L*sinθ. Ιn this problem θ =90 degrees.
     
  5. Jul 27, 2009 #4
    thanks guys, but so to find the net force on the loop, do i need to find both magnetic forces and electric forces and combine them? i can see now how to find the magnetic force, but from the information given, im not sure there is an electric force?
     
  6. Jul 27, 2009 #5

    rl.bhat

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    There is no electric force between the current carrying conductors.
     
  7. Jul 27, 2009 #6
    o ok...so this is what i have: there are going to be two forces, one for the parallel wires 3cm apart (attracting), and the one for the parallel wires 8 cm apart (repulsing). the first force is equal to I*L*sin90*B = [2.5 amps]*[sin90]*[10^-7(telsa*m^2)/amp*m]*[2(2.5amps)/.03m]*L. I would do the same for the parallel wires farther apart, except use the distance .08m.
    The thing is I am not sure how to get the measurement for L, since it is not on the diagram and the diagram is not set to scale.
     
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