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प्रश्न
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?
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उत्तर
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:
Hh = 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
∴ Hh = 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:
Hv = 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:
Hh = H cos δ + B
= 0.39 cos 35° + 0.2
= 0.52 G
Vertical component of earth’s magnetic field:
Hv = 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
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