The Total solution for NCERT class 6-12
A circular coil of 16turns and radius 10 cm carrying a current of 0.75 A rests with its planenormal 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 fielddirection. When the coil is turned slightly and released, it oscillates aboutits stable equilibrium with a frequency of 2.0 s−1. What is themoment of inertia of the coil about its axis of rotation?
Number of turnsin the circular coil, N = 16
Radius of thecoil, r = 10 cm = 0.1 m
Cross-section of thecoil, A = πr2 = π ×(0.1)2 m2
Current in thecoil, I = 0.75 A
Magnetic fieldstrength, B = 5.0 × 10−2 T
Frequency ofoscillations of the coil, v = 2.0 s−1
= 16 × 0.75× π × (0.1)2
= 0.377 J T−1
Where,
I = Moment of inertia of the coil
Hence, the moment of inertia of the coil about its axis of rotationis