Class 12 Physics
CBSE and UP Board
Magnetism and Matter
Chapter-5 Exercise 5.19
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 5 Question 19
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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 x 10_4T
Angle of dip at the location, δ = 35°
Angle of declination, 0 ~ 0°
For 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:
Hh = H cos δ – B
Where,
B = Magnetic field at 4 cm due to current I in the four wires = 4 x μ0 I/2πR
μ0 = Permeability of free space = 4πx 10-7 Tm A-1
Therefore , B = 4 x (4πx 10-7 x I)/ (2π x .04)
= 0.2 x 10-4 T = 0.2 G
Therefore , Hh = 0.39 cos 35° – 0.2 = 0.39 x 0.819 – 0.2 ≈ 0.12 G
The vertical component of earth’s magnetic field is given as: Hv = H sinδ = 0.39 sin 35° = 0.22 G The angle made by the field with its horizontal component is given as:
0 = tan⁻¹Hv/Hh
= tan⁻¹ 0.22/0.12= 61.39º
The resultant field at the point is given as:
H1 = √ [(H)² + (Hh)²]
=√(0.22)²+ (0.12)² = 0.25 G
For a point 4 cm above the cable:
Horizontal component of earth’s magnetic field: Hh = Hcosδ + B = 0.39 cos 35° + 0.2 = 0.52 G Vertical component of earth’s magnetic field: Hv = Hsinδ = 0.39 sin 35° = 0.22 G
Angle, 0 = tan-1 Hv/Hh = tan-1 0.22/0.52= 22.9°
And resultant field:
H2= √ [(H)² + (Hh)²]
= √(0.22)²+ (0.52)2 = 0.56 T