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 magnitude5.0 × 10–2 T. The coil is free to turn about an axis in its planeperpendicular to the field direction. When the coil is turned slightlyand released, it oscillates about its stable equilibrium with afrequency of 2.0 s–1. What is the moment of inertia of the coil aboutits axis of rotation?
Given: Number of turns in the coil , N = 16 ; radius of the coil, r = 10 cm ; current passing through the coil, I = 0 .75 A ; magnetic field, B = 5 × 10 − 2 T , and frequency of the oscillation, ν = 2 s − 1
The magnetic moment is given as,
M = N I A
By substituting the given values, we get,
M = 16 × 0.75 × ( π × 0.1 2 ) = 0.377 JT -1
Frequency of the oscillation is given as,
ν = 1 2 π M B I
By substituting the values of M, B , ν in the above equation, we get,
2 = 1 2 π 0.377 × 5 × 10 − 2 I ( 4 π ) 2 = 0.377 × 5 × 10 − 2 I I = 1.2 × 10 − 4 kgm 2
Thus, the moment of inertia is 1.2 × 10 − 4 kgm 2 .
Given: Number of turns in the coil ,
The magnetic moment is given as,
By substituting the given values, we get,
Frequency of the oscillation is given as,
By substituting the values of
Thus, the moment of inertia is
