The Total solution for NCERT class 6-12
A long straighthorizontal cable carries a current of 2.5 A in the direction 10º south of westto 10° north of east. The magnetic meridian of the place happens to be 10ºwest of the geographic meridian. The earth’s magnetic field at the location is0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignorethe thickness of the cable). (At neutral points, magnetic field dueto a current-carrying cable is equal and opposite to the horizontal componentof earth’s magnetic field.)
Current in thewire, I = 2.5 A
Earth’s magneticfield, H = 0.33 G = 0.33 × 10−4 T
The horizontal componentof earth’s magnetic field is given as:
The magnetic field atthe neutral point at a distance R from the cable is given bythe relation:
Where,
= Permeability of free space =
Hence, a set ofneutral points parallel to and above the cable are located at a normal distanceof 1.51 cm.