Initially, our solenoid without the iron core has self-inductance given by:`L = mu_0 N^2 pi R^2 l =( mu_0 N^2 A)/l` Where `mu_0` is the vacuum permeability (for air, we can assume the same value for most practical uses), `N` is the number of windings of our solenoid, `R` its radius, `A` the cross-sectional area and `l` its length.Suppose that we now insert an iron core inside our solenoid. Let's assume that the...
Initially, our solenoid without the iron core has self-inductance given by:
`L = mu_0 N^2 pi R^2 l =( mu_0 N^2 A)/l`
Where `mu_0` is the vacuum permeability (for air, we can assume the same value for most practical uses), `N` is the number of windings of our solenoid, `R` its radius, `A` the cross-sectional area and `l` its length.
Suppose that we now insert an iron core inside our solenoid. Let's assume that the iron core has the same cross-sectional area as our solenoid.
Now, we know that iron has a permeability `mu` higher than the vacuum permeability `mu_0`. So, the inductance of our "new" solenoid becomes:
`L' = (kappa mu_0 N^2 A)/l = kappa L > L`
Where `kappa = mu/mu_0` . Since `kappa gtgt 1` , we see that the self-inductance of our solenoid has increased!
To see the direct effect of increasing the self-inductance, suppose that the solenoid was connected to a circuit consisting of an AC (alternating current) voltage source and a lamp. Suppose also that the lamp is initially bright for our configuration. For AC circuits, inductors (like our solenoid) act as resistances (the self-inductance is a resistance to the change of current passing through). So when we add an iron core to our solenoid, the inductance increases and so the lamp dims (since the permeability of iron is many times greater than that of the vacuum and of air, the lamp might even turn off)!
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