Self-Inductance of a Solenoid

Setting up the solenoid. Consider a long solenoid of length $l$, cross-sectional area $A$, wound with a total of $N$ turns, so that it has $n=N/l$ turns per unit length. When it carries a current $I$, the magnetic field WELL INSIDE a long solenoid (far from its two ends, so edge effects can be neglected) is uniform and given by

directed along the solenoid's axis, where $\mu_0=4\pi\times10^{-7}\ \text{T m/A}$ is the permeability of free space.

Flux linkage. The flux through ONE turn of the solenoid is simply this field times the cross-sectional area, $\Phi_B=BA=\mu_0 nIA$. Since ALL $N=nl$ turns link essentially this same flux (the field is uniform along the solenoid's interior), the TOTAL flux linkage is

Extracting $L$. Comparing this with the defining relation $N\Phi_B=LI$ from Section 6.7 gives, directly,

Reading the formula. This result depends ONLY on the solenoid's geometry -- how tightly it is wound ($n$), its cross-sectional area ($A$), and its length ($l$) -- and on the medium filling its core; it does not depend on the current $I$ at all, exactly as an inductance should not (inductance is a property of the COIL, defined so that it stays fixed while the current varies). Writing $n=N/l$ shows the equivalent, and often more directly useful, form

which makes explicit that $L$ grows with the SQUARE of the number of turns -- doubling the number of turns on a given solenoid (keeping its length and area fixed) quadruples its self-inductance, since each of the doubled turns both produces twice the field AND links twice as many turns. …