5.19 A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?

 
Number of horizontal wires in the telephone cable, n =4
Current in each wire, I= 1.0 A
Earth's magnetic field at a location, H=0.39G=0.39×104T
Angle of dip at the location, õ = 35°
Angle of declination, θ-0°
Fora point 4 cm below the cable
Distance, r = 4c m, 0.04 m
The horizontal component of the earth's magnetic field can be written as:
Hh=H cosõ-B
Where, 

B=Magnetic field at 4 cm due ot current I in the four wires=4×μ0I2πr
μ0=Permeability of free=4π×10-7Tm A-1
B=4×4π×107×12π×0.04=0.2×104T=0.2GHh=0.39cos350.2=0.39×0.8190.20.12G
The vertical component of earth's magnetic field is given as: 
Hv=H sin δ=0.39 sin 35°=0.22 G
The angle made by the field with its horizontal component is given as
θ=tan1HvHh=tan10.220.12=61.39
The resultant field at the point is given as
H1=(Hv)2+(Hh)2=(0.22)2+(0.12)2=0.25G
For a point 4 cm above the cable
Horizontal component of earth's magnetic field:
Hh=H cos δ + B=0.39 cos 35°+0.2=0.52 G
Vertical component of earth's magnetic field:
Hv=H sin δ=0.39 sin 35°=0.22 G
Angle, θ=tan-1HvHh=tan-10.220.52=22.9°
And a resultant field:
H2=(Hv)2+(Hh)2=(0.22)2+(0.52)2=0.56 T