Class 12 Physics
CBSE and UP Board
Moving Charges and Magnetism
Chapter-4 Exercise 4.9
NCERT Solutions Class 12 Physics Chapter 4
A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
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Length of a side of the square coil, 1 = 10 cm = 0.1 m
Current flowing in the coil, I = 12 A
Number of turns on the coil, n = 20
Angle made by the plane of the coil with magnetic field, 0 = 30°
Strength of magnetic field, B = 0.80 T
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,
τ = n BIA sin0
Where, A = Area of the square coil = 1×1 = 0.1 x 0.1 = 0.01m²
So,
τ = 20 x 0.8 x 12 x 0.01 x Sin30°
= 0.96 N m
Hence, the magnitude of the torque experienced by the coil is 0.96 N m.