An iron core solenoid of length `l` and cross-sectional area `A` having `N` turns on it is connected to a battery through a resistance as shown in the figure. At instant `t=0`, the iron rod of permeability `mu` from the core is abruptly removed. Find the current as a function of time.

Explanation

Self inductance of the solenoid
`L=(mu_(0)muN^(2)A)/(l)`
and initial current `i_(0)=(epsilon)/(R )` because circuit is in the steady state.
When iron rod is abruptly removed from the core the number of flux linkages abruptly do no change.
`Nphi=Li_(0)=L'i_(0)'`
`((mu_(0)muN^(2)A)/(l))i_(0)=((mu_(0)N^(2)A)/(l))i'_(0)`
`:. ` Just after `t=0`, initial current `i_(0)'=mu i_(0)=mu((epsilon)/(R ))`
At instant, if the current in the circuit is `i`, then apply `KVL` in the loop, we get
`L(di)/(dt)+iR=epsilon`
`-L(di)/((epsilon-iR))=dt`
On integration we get,
`(L)/(-R) ln(epsilon-iR)=t+c_(1)`
At `t=0`, `i=i_(0)'=(mu(epsilon)/(R ))`
`rArr ((L)/(-R))ln(epsilon-i_(0)'R)=C_(1)`
`:. ln((epsilon-iR)/(epsilon-i_(0)'R))=-(tR)/(L)rArr((epsilon-iR)/(epsilon-i_(0)'R))=e^(-(t R)/(L))`
`rArr((epsilon-iR)/(epsilon-(mu(epsilon)/(R))R))=e^(-(tR)/(L))`
On rearraging the equation, we get
`i=(epsilon)/(R )[1+(mu-1)e^(-tR//L)]`