Three infinite straight wires are fixed in place and aligned parallel to the z-a

Question

Three infinite straight wires are fixed in place and aligned parallel to the z-axis as shown. The wire at (x,y) = (-14 cm, 0) carries current I1 = 2.9 A in the negative z-direction. The wire at (x,y) = (14 cm, 0) carries current I2 = 0.8 A in the positive z-direction. The wire at (x,y) = (0, 24.2 cm) carries current I3 = 7.5 A in the positive z-direction.

1)What is Bx(0,0), the x-component of the magnetic field produced by these three wires at the origin?

2)What is By(0,0), the y-component of the magnetic field produced by these three wires at the origin?

3)What is Fx(1), the x-component of the force exerted on a one meter length of the wire carrying current I1?

4)What is Fy(1), the y-component of the force exerted on a one meter length of the wire carrying current I1?

5)What is Fx(2), the x-component of the force exerted on a one meter length of the wire carrying current I2?

Explanation / Answer

1) H = 0.121 m

Bx = (uo*I3)/(2pi*r) = (2*10^-7*7.5)/(0.121) = 1.24*10^-5 T (+x-direction)

2) H = 0.121 m

Bx = (uo/2pi*r)(I1 + I2) = (2*10^-7*3.7)/(0.07) = 1.06*10^-5 T (-y-direction)

3) F = I(L x B)

Fx1 = F21 + F31*cos(theta)

= [(I1*I2*uo*L)/(2pi*d)] + [(I1*I3*uo*L)*cos(theta)/(2pi*d)]

= [(I1*uo*L)/(2pi*d)]*[I2 + I3*cos(theta)]

4) Fy1 = (I1*I3*uo*L*sin60)/(2pi*d)

5) Fy2 = F12 – F32*cos(theta)

     = [(I1*I2*uo*L)/(2pi*d)] – [(I2*I3*uo*L*cos(theta))/(2pi*d)]

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