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I need help in this question: Four Wires

  1. Jun 24, 2009 #1
    1. The problem statement, all variables and given/known data


    The four wires that lie at the corners of a square of side a= 3.50 cm are carrying equal currents i= 2.10 A into (+) or out of (-) the page, as shown in the picture.

    Part A: Calculate the y component of the magnetic field at the center of the square. ( I already got the answer for this one) 4.80×10-5 T

    Part B: Calculate the x-component of the force on a 1.0-cm long piece of the lower right-hand wire, due to the other three wires.

    2. Relevant equations

    mu=4*pi*a0^-7 T*m/A

    3. The attempt at a solution

    I need help in the second part. I don't know where to start.
  2. jcsd
  3. Jun 24, 2009 #2


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  4. Jun 28, 2009 #3
    Okay, so I know that B is always perpendicular to Force so B2 is zero in the y direction, F2 is zero in the x direction.
    then the superposition is:

    B total = B1y + B3y
    B total = μI/(2π0.0350)*1/√2 + μI/(2π0.0350)
    B total = 2.0485*10^-5 T

    Using the force formula,

    F = BIL
    =2.0485*10^-5 *2.1*0.01= 4.3019*10^-7 N

    I did all of this and I got the wrong answer, what am I missing??

    P.S: sorry for the delay I had a busy week.
  5. Jun 28, 2009 #4


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    What you want is the x-directed force component.

    The B-Field is a vector field, and the force component from the diagonal has a cos45 factor still needed to normalize it to x.

    Btotal = 1.2 * 10-5 + (.707)(.8 * 10-5)
  6. Jun 28, 2009 #5
    Thank you!
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