The Inductance of a Solenoid

A solenoid resists changes in current because changing current changes the magnetic flux through its own coils, and that changing flux induces a back voltage. Inductance L measures how strongly a device does this: L = (flux linkage)/(current) = N phi / I. For a long solenoid the field inside is uniform, B = mu0 n I, where n = N/l is turns per unit length. The flux through one loop is phi = B A = mu0 (N/l) I A. Multiply by N turns to get the total flux linkage, then divide by I, and the I cancels: L = mu0 N² A / l.

That N² is the whole story of the intuition question. Doubling N doubles B, since there are twice as many loops per length driving the field. It also doubles how many loops that stronger field links. Two factors of two multiply to four. Nothing else in the formula has that compounding effect: A and l enter only once, linearly, because they affect either the field or the linkage, not both.

This formula sets up everything that follows in the unit. Once you can compute L, you can treat a solenoid as a circuit element with voltage V = L dI/dt, the inductive twin of V = IR. Slide an iron core inside and mu0 gets replaced by a much larger permeability mu, since iron concentrates field lines far more effectively than air or vacuum, boosting L by a factor of hundreds or thousands. That same L shows up next in LC and RL circuits, where it trades energy with capacitors and resistors instead of just sitting there.