(A) When the solenoid is tightly wound,the magnetic field lines are confined within the solenoid,resulting in a high self-inductance $(L)$.
When the coil is stretched,the gaps between the turns increase,which causes the magnetic flux linkage per turn to decrease,leading to a decrease in the self-inductance $(L)$ of the solenoid.
According to the relation $V = L(di/dt) + iR$,for a fixed $DC$ voltage source,the steady-state current is determined by $I = V/R$.
However,during the stretching process,the change in self-inductance induces an electromotive force (emf) $\varepsilon = -L(di/dt) - i(dL/dt)$.
Since the self-inductance $L$ decreases $(dL/dt < 0)$,the induced emf opposes the change,effectively reducing the back-emf. Consequently,the current in the circuit increases.