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, $H=0.39G=0.39×{10}^{-4}T$
Angle of dip at the location, õ = 35°
Angle of declination, $\mathrm{\theta }$-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:
${\mathrm{H}}_{\mathrm{h}}$=H cosõ-B
Where,


$\begin{array}{l}\therefore B=4×\frac{4\pi ×{10}^{-7}×1}{2\pi ×0.04}\\ =0.2×{10}^{-4}\mathrm{T}=0.2\mathrm{G}\\ \therefore {\mathrm{H}}_{\mathrm{h}}=0.39\mathrm{cos}{35}^{\circ }-0.2=0.39×0.819-0.2\approx 0.12\mathrm{G}\end{array}$

The angle made by the field with its horizontal component is given as
$\begin{array}{rl}\theta & ={\mathrm{tan}}^{-1}\frac{{H}_{v}}{{H}_{h}}\\ & ={\mathrm{tan}}^{-1}\frac{0.22}{0.12}={61.39}^{\circ }\end{array}$
The resultant field at the point is given as
$\begin{array}{rl}{H}_{1}& =\sqrt{{\left({H}_{v}\right)}^{2}+{\left({H}_{h}\right)}^{2}}\\ & =\sqrt{\left(0.22{\right)}^{2}+\left(0.12{\right)}^{2}}=0.25\mathrm{G}\end{array}$
For a point 4 cm above the cable
Horizontal component of earth's magnetic field:
Vertical component of earth's magnetic field:

And a resultant field: