A circular coil of 16 turns and radius 10cm carrying a current of 0.75A rests with its plane normal to an external field of magnitude `5*0xx10^-2T`. 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*0s^-1`. What is the moment of inertia of the coil about its axis of rotation?

Explanation

Number of turns in the circular coil, N=16
Radius of the coil, r=10cm=0.1m
Cross-section of the coil, A=`nr^(2)=nxx(0.1)^(2)m^(2)`
Current in the coil, I=0.75A
Magnetic field strength, B=5.0`xx10^(-2)T`
Frequency of oscillations of the coil, `v=2.0s^(-1)`
`therefore` M agnetic moment, M=NIA=`NIpir^(2)`
`=16xx0.75xxnxx(0.1)^(2)`
`=0.377 J T^(-1)`
Frequency is given by the relation:
`v=(1)/(2pi)sqrt((MB)/(l))`
Where, ltbr gt I=Moment of inertia of the coil
`therefore I=(MB)/(4pi^(2) V^(2))`
`=(0.377x x5xx10^(-2))/(4pi^(2)xx(2)^(2))`
`=1.19xx10^(-4)kg m^(2)`
Hence, the moment of inertia of the coil about its axis of rotation is `1.19xx10^(-4)kg m^(2)`