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 $35^{\circ} .$ The magnetic declination is nearly zero. What are the resultant magnetic fields at points $4.0\; cm$ below and above the cable?

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, $H=0.39 \,G=0.39 \times 10^{-4} \,T$

Angle of dip at the location, $\delta=35^{\circ}$

Angle of declination, $\theta \sim 0^{\circ}$ Eor 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 $H_{h}=H \cos \delta-B$

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 $=4 \pi \times 10^{-7} \operatorname{Tm} \,A ^{-1}$

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

$=0.2 \times 10^{-4} \,T =0.2 \,G$

$\therefore H_{h}=0.39 \cos 35^{\circ}-0.2$

$=0.39 \times 0.819-0.2 \approx 0.12 \,G$

The vertical component of earth's magnetic field is given as

$H_{v}=H \sin \delta$

$=0.39 \sin 35^{\circ}=0.22\, G$

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:

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

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

For a point $4 \,cm$ above the cable Horizontal component of earth's magnetic field:

$H_{h}=H \cos \delta+B$

$=0.39 \cos 35^{\circ}+0.2=0.52\, G$

Vertical component of earth's magnetic field

$H_{v}=H \sin \delta$

$=0.39 \sin 35^{\circ}=0.22 \,G$

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

And resultant field:

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

$=\sqrt{(0.22)^{2}+(0.52)^{2}} \approx 0.56\, T$