Question

An iron core solenoid of length $l$ and cross-section 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. Current in the circuit just after removal of the rod is

• A. $\frac{ϵ}{R}$
• B. $\frac{\mu ϵ}{R}$
• C. zero
• D. $\frac{ϵ}{\mu R}$

Answer 1

The correct option is B $\frac{\mu ϵ}{R}$

Let $L$ be inductance of selenoid before removing core
$L^{\prime }$ be inductance of selenoid after removing core
$\Rightarrow L^{\prime }=\frac{L}{\mu }$ ....(i)

Flux linking with the coil does not change suddenly.
Hence,
$\varphi =L_1{i}_1=L_2{i}_{2}L.\frac{\epsilon }{R}=L^{\prime }.I^{\prime }$
(As current before removal of core $=\frac{ϵ}{R}$)
using eq. (i),
$\therefore \frac{\epsilon }{R}=\frac{I^{\prime }}{\mu }\Rightarrow I^{\prime }=\frac{\mu ϵ}{R}$