05 · Advanced

Inductance and magnetic energy

Define oriented magnetic flux linkage and self/mutual inductance; derive reversible U=∫I dλ and ½LI² for fixed linear geometry, compute an ideal long-solenoid inductance and field-energy cross-check, obtain a rectangular-cross-section toroid's logarithmic result, and distinguish reciprocity, passivity, loss, nonlinear magnetic media and dispersion.

270–330 minAdvanced
In this lesson
  • Define oriented flux Φ=∫B·dS and linkage λ=NΦ, with sign tied to winding orientation.
  • Derive λ=LI for a fixed linear magnetic system and state why L depends on geometry and constitutive assumptions.
  • Derive dU=I dλ during reversible quasistatic charging and integrate U=½LI² for constant L.
  • Derive B≈μ₀NI/ℓ and L≈μ₀N²A/ℓ for a long ideal tightly wound solenoid with negligible fringing.
  • Cross-check solenoid circuit energy against vacuum field-energy density u=B²/(2μ₀) and control units.
  • Derive B(r)=μ₀NI/(2πr) and L=μ₀N²h/(2π)ln(b/a) for an ideal vacuum toroid of rectangular cross-section.
  • Define λ₁=L₁I₁+M₁₂I₂ and λ₂=M₂₁I₁+L₂I₂ and give conditions for reciprocal M₁₂=M₂₁.
  • Derive U=½L₁I₁²+½L₂I₂²+MI₁I₂ and the passivity bound |M|≤√(L₁L₂) for a linear reciprocal passive system.
  • Explain why ideal zero-radius filament self-inductance can be singular and why finite cross-sections and return paths matter.
  • State the assumptions behind u=½B·H and reversible nonlinear u=∫H·dB and caution against direct complex-μ(ω) substitution in dispersive/lossy media.

Flux linkage makes the magnetic field of a stationary current into a reproducible circuit quantity. The familiar ½LI² energy is a result of reversible charging in a fixed LINEAR system, not a universal law for arbitrary geometry, hysteretic magnetic media or frequency-dependent response. We derive the field and circuit descriptions side by side and check their common assumptions.

0105.1 — Signed flux, turns, and self-inductance

For each closed winding orient its spanning surface by the right-hand rule relative to the chosen positive current direction. Its magnetic flux is Φ=∫B·n̂dS; reversing the winding orientation reverses signed Φ. With N identical tightly coupled turns, flux linkage is λ=NΦ; for unequal turns use λ=ΣkΦk.

For fixed geometry, linear vacuum magnetostatics implies B∝I, hence λ=LI and L=λ/I for nonzero I. Inductance describes that complete current-return geometry and surrounding material, not an isolated undefined open segment. In general a multiconductor structure requires an inductance matrix.

Φ=∫S B·n̂dS; λ=Σk Φk = NΦ (equal turns)
λ=LI; [L]=Wb/A=H=V·s/A

0205.2 — Derive energy from reversible quasistatic charging

Faraday's induced emf opposes the change in linked flux: ε_ind=−dλ/dt for a fixed oriented winding. To build the current quasistatically, an external source must supply power I dλ/dt against that back-emf, neglecting resistive dissipation, motion and radiation. Therefore incremental STORED magnetic energy is dU=I dλ.

With fixed linear L, λ=LI, so U=∫₀ᴵ i L di=½LI². The ½ arises from the linear growth of flux linkage during charging. If L varies with current, geometry or magnetic history, return to the work integral and explicitly account for mechanical/source work.

εind=−dλ/dt; dU=I dλ (reversible, fixed geometry)
U=∫₀ᴵ i L di=½LI2 (constant L)
[U]=H·A2=Wb·A=J

0305.3 — Long ideal solenoid: derive B and L

Long solenoid: current orientation and approximate axial BCross-sectional projection of a long solenoid with its axis along positive x. Five turns appear as wire sections above and below the central region; dots on the top show conventional current out of page (+z), crosses on the bottom show current into page (−z). The approximately uniform interior magnetic field B points right along positive x.B : +x →I : +z ⊙I : −z ⊗
n = N/ℓ
B ≈ μ0nI
L ≈ μ0N2A/ℓ
ℓ ≫ √A
Long solenoid: current orientation and approximate axial B

With winding current +z out of page on TOP and −z into page at the BOTTOM, the right-hand rule yields B∥+x along the solenoid interior. For N tightly wound turns across length ℓ, n=N/ℓ, the LONG ideal vacuum-solenoid approximation gives Binside≈μ0nI and L≈μ0N2A/ℓ, with nearly uniform interior flux and negligible end/fringe fields. These are physical wire cross sections, not stretchable inductor circuit symbols.

