Question

A long horizontal wire carries 14.7 A of current due north. What is the magnitude of the net magnetic field 15.4 cm due west of the wire if the Earth's field there points downward, 39.5° below the horizontal, and has magnitude 4.97E-5 T? Calculate the angle relative to the horizontal with up as the positive direction.

Answer 1

given
Bh = 4.97*10^-5 T

magnetic field due to wire at the given point,

B = mue*I/(2*pi*r)

= 4*pi*10^-7*14.7/(2*pi*0.154)

= 1.91*10^-5 T (towards up)

let up is +y axis and North is z axis

Bnety = B - Bh*sin(39.5)

= 1.91*10^-5 - 4.97*10^-5*sin(39.5)

= -1.25*10^-5 T

Bnetz = Bh*cos(39.5)

= 4.97*10^-5*cos(39.5)

= 3.83*10^-5 T

|Bnet| = sqrt(Bnety^2 + Bnetz^2)

= sqrt(1.25^2 + 3.83^2)*10^-5

= 4.03*10^-5 T <<<<<<<<<<-----------------Answer

direction : theta = tan^-1(Bnety/Bnetz)

= tan^-(1.25/3.83)

= 18.1 degrees below horizontal

= -18.1 degrees   <<<<<<<<<<-----------------Answer