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

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