Consider a tightly wound, long vacuum solenoid of length ℓ, N turns, transverse cross-section area A and current I. Assume ℓ much greater than its transverse size, uniform turn density n=N/ℓ and negligible end/fringe field. Ampère's rectangular path with one long leg inside and the return outside then gives approximately Binside=μ₀nI along the solenoid axis and Boutside≈0.

If all N turns intercept approximately the same interior flux Φ≈BA, the linkage is λ=NBA. Dividing by I gives L≈μ₀N²A/ℓ. This approximation is not exact for finite solenoids or tightly packed coils with strong external/return fields.

n=N/ℓ; Binside≈μ0nI=μ0NI/ℓ
Φper turn≈BA; λ=NΦ≈μ0N2IA/ℓ
L≈μ0N2A/ℓ

0405.4 — Vacuum magnetic field energy and the solenoid cross-check

For localized finite-energy stationary currents in vacuum, the magnetic field energy can be written U=(1/(2μ₀))∫all space B²d³r, with boundary terms and wire self-field singularities treated correctly. The ideal solenoid assumes field concentrated in interior volume Aℓ, so its integral becomes U≈(B²/(2μ₀))Aℓ.

Substituting B≈μ₀NI/ℓ gives U≈μ₀N²AI²/(2ℓ)=½LI². These are TWO REPRESENTATIONS of the SAME energy and must not be added. A real coil has finite exterior/fringing and finite wire fields; its total-energy integral includes those regions.

uB(vacuum)=B2/(2μ0); U=∫uB d3r
Usolenoid≈(B2/(2μ0))Aℓ≈μ0N2AI2/(2ℓ)=½LI2

0505.5 — Numerical solenoid check: flux, inductance and energy

Take N=1000 turns, ℓ=0.500 m, A=1.00×10⁻³ m² and I=0.200 A. The long-solenoid approximation gives B≈5.03×10⁻⁴ T, L≈2.51×10⁻³ H, and U=½LI²≈5.03×10⁻⁵ J.

The interior volume Aℓ=5.00×10⁻⁴ m³ and B²/(2μ₀) reproduce the same ≈50.3 μJ. The unit check [B²/μ₀] is J/m³, since 1 T=A-compatible N/(A·m) and μ₀ has N/A².

B≈0.503 mT; L≈2.51 mH
Ucircuit≈50.3 μJ; Ufield≈50.3 μJ

0605.6 — Concentric toroid with rectangular cross-section

Consider an ideal densely wound toroidal vacuum coil with N turns, inner radius a, outer radius b>a and axial core height h. With rotational symmetry and an Ampèrian circle at radius r within its annular cross-section, Bφ(r)(2πr)=μ₀NI. Treat external leakage and finite winding gaps as negligible.

The flux per turn through the rectangular cross-section is Φ=∫aᵇBφ(r)h dr=μ₀NIh/(2π)ln(b/a). Multiply by N and divide by I to obtain L=μ₀N²h/(2π)ln(b/a). Using a single B across a broad annulus instead of integrating its 1/r variation is generally wrong.

a<r<b: Bφ(r)=μ0NI/(2πr)
Φper turn=μ0NIh/(2π) ln(b/a)
L=μ0N2h/(2π) ln(b/a)

0705.7 — Two coupled circuits, reciprocity and energy positivity

Choose the winding orientations and currents once. For a fixed linear two-coil system write λ₁=L₁I₁+M₁₂I₂ and λ₂=M₂₁I₁+L₂I₂. In a passive linear reciprocal magnetic environment with compatible orientations, M₁₂=M₂₁=M, which may be positive or negative depending on the winding convention.

Incremental reversible work dU=I₁dλ₁+I₂dλ₂ then integrates to U=½L₁I₁²+½L₂I₂²+MI₁I₂. Nonnegative field energy for all current pairs requires L₁≥0, L₂≥0 and M²≤L₁L₂. A nonreciprocal, hysteretic or time-modulated material cannot be presumed to satisfy the same scalar-M model without checking its assumptions.

