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.

Solution:

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 = 0.39 G = 0.39 × 10−4 T

Angle of dip at the location, δ = 35°

Angle of declination, θ ∼ 0°

For a point 4 cm below the cable:

Distance, r = 4 cm = 0.04 m

The horizontal component of earth’s magnetic field can be written as:

Hh = Hcosδ − B

Where,

B = Magnetic field at 4 cm due to current I in the four wires

$=4 \times \frac{\mu_{0} I}{2 \pi r}$

$\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}$

$\therefore B=4 \times \frac{4 \pi \times 10^{-7} \times 1}{2 \pi \times 0.04}$

= 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 component of earth’s magnetic field is given as:

Hv = Hsinδ

= 0.39 sin 35° = 0.22 G

The angle made by the field with its horizontal component is given as:

$\theta=\tan ^{-1} \frac{H_{v}}{H_{h}}$

$=\tan ^{-1} \frac{0.22}{0.12}=61.39^{\circ}$

The resultant field at the point is given as:

$H_{1}=\sqrt{\left(H_{v}\right)^{2}+\left(H_{h}\right)^{2}}$

$=\sqrt{(0.22)^{2}+(0.12)^{2}}=0.25 \mathrm{G}$

For a point 4 cm above the cable:

Horizontal component of earth’s magnetic field:

Hh = Hcosδ + B

= 0.39 cos 35° + 0.2 = 0.52 G

Vertical component of earth’s magnetic field:

Hv = Hsinδ

= 0.39 sin 35° = 0.22 G

Angle, $\theta=\tan ^{-1} \frac{H_{v}}{H_{h}}=\tan ^{-1} \frac{0.22}{0.52}=22.9^{\circ}$

And resultant field:

$H_{2}=\sqrt{\left(H_{v}\right)^{2}+\left(H_{h}\right)^{2}}$

$=\sqrt{(0.22)^{2}+(0.52)^{2}}=0.56 \mathrm{~T}$