Three infinite straight wires are fixed in place and aligned

Three infinite straight wires are fixed in place and aligned parallel to the z-axis as shown in the left figure below. The wire at (z,y) = (-15.5, 0) cm cam s current i-3.4 A in the negative z- direction. The wire at (z,y)-(15.5,0) cm carries curren 32 0.5 A in the positive z-direction. The wire at (z,v) = (0,268) an carries current is = 5.2 A in the positive z-direction. la Figure 1: (a) What are the r- and y-components of the magneic field produced by these three wires (b) What are the z- and y-components of the force exerted on a one meter length of the wire (e) What is the a-component of the force exerted n a one meter lenjth of the wire carrying (d) A small square loop with area 1.0 cm2 and current i = 1.0 A is now placed at the origin at the origin? (4pt) carrying current i? (4pt) current i2? (2pt) with the plane of the loop perpendicular to the zy-plane and inclined at 30° with respect to the z-axis as shown in the middle figure above. What are the z- and y-components of the magnetic dipole moment of the loop? (4pt) the magnetic field is constant across the area of the loop and equal to the value in part (a), what is the vector of the torque acting on the loop? (3pt) (f) Another wire is now added (the loop is removed), also aligned with the z-axis, at (z,y) (0, -26.8) cm as shown in the right figure above. This wire carries current i. What should be the value and direction of i to make the z-component of the magnetic feld equal to zero at the origin? (3pt)

Solution

given three infinite wires

distacne between wires d = 26 cm

given currents

i1 = 3.4 A (into the page)

i2 = 0.5 A (out of the page)

i3 = 5.2 A (out of the page)

a. consider magnetic field at the origin

x component = 2kI3/dsin(60) = 2*10^-7*5.2/0.26*sin(60) = 4.6188*10^-6 T

y component = -4k[i1 + i2]/d = -4*10^-7[3.9]/0.26 = -6*10^-6 T

b. for a wire carrying current i1, and length l

y component of force = -Bx*i1 = -15.70392*10^-6 N

x component of force = -By*i1 = -20.4*19^-6 N

c. for a wire carrying current i2 and length 1 m

x component of force = By(i2) = 3*10^-6 N

d. square loop of area A = 1*10^-4 m^2

current i = 1 A

angle of the plane f the loop with x axis = 30 deg

magnetic dipole = i*A = 10^-4 A m^2

x component = iA*cos(60) = 0.5*10^-4 A m^2

y compoennt = iAsin(60) = 0.766*10^-4 A m^2