A solenoid of resistance R and inductance L has a piece of soft iron inside it. A battery of emf E and of negligible internal resistance is connected across the solenoid as shown in Fig. At any instant, the piece of soft iron is pulled out suddenly so that inductance of the solenoid decrease to `etaL(eta lt 1)` with battery remaining connected.

Assume t = 0 is the instant when iron piece has been pulled out, the current as a function of time after this is

Options

A. `i=E/R [1-(1-(1)/(eta))e^(-t/(lambda))]`
B. `i=E/R [1+(1+(1)/(eta))e^(-t/(lambda))]`
C. `i=E/R [1-(1+(1)/(eta))e^(-t/(lambda))]`
D. `i=E/R [1+(1-(1)/(eta))e^(-t/(lambda))]`

Answer

A

Explanation

As current increase on pulling out the iron piece, the current will decrease to achieve its steady state value `E//R` again. At any time:
`E=IR+etaL(dI)/(dt)implies int_(I_2)^(i) (dI)/(E-IR)= int_(0)^(t)(dt)/(etaL)`
Solve to get: i=(E)/(R)[1-(1-(1)/(eta))e^(-t//lambda)]`.