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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. |
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Answer» `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)]` 
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