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Four very long, current carrying wires in the same plane

  1. Mar 19, 2013 #1
    1. The problem statement, all variables and given/known data
    Four very long, current-carrying wires in the same plane intersect to form a square with side lengths of 31.0 cm, as shown in the figure (Figure 1) . The currents running through the wires are 8.0 A, 20.0 A, 10.0 A, and I.

    Find the magnitude of the current that will make the magnetic field at the center of the square equal to zero.

    Picture here:
    http://session.masteringphysics.com/problemAsset/1003626/14/yf_Figure_28_35.jpg

    2. Relevant equations

    B = μ0I/2πr

    3. The attempt at a solution

    I know the steps and solution, but what I'm not clear about is how the right hand rule is used to determine the direction of the magnetic field.

    Where exactly do you position your hand? On top of the wire? To the left of it? To the right of it? How much do you curl? All the way around?

    Here is the solution:

    μ0I/2πr where r = distance of B from the wire.

    r = .5 ( 31cm) = .155m
    μ0=4π*10^-7

    ƩB = 0 = μ0/2πr * (I - 10 - 8 + 20)
    0 = I - 10 - 8 + 20 = I + 2
    I = -2A

    Magnitude of I = abs(I) = 2A

    So using the right hand rule, why is 20 positive whereas 10 and 8 are negative? Assuming that into the page/monitor is negative, shouldn't the B on the 20A also go into the page/monitor via RHR?
     
  2. jcsd
  3. Mar 19, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi penguinnnnnx5! :smile:
    your thumb points along the current

    your fingers curl round the wire in the same direction that the circular magnetic field lines curl round the wire :wink:
    the point is to the left of the 20 but to the right of the 10 and 8

    so your fingers are coming out of the page through that point for the 20, but going into the page for the 10 and 8 :smile:
     
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