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 `5.0xx 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 `2.0s^(-1)`. What is the moment of inertia of the coil about its axis of rotation ?

Explanation

R = 10 cm = 0 .10 m , `I = 0.75 A, B = 5.0 xx 10^(-2) T , f = 2.0 s^(-1). I` = ?
`T = (1)/(2) = 0.5s , N = 16 , A = pi R^(2) = 3.142 xx (0.1)^(2) = 3.142 xx 10^(-2) m^(2)`
Magnetic moment of the coil ,
,m = NIA
`= 16 xx 0.75 xx 3.14 xx 10^(-2)`
`= 37.704xx 10^(-2)`
`= 0.377 "Am"^(2) "or" JT^(-1)`
Period of oscillation , ` T = 2pi sqrt((I)/(mB))`
0.5 = 45 .77 `sqrt(I)`
i.e., `" " sqrt(I) = 0.01092`
or `" " I = 1.193 xx 10^(-4) "kgm"^(2)`
`:.` M.I. of the coil about an axis passing through the centre and perpendicular to the plane of the coil = `1.193 xx 10^(-4) "kgm"^(2)`.