Question

Question 5.9: 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?

Class 12 - Physics - Magnetism And Matter Magnetism And Matter Page-201

Answer 1

number of turns, N = 16
radius of circular coil , r = 10cm= 0.1 m
current, I = 0.75A
magnetic field , B = 5 × 10^-2 T
frequency,f = 2Hz
so, the circular coil carries a dipole moment
M = NIA = 16 × 0.75 × π(0.1)² = 0.3768 Am²

time period of oscillation, $\bf{T=2\pi\sqrt{\frac{I}{MB}}}$
so, frequency ,$\bf{f=\frac{1}{2\pi}\sqrt{\frac{MB}{I}}}$
so, moment of inertia , $I=\frac{1}{4\pi^2}\frac{MB}{f^2}$
I = (1 × 0.3768 × 5 × 10^-2 )/(4 × 3.14^2 × 2^2)
= 11.9 × 10^-5 kgm²