Part 3 · Chapter 14

Inductance and Magnetic Energy

Every coil remembers the current that made it. Inductance is that memory written as a number — a circuit's reluctance to let its own magnetic field change — and the energy it stores lives not in the wire but in the empty space the field fills. From flux linkage to the henry, this is where geometry becomes a circuit element.

Electromagnetic Field Theory Prof. Mithun Mondal Reading time ≈ 50 min
i What you'll learn
  • Self-inductance \(L\) as flux linkage per ampere — and why it is purely geometric.
  • Mutual inductance \(M\), the reciprocity \(M_{12}=M_{21}\), and the coupling coefficient \(k\).
  • Inductance formulas for the solenoid, toroid, and coaxial line.
  • The split into internal and external inductance, and what skin effect does to it.
  • Energy stored in an inductor, \(\tfrac12 LI^2\), and the field energy density \(B^2/2\mu\).
  • Combining coupled coils in series and the energy method for finding \(L\).
Section 14-1

Self-Inductance

A current \(I\) in a circuit sets up a magnetic flux \(\Phi\) that threads the circuit itself. If the loop is wound \(N\) times, each turn catches that flux, and the total flux linkage is \(\Lambda = N\Phi\). The ratio of linkage to the current that produced it is the self-inductance:

Self-inductance (henrys)
\[ L = \frac{\Lambda}{I} = \frac{N\Phi}{I} \]

Because \(B\) is proportional to \(I\), the current cancels: \(L\) depends only on the geometry of the circuit and the permeability of the medium — never on how much current flows (for linear materials). It is the magnetic counterpart of capacitance from Chapter 8, where \(C=Q/V\) was likewise pure geometry. The unit is the henry (Wb/A = V·s/A).

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Inductance is geometry, not current
\[ L = \frac{N\Phi}{I} = \frac{1}{I}\int_S \vec{B}\cdot d\vec{S}\times N \]

Double the current and the flux doubles too, so \(L\) is unchanged. The henry is enormous: practical inductors run from nanohenries (a straight wire) to henries (an iron-cored choke). Inductance only reveals