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

The magnetic dipole moment of the coil of number of N, area A and carrying current I is given by
M=NIA
Here, N = 16 and I = 0.75 A`, A = pir^2 = pi(0.10)^2 m^2`
` rArr M = 16 xx 0.75 xx pi (0.10)^2 Am^2 = 0.377 Am^2`
When the coil is slightly rotated by an angle then the torque acting on it can be written as follows:
`MB sin theta = I alpha `
Here, I is moment of inertia of coil and `alpha` is angular acceleration. For small we can write the above equation as follows:
`alpha = (MB)/(I) theta `
Since it is restoring torque, it is a condition of SHM.
` |apha | = omega^2 theta`
where `omega = sqrt((MB)/(I))`
Time period is given by`T = 2pi//omega `
` rArr T = 2pi sqrt( (I)/(MB))`
Frequency is
`v = (1)/(2pi) sqrt((MB)/(I))`
` therefore` The moment of inertia of the coil about the axis of rotation is:
`I = (MB)/(4pi^2 v^2)`
here , `B = 5.0 xx 10^(-2) T`
` v = 2s^(-1)`
` rArr I = (0.377 xx 5.0 xx 10^(-2))/(4 xx (3.14)^2 xx 2^2) `
` = 1.194 xx 10^(-4) Kg m^2`