A long air-core solenoid has cross-sectional area $A$ and $N$ loops of wire on its length $d$. (a) Find its self-inductance. (b) What is its inductance if the core material has a permeability of $\mu^{\prime}$ ?
(a) We can write
$$
|\varepsilon|=N\left|\frac{\Delta \Phi_s}{\Delta t}\right| \quad \text { and } \quad|\varepsilon|=L\left|\frac{\Delta i}{\Delta t}\right|
$$
Equating these two expressions for $|\epsilon|$ yields
$$
L=N\left|\frac{\Delta \Phi_M}{\Delta i}\right|
$$
If the current changes from zero to $I$, then the flux changes from zero to $\Phi_M$. Therefore, $\Delta i=I$ and $\Delta \Phi_M=\Phi_M$ in this case. The self-inductance, assumed constant for all cases, is then $$
L=N \frac{\Phi_M}{I}=N \frac{B A}{I}
$$
But for an air-core solenoid, $B=\mu_0 A I=\mu_0(N / d) I$. Substitution gives $L=\mu_0 N^2 A / d$.
(b) If the material of the core has permeability $\mu$ instead of $\mu_{\mathrm{r}}$, then $B$, and therefore $L$, will be increased by the factor $\mu / \mu_0$. In that case, $L=\mu N^2 A / d$. An iron-core solenoid has a much higher self-inductance than an air-core solenoid has.