An oscillating circuit consists of an inductance coil `L` and a capacitor with capacitance `C`. The resistance of the coil and the lead wires is negligible . The coil is placed in a permanent magnetic filed so that the total flux passing through all the turns of the coild is equal to `Phi` . At the moment `t=0` the magnetic field was switched off. Assuiming the switching off time to be negligible compared to the natural oscillation paeriod of the cirucuit, find the circuit current as a fuction of time `t`

Explanation

The flux in the coil is
`Phi(t)={(Phi,t lt 0),(0, tgt0):}`
The equation of the currenct is `-L(dI)/( dt)=(int_(0)^(t)Idt)/(C) ...(1)`
This mean that `LC(d^(2)I)/( dt^(2))+I=0`
or with `omega_(0)^(2)=(1)/(LC)` `I=I_(0)sin ( omega_(0)t+ alpha)`
Putting in `(1)` `-LI_(0)omega_(0) cos ( omega_(0)t+ alpha)=-(I_(0))/( omega_(0)C)[cos( omega_(0)t+ alpha)- cos alpha]`
This implies `cos alpha=0 :. I =+- I_(0) cos omega_(0) t`. From Farayday's law
`epsilon=-(dPhi)/( dt)=-L(dI)/( dt)`
on initegrating from `t=- epsilon` to `-epsilon` where `e rarr0`
`Phit=LI_(0) ` with `+` sign I n `I`
so, `I=(Phi)/(l) cos omega_(0) t`