Q.A wire in the form of a tightly wound solenoid is connected to a DC source, and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil, will the current increase or decrease? Explain.

In DC steady state the inductor plays no role: with $dI/dt=0$ the back-emf $L\,dI/dt$ is zero, so the solenoid behaves as a plain resistor and

Spreading the turns of the coil apart (creating gaps) merely changes the pitch/geometry of the helix; it does not cut or elongate the wire, so the wire's own resistance $R$ is unchanged. Hence the steady current is essentially unchanged. (While you are actually stretching it, $L$ falls and a brief transient …

Why inductance drops out in DC steady state

Apply Kirchhoff's voltage law to the coil (resistance $R$, self-inductance $L$) on a DC source of emf $V$:

Once the current is steady, $\dfrac{dI}{dt}=0$, so the inductive term vanishes and

The value of $L$ never appears in the steady current. So even though stretching the solenoid lowers its inductance ($L=\mu_0 N^2 A/l$ decreases as $l$ grows), that has no effect on the final DC current.

Does the resistance change?

The steady current depends only on $R$. Creating gaps between successive turns spreads the helix out along its axis, but the wire itself is the same piece of wire — same material, same length, same cross-section. You are changing the coil's pitch, not lengthening or thinning the conductor. Therefore

is unchanged, and so is $I=V/R$.

The only real effect: a brief transient …

Method: Steady-State DC Circuit Analysis With an Inductor — When Does $L$ Matter?

Many questions describe a coil/solenoid connected to a DC source and ask how some geometric change affects the current. The key method insight: in steady state, self-inductance drops out of the current equation entirely — so you must first identify whether the question is asking about the transient or the final steady value.

Steps

Step 1: Write the full circuit equation including the inductor's back-emf

For a coil of resistance $R$ and self-inductance $L$ driven by a DC source of emf $V$:

Step 2: Identify which regime the question is asking about

• Steady state ("what is the current" without reference to "just after," "immediately," or "at the instant of"): the current has settled and is no longer changing, so $dI/dt = 0$.
• Transient (explicitly "at the instant," "immediately after," "just as"): $dI/dt \neq 0$ and $L$ genuinely matters.

Step 3: For the steady-state case, drop the inductive term and solve

Notice $L$ has completely disappeared from this equation — whatever the coil's inductance is, it plays NO role in the final steady current.

Step 4: Ask separately whether $R$ has changed — this is the only thing that CAN affect $I$ …