Class 12 Physics
CBSE and UP Board
Magnetism and Matter
Chapter-5 Exercise 5.20
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 5 Question 20
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Number of turns in the circular coil, N = 30,
Radius of the circular coil, r = 12 cm = 0.12 m
Current in the coil, I = 0.35 A, Angle of dip, δ = 45°
Ans (a).
The magnetic field due to current I, at a distance r, is given as:
B= μ0 /4π 2πNI/r
Where, μ0 = Permeability of free space = 4π x 10⁻7 T m A-1 .
B = ( 4π x 10⁻7 x 2π x 30 x 0.35)/ (4π x 0.12) = 5.49 x 10⁻⁵
The compass needle points from West to East. Hence, the horizontal component of earth’s magnetic field is given as: BH = B sinδ = 5.49 x 10⁻⁵ sin 45° = 3.88 x 10⁻⁵ T = 0.388 G
Ans (b).
When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of 90 °, the needle will reverse its original direction. In this case, the needle will point from East to West.