Q.A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain.

The key idea here is Self-Inductance — specifically, the self-inductance of the solenoid changes when the iron core is inserted.

1. Inserting an iron core increases the magnetic permeability ($\mu$) of the solenoid's interior. Since the self-inductance $L$ of a solenoid is proportional to $\mu$, $L$ increases.
2. The solenoid is connected to a battery, so it has a constant applied voltage $V$. The steady current $I$ is given by $V = IR$ (Ohm's law for DC steady state), where $R$ is the solenoid's resistance.
3. In steady DC, the inductive reactance ($\omega L$) is zero — inductance only affects transients, not the final steady current. The battery sees only the resistance $R$ of the wire. …

The Core Idea: Inductance and Opposition to Change

When you insert an iron core into a solenoid, you dramatically increase its self-inductance $L$. The inductance of a solenoid is given by:

where $\mu_r$ is the relative permeability of the core material. For iron, $\mu_r$ can be thousands — so $L$ jumps by a factor of 1000 or more.

But here’s the key: inductance only matters when the current is changing. A steady current produces no induced emf. So the real question is: what happens during the moment you insert the core?

Step-by-Step Reasoning

1. The initial condition: steady current

Before inserting the core, the solenoid carries a steady current $I_0 = V/R$, where $V$ is the battery voltage and $R$ is the total resistance of the circuit (solenoid wire + battery internal resistance). No induced emf exists because $dI/dt = 0$.

2. What happens when you insert the core?

As you push the iron core in, the magnetic flux through each turn of the solenoid increases sharply — because iron concentrates the magnetic field lines. This change in flux induces an emf in the solenoid itself (self-induction).

3. The induced emf opposes the change

By Lenz’s law, the induced emf opposes the increase in flux. Since the flux is increasing (due to the core), the induced emf acts to reduce the current. So momentarily, the current drops below $I_0$.

4. The battery fights back

The circuit now has:

The battery tries to maintain the current. After a brief transient (lasting milliseconds), a new steady state is reached where $dI/dt = 0$ again. In steady state, the inductor acts like a short (zero reactance for DC), so:

$I_{\text{final}} = \frac{V}{R} = I_0$ …

Method: Steady-State vs Transient Analysis in an Inductive DC Circuit

This method applies whenever a question changes some property of an inductor (inserting a core, changing turns, etc.) in a DC circuit and asks what happens to the current.

Steps

Step 1: Identify what the source is holding constant

In a circuit driven by a battery, the applied EMF/voltage $V$ is fixed regardless of what happens inside the circuit. This is the anchor for the whole analysis.

Step 2: Separate the "transient" moment from the final "steady state"

Any change to $L$ (like inserting a core) matters only WHILE the current is changing — during that brief transient, the inductor develops an emf

which, by Lenz's law, opposes the change. Once the current settles again ($dI/dt = 0$), this term vanishes.

Step 3: Apply Ohm's law for the steady state

In steady state, the loop equation reduces to Ohm's law for the DC circuit alone:

$V = IR$ …