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 `5.0 xx 10^(-2)T.` The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coll is turned slightly and released, it oscillates about its stable equilibrium with a frequency of `2.0 s^(-1).` What is the moment of inertia of the coil about its axis of rotation?

Explanation

Magnetic moment of a current carrying coil is,
`m= NIA`
`= NI (pi R^(2) )`
`= (16) (0.75) (3.14) (0.1)^(2)`
`therefore m= 0.3768 "Am"^(2)` ( or `JT^(-1)` )
Periodic time of oscillation of above coil,
`T= 2pi sqrt((I)/(mB) )` (Where, `I=` moment of inertia of given coil about axis is rotation)
Frequency of oscillation is,
`f- (1)/(T) = (1)/(2pi )sqrt((mB)/( I) )`
`therefore f^(2) = (1)/( 4pi^(2) ) ((mB)/( (I)))`
`therefore I= (mB)/( 4 pi^(2) f^(2) )`
`therefore I= ((0.3768) (5 xx 10^(-2) ) )/( (4) (3.14)^(2) (2)^(2) )`
`therefore I= 1.194 xx 10^(-4) (kg) (m^(2) )`