A ring of thin wire with active resistance `R` and inductance `L` rotates with constant angular velocity `omega` in the external uniform magneitce field perpendicular to the rotation axis. In the process , the flux of magnetic induction of external field across the ring varies with time as `Phi=Phi_(0)cos omega t`. Demonstrate that
(a) the inductive current in the ring varies with time as `I=I_(m) sin ( omegat-varphi)`, where `I_(m)=._(m) Phi//sqrt(R^(2)+ omega^(2)L^(2))` with `tan varphi=omegaL//R`,
(b) the mean mechanical is defined by the formula `P=1//2 omega^(2) Phi_(0)^(2)R//(R^(2)+omega^(2)L^(2))`.

Explanation

`(a)` We have
`epsilon=-(dPhi)/( dt)=omegaPhi_(0)sin omegat=Ldot(I)+RI`
Put `I=I_(m) sin ( omegat-varphi)`. Then
`omegaPhi_(0)sin omegat =omegaPhi_(0){sin ( omega t-varphi)cos varphi+cos (o omega-varphi)sin varphi}`
`=LI_(m) omegacos ( omegat-varphi)+RI_(m) sin (omegat-varphi)`
so `RI_(M) = omega Phi_(0) cos varphi` and `LI_(m) = Phi _(0) sin varphi`
or ` I_(m)=( omega Phi_(0))/( sqrt(R^(2)+omega^(2)L^(2)))` and `tan varphi=( omegaL)/( R)`
`(b)` Mean mechanical power required to maintain rotation `=` energy loss per unit time
`=(1)/(T)int_(0)^(T)RI^(2)dt=(1)/(2) RI_(m)^(2)=(1)/(2)(omega^(2) Phi_(0)^(2)R)/( R^(2)+omega^(2)L^(2))`