I don't understand what inductors do

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Introducing a metal piece inside an inductor coil affects the inductance, which influences how quickly the current builds up in the circuit. While the resistance remains constant, the inductance increases with the ferrous insert, causing the current to rise more slowly. In steady-state DC conditions, both scenarios ultimately result in the same current flow and light intensity. However, the presence of the metal piece allows for slightly more energy storage in the magnetic field. A diode connected across the circuit can demonstrate the effects of the collapsing field when the switch is opened, producing a brief pulse of EMF.
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Hi! Here is probably a simple problem but I can't solve it:

We have an inductor (just a coiled wire) in series with a light bulb, connected to an ideal battery. The light bulb is shining and we measure the intensity of light. Now we introduce a metal piece inside the coil. AFTER the piece is placed inside, we measure the intensity of the light again. Will there be any difference in the intensity of light and why?

Help me please!
 
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Inductance is the ability to store energy during field growth (field moment).

In either case, the wire itself has the same amount of resistance. So, in the end, BOTH cases show the same amount of current flow.

BUT, in the first case, the air-core in the coil allows a faster field moment (lower inductance).
In the second case (with ferrous insert), the circuit current will grow more slowly as the now LARGER inductance takes more time to build it's field.

Except for a slightly larger amount of energy stored in fields, and slightly slower rising of current at switch closure, there is no difference in the circuits from the final PURE DC condition.

If you connect a diode reverse-biased across the far end of the winding and the opposite bulb terminal you can also see a small flicker when you open the switch and the field collepses. The field energy is released a pulse of emf.
 
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