$A$ telephone cable at a place has four long straight horizontal wires carrying a current of $1.0 \; A$ in the same direction from east to west. The earth's magnetic field at the place is $0.39 \; G$ and the angle of dip is $35^{\circ}$. The magnetic declination is nearly zero. What are the resultant magnetic fields at points $4.0 \; cm$ below and above the cable?

(A) Number of horizontal wires in the telephone cable,$n = 4$.
Current in each wire,$I = 1.0 \; A$.
Earth's magnetic field at the location,$H = 0.39 \; G = 0.39 \times 10^{-4} \; T$.
Angle of dip at the location,$\delta = 35^{\circ}$.
Angle of declination,$\theta \approx 0^{\circ}$.
For a point $4 \; cm$ below the cable:
Distance,$r = 4 \; cm = 0.04 \; m$.
The magnetic field $B$ due to the current in the four wires is $B = 4 \times \frac{\mu_{0} I}{2 \pi r} = 4 \times \frac{4 \pi \times 10^{-7} \times 1}{2 \pi \times 0.04} = 0.2 \times 10^{-4} \; T = 0.2 \; G$.
The horizontal component of the earth's magnetic field is $H_{h} = H \cos \delta - B = 0.39 \cos 35^{\circ} - 0.2 = 0.39 \times 0.819 - 0.2 \approx 0.12 \; G$.
The vertical component of the earth's magnetic field is $H_{v} = H \sin \delta = 0.39 \sin 35^{\circ} \approx 0.22 \; G$.
The resultant magnetic field is $H_{1} = \sqrt{H_{v}^{2} + H_{h}^{2}} = \sqrt{(0.22)^{2} + (0.12)^{2}} \approx 0.25 \; G$.
For a point $4 \; cm$ above the cable:
The horizontal component of the earth's magnetic field is $H_{h} = H \cos \delta + B = 0.39 \cos 35^{\circ} + 0.2 = 0.319 + 0.2 = 0.52 \; G$.
The vertical component remains $H_{v} = 0.22 \; G$.
The resultant magnetic field is $H_{2} = \sqrt{H_{v}^{2} + H_{h}^{2}} = \sqrt{(0.22)^{2} + (0.52)^{2}} \approx 0.56 \; G$.