A circular coil of 16 turns and radius 10 cm carrying a current
Question:

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?

Solution:

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 × (0.1)2 m2

Current in the coilI = 0.75 A

Magnetic field strength, B = 5.0 × 10−2 T

Frequency of oscillations of the coil, v = 2.0 s−1

$\therefore$ Magnetic moment, $M=N I A=N I \pi r^{2}$

$=16 \times 0.75 \times \pi \times(0.1)^{2}$

$=0.377 \mathrm{~J} \mathrm{~T}^{-1}$

Frequency is given by the relation:

$v=\frac{1}{2 \pi} \sqrt{\frac{M B}{I}}$

Where,

I = Moment of inertia of the coil

$\therefore I=\frac{M B}{4 \pi^{2} v^{2}}$

$=\frac{0.377 \times 5 \times 10^{-2}}{4 \pi^{2} \times(2)^{2}}$

$=1.19 \times 10^{-4} \mathrm{~kg} \mathrm{~m}^{2}$

Hence, the moment of inertia of the coil about its axis of rotation is $1.19 \times 10^{-4} \mathrm{~kg} \mathrm{~m}^{2}$.