Question

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?

Answer 1

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 the earth’s magnetic field can be written as:

H_h = H cos δ − B

Where,

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

= ${\displaystyle 4\times \frac{{\mu }_0\text{I}}{2\pi \text{r}}}$

μ₀ = Permeability of free space = 4π × 10⁻⁷ T mA⁻¹

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

= 0.2 × 10⁻⁴ T

= 0.2 G

∴ H_h = 0.39 cos 35° − 0.2

= 0.39 × 0.819 − 0.2 ≈ 0.12 G

The vertical component of the earth’s magnetic field is given as:

H_v = H sin δ

= 0.39 sin 35°

= 0.22 G

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

θ = ${\displaystyle {\tan }^{-1} \frac{{\text{H}}_{\text{v}}}{{\text{H}}_{\text{h}}}}$

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

= 61.39°

The resultant field at the point is given as:

H₁ = ${\displaystyle \sqrt{{\left({\text{H}}_{\text{v}}\right)}^2+{\left({\text{H}}_{\text{h}}\right)}^2}}$

= ${\displaystyle \sqrt{{\left(0.22\right)}^2+{\left(0.12\right)}^2}}$

= 0.25 G 

For a point 4 cm above the cable:

Horizontal component of earth’s magnetic field:

H_h = H cos δ + B

= 0.39 cos 35° + 0.2

= 0.52 G

Vertical component of earth’s magnetic field:

H_v = H sin δ

= 0.39 sin 35°

= 0.22 G

Angle, θ = ${\displaystyle {\tan }^{-1} \frac{{\text{H}}_{\text{v}}}{{\text{H}}_{\text{h}}}}$

= ${\displaystyle {\tan }^{-1} \frac{0.22}{0.52}}$

= 22.9°

And resultant field:

H₂ = ${\displaystyle \sqrt{{\left({\text{H}}_{\text{v}}\right)}^2+{\left({\text{H}}_{\text{h}}\right)}^2}}$

= ${\displaystyle \sqrt{{\left(0.22\right)}^2+{\left(0.52\right)}^2}}$

= 0.56 T