A long straight horizontal cable carries a current of 2.5 A in the direction 10º south of west to 10° north of east.

Current in the wire, I = 2.5 A

Angle of dip at the given location on earth, $\delta=0^{\circ}$

Earth's magnetic field, $H=0.33 \mathrm{G}=0.33 \times 10^{-4} \mathrm{~T}$

The horizontal component of earth's magnetic field is given as:

$H_{H}=H \cos \delta$

$=0.33 \times 10^{-4} \times \cos 0^{\circ}=0.33 \times 10^{-4} \mathrm{~T}$

The magnetic field at the neutral point at a distance R from the cable is given by the relation:

$H_{H}=\frac{\mu_{0} I}{2 \pi R}$

Where,

$\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}$

$\therefore R=\frac{\mu_{0} I}{2 \pi H_{H}}$

$=\frac{4 \pi \times 10^{-7} \times 2.5}{2 \pi \times 0.33 \times 10^{-4}}=15.15 \times 10^{-3} \mathrm{~m}=1.51 \mathrm{~cm}$

Hence, a set of neutral points parallel to and above the cable are located at a normal distance of 1.51 cm.