λ1=L1I1+MI2; λ2=MI1+L2I2
U=½L1I1²+½L2I2²+MI1I2
passive reciprocal linear case: |M|≤√(L1L2)

0805.8 — Why a zero-radius filament is not a finite self-inductor

An ideal filament produces B≈μ₀I/(2πs) near its axis. Its magnetic energy per unit length in an annulus scales as ∫(B²/(2μ₀))2πs ds∝∫ds/s and diverges logarithmically as the inner radius tends to zero. An infinite isolated wire also creates an unbounded far-field energy per length when no return path sets a finite outer scale.

Physical conductors have finite cross-sections, spatial current distributions and return-current geometry. A numerical self-inductance calculation must state those dimensions and its high-frequency skin-current assumptions. Do not identify a divergent ideal self-field with a measured finite coil inductance.

Bfilament∝1/s ⇒ Uper length∝∫(ds/s)
finite cross-section + return path + field distribution are required

0905.9 — Magnetic materials and high-frequency energy limits

In a local, linear, lossless, nondispersive medium under reversible quasistatic excitation, B=μH and u=½B·H. If the magnetization curve is nonlinear but reversible, the charging-work density follows u=∫₀ᴮ H(B′)·dB′ along the chosen path. Hysteresis can make the work path dependent and include dissipated heat.

A frequency-dependent complex permeability μ(ω) cannot simply be substituted into ½B·H as a generic THz stored-energy formula. Dispersion, loss, kinetic/material degrees of freedom and electromagnetic power flow require a broader treatment; neither the static L nor the static u is automatically an exact high-frequency device parameter.

linear reversible static: uB=½B·H
nonlinear reversible quasistatic: uB=∫₀ᴮ H(B′)·dB′

05.10 — Six mastery exercises with worked solutions

  1. Derive L≈μ₀N²A/ℓ for a long vacuum solenoid from B≈μ₀NI/ℓ and oriented N-turn flux linkage.
  2. For N=1000, ℓ=0.500 m, A=1.00×10⁻³ m², I=0.200 A, calculate B, L, U and verify field/circuit energy equality.
  3. For an ideal vacuum toroid with N=500, a=0.020 m, b=0.040 m and h=0.010 m, compute L from the logarithmic cross-section integral.
  4. Given L₁=4.00 mH and L₂=9.00 mH, what is the allowed |M| for a passive reciprocal linear system? Does M=7.00 mH satisfy the energy requirement?
  5. For L₁=4.00 mH, L₂=9.00 mH, M=+3.00 mH, I₁=1.00 A and I₂=+0.500 A, calculate U; repeat for I₂=−0.500 A.
  6. Explain why the self-energy of a mathematical zero-radius current filament diverges and why using a complex μ(ω) in static ½B·H is not a general energy calculation.
Worked solutions

1. Φ≈BA≈μ₀NIA/ℓ, λ=NΦ≈μ₀N²IA/ℓ and L=λ/I≈μ₀N²A/ℓ, with the long-coil/fringe-free assumptions.

2. B≈0.503 mT, L≈2.51 mH, U≈50.3 μJ. The field density B²/(2μ₀) integrated over volume Aℓ≈5.00×10⁻⁴ m³ gives the same ≈50.3 μJ.

3. L=μ₀N²h/(2π)ln(b/a)=2×10⁻⁷×(500²)×0.010×ln2≈3.47×10⁻⁴ H=0.347 mH, assuming ideal toroid.

4. |M|≤√(4.00×9.00) mH=6.00 mH. M=7.00 mH violates the nonnegative quadratic energy condition for this passive reciprocal model.

5. U=½(0.004)(1)²+½(0.009)(0.500)²+(0.003)(1)(0.500)=0.004625 J=4.625 mJ. Reversing I₂ changes only the mutual term, giving 0.001625 J=1.625 mJ.

6. For B∝1/s the energy integral near the wire is ∫B²s ds∝∫ds/s, divergent at zero radius. A lossy/dispersive μ(ω) requires material and loss dynamics in the energy balance, not a static algebraic replacement.

EM-02 now spans force laws, steady-current source integrals, Ampère's circulation, vector potential and magnetic energy. EM-03 will derive the complete time-dependent Maxwell system, including Faraday induction, displacement current, continuity and electromagnetic power flow.