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

= `4 xx (μ_0"I")/(2π"r")`

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

∴ B = `4 xx (4π xx 10^-7 xx 1)/(2π xx 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 = `tan^-1  "H"_"v"/"H"_"h"`

= `tan^-1  0.22/0.12`

= 61.39°

The resultant field at the point is given as:

H₁ = `sqrt(("H"_"v")^2 + ("H"_"h")^2)`

= `sqrt((0.22)^2 + (0.12)^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 = `tan^-1  "H"_"v"/"H"_"h"`

= `tan^-1  0.22/0.52`

= 22.9°

And resultant field:

H₂ = `sqrt(("H"_"v")^2 + ("H"_"h")^2)`

= `sqrt((0.22)^2 + (0.52)^2)`

= 0.56 T