A long straight horizontal cable carries a current of 2.5 A in the direction 10º south of west to 10º north of east. The magnetic meridian of the place happens to be 10º west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable). (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)

Current in the wire, I = 2.5 A

Angle of dip at the given location on earth, δ = 0°

Earth’s magnetic field, H = 0.33 G = 0.33 x 10^-4 T

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

H_H = H cos δ

= 0.33 x 10^-4 x cos 0° = 0.33 x 10^-4 T

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

H_H =μ0 I /2πR

Where, μ0 = Permeability of free space = 4π x 10^-7 T m A^-1

Therefore R = μ0 I /2πH_H

=  (4π x 10^-7 x 2.5 )/ (2π x 0.33 x 10^-4)  = 15.15 x 10⁻³ = 1.51 cm

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