# 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.

**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 35º. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?

**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:

*H**h* = *H*cos*δ* − *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

∴ *H**h* = 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:

*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:

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

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

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:

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

And resultant field:

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

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

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