Three long straight wires are seen endon in the figure below

Three long, straight wires are seen end-on in the figure below. The distance between the wires is r = 0.250 m. Wires A and B carry current I_A = I_B = 1.72 A,and wire C carries current I_C = 3.34 A.

Assume (for example) the only forces exerted on wire A are due to wires B and C. Find the force per unit length exerted on the following. (Express your answers in vector form.)

Please express answer in vector form i and j

(b) wire B

Solution

a)
the force is,

   FAC = u_o*I_A*I_C / (2*pi*r) = (1.2566*10^-6* 1.72*3.34) / (2*pi*0.250) =4.6*10^-6 N/m
in vector form,
    FAC = 4.6*10^-6 * (cos 60 i - sin 60 j)
           = [2.3*10^-6 i - 3.98*10^-6 j] N/s

the force is,
   FAB = u_o*I_A*I_B / (2*pi*r)= (1.2566*10^-6* 1.72*1.72) / (2*pi*0.250) = 2.366*10^-6 N/m
in vector form,
FAC = 2.366*10^-6 * ( - cos 60 i - sin 60 j)
      = -1.183*10^-6 i - 2.049*10^-6 j N/s

hence,

    FA =[2.3*10^-6 i - 3.98*10^-6 j] -1.183*10^-6 i - 2.049*10^-6 j
         = (1.117*10^-6i - 6.029*10^-6 j) N/m

magnitude: 6.13*10^-6 N/m

b)
the force is,

   F_BC = u_o*I_A*I_C / (2*pi*r) = (1.2566*10^-6* 1.72*3.34) / (2*pi*0.250) = 4.6*10^-6 N/m

in vector form,
   F_BC = 4.6*10^-6 i N/m

the force is,

    FBA = u_o*I_A*I_B / (2*pi*r) = (1.2566*10^-6* 1.72*1.72) / (2*pi*0.250) = 2.366*10^-6 N/m

in vector form,
    FBA = 2.366*10^-6 * ( cos 60 i + sin 60 j)
           = 1.183*10^-6 i + 2.049*10^-6 j N/s

the force is,

    FB = 4.6*10^-6 i + 1.183*10^-6 i + 2.049*10^-6 j
         = (5.783*10^-6i + 2.049*10^-6 j) N/m

magnitude: 6.135*10^-6 N/m