Class 12 Physics
CBSE and UP Board
Magnetism and Matter
Chapter-5 Exercise 5.9
NCERT Solutions for Class 12 Physics Chapter 5 Question 9
A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 × 10⁻² T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s⁻¹. What is the moment of inertia of the coil about its axis of rotation?
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Number of turns in the circular coil, N = 16
Radius of the coil, r = 10 cm = 0.1 m
Cross-section of the coil, A =πr2 = π x (0.1)2 m2
Current in the coil, I = 0.75 A
Magnetic field strength, B = 5.0 x 10-2 T
Frequency of oscillations of the coil, v = 2.0 s-1
Therefore , Magnetic moment, M = NIA = NJπr2
= 16 x 0.75 x n x (0.1)2 = 0.377 J T-1
Frequency is given by the relation:
v =( 1/2 π) √ (MB/I)
Where,
I = Moment of inertia of the coil
Therefore I = MB /(4π²v²) = (0.377 x 5 x 10⁻²)/( 4π² x 2² )
= 1.19 x 10⁻4 kg m²
Hence, the moment of inertia of the coil about its axis of rotation is 1.19 x 10⁻4 kg m²