Derive inductance from flux linkage and energy

Inductance describes a response to changing current

A current produces magnetic flux linked to a circuit. For fixed linear geometry, flux linkage is proportional to current: N Phi = L I. The induced emf is −d(N Phi)/dt = −L dI/dt. The minus sign describes opposition to change, not opposition to every current. An inductor can sustain an existing current while returning its stored energy to another element.

Derive the long-solenoid geometry

For an ideal long solenoid of length ell, area A, N turns and constant core permeability mu, B = mu N I/ell. Multiplying its per-turn flux B A by N gives L = mu N² A/ell. Doubling turns at fixed length and area multiplies inductance by four. Doubling length at fixed total turns halves it. Keeping turns per length fixed while changing length is a different constraint and gives a different scaling.

Derive energy rather than memorize a second formula

Under passive sign convention the terminal voltage is vL = L dI/dt, so absorbed power is vL I. Integrating from zero current to I gives U = integral L I dI = L I²/2 for constant L. The ideal field-energy density B²/(2mu), multiplied by A ell, gives the same result. A nonlinear or hysteretic magnetic core cannot silently use a constant-L energy formula through every state.

Separate circuit approximations from physical universals

Connecting wires are often modeled with zero inductance, while a wound element is assigned significant inductance. Real wires and their return paths do possess inductance. Long-solenoid uniformity and negligible exterior field are approximations; finite end fields, winding resistance and changing core temperature can matter. Stored energy can later become resistor heat or capacitor electric energy, and a proper account must include every destination.

Worked original case

For N = 200, A = 0.01 m², ell = 1 m and the stipulated mu0 ≈ 4pi×10⁻⁷, L ≈ 0.000502655 H. At I = 2 A, B ≈ 0.000502655 T and U ≈ 0.00100531 J. A positive 1 A/s ramp produces induced emf −0.000502655 V, while the passive voltage drop has the opposite sign.

Investigation

Derive flux linkage, inductance and energy from geometry. Compare fixed-total-turns and fixed-turn-density length changes, and retain the field-energy cross-check. Explain how a nonlinear core or a resistive winding would invalidate one of the ideal reductions without implying energy nonconservation.

Changed-case practice

Try a new practice case

If total turns double and length doubles while area and permeability stay fixed, what is new L divided by old L?

Numerical result in the requested units Check this practice result

This public, unsaved check is not a secure assessment. Explain the source integral, symmetry or charge boundary in the retained course evidence workspace.

Sources and boundaries

Original teaching checked 31 October 2026 against current official Unit 13 and the relevant OpenStax discussion. No source exercises or diagrams are reproduced. All records and apparatus are virtual, supplied or generated; no real wiring, switching, magnetic braking or electrical measurement is requested. Public feedback and retained work do not establish independent mastery, supervised practical participation or AP authorization.

The circuit and field-energy comparison is also checked against magnetic field energy.