Question

A long horizontal wire carries 14.0 A of current due north. What is the magnitude of the net magnetic field 23.9 cm due west of the wire if the Earth's field there points downward, 34.4° below the horizontal, and has magnitude 5.11E-5 T? Calculate the angle relative to the horizontal with up as the positive direction.

Answer 1

given,

current in the wire = 14 A

earth's magnitic field is 34.4 degree below the horizontal

magnitude of earth's magnetic field = 5.11 * 10^-5 T

The field from the wire is clockwise about the current direction, or upward at a point west of the wire.

magnetic field at distance 23.9 cm from wire is

B_wire = mu0 * I / (2pi * d)

B_wire = 2 * 10^-7 * 14 / 0.239

B_wire = 1.171548 * 10^-5 T

Net field:

B_up = 5.11 * 10^-5 * sin(-34.4) + B_wire

B_up = 5.11 * 10^-5 * sin(-34.4) + 1.171548 * 10^-5

B_up = -1.171543 * 10^-5 T

B_north = 5.11 * 10^-5 * cos(-34.4)

B_north = 4.216 * 10^-5 T

B = sqrt(B_up^2 + B_north^2)

B = sqrt((-1.171543 * 10^-5)^2 + (4.216 * 10^-5)^2)

B = 4.375 * 10^-5 T

net magnetic field = 4.375 * 10^-5 T

relative angle = tan^-1(B_up / B_north)

raltive angle = tan^-1((-1.171543 * 10^-5) / (4.216 * 10^-5))

relative angle = -15.53 degree