Class 12 Physics
CBSE and UP Board
Magnetism and Matter
Chapter-5 Exercise 5.18
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 5 Question 18
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.)
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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:
HH = 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:
HH =μ0 I /2πR
Where, μ0 = Permeability of free space = 4π x 10-7 T m A-1
Therefore R = μ0 I /2πHH
= (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.