Question -
Answer -
Number of horizontalwires in the telephone cable, n = 4
Current in eachwire, I = 1.0 A
Earth’s magnetic fieldat a location, H = 0.39 G = 0.39 × 10−4 T
Angle of dip at thelocation, δ = 35°
Angle ofdeclination, θ ∼ 0°
For a point 4 cm below the cable:
Distance, r =4 cm = 0.04 m
The horizontal componentof earth’s magnetic field can be written as:
Hh = Hcosδ − B
Where,
B = Magnetic field at 4 cm due tocurrent I in the four wires
= Permeability offree space = 4π × 10−7 Tm A−1
= 0.2 × 10−4 T= 0.2 G
∴ Hh =0.39 cos 35° − 0.2
= 0.39 × 0.819 −0.2 ≈ 0.12 G
The vertical componentof earth’s magnetic field is given as:
Hv = Hsinδ
= 0.39 sin 35° = 0.22 G
The angle made by thefield with its horizontal component is given as:
The resultant field atthe point is given as:
For a point 4 cm above the cable:
Horizontal component ofearth’s magnetic field:
Hh = Hcosδ + B
= 0.39 cos 35° + 0.2 =0.52 G
Vertical component ofearth’s magnetic field:
Hv = Hsinδ
= 0.39 sin 35° = 0.22 G
Angle, θ 
= 22.9°
And resultant field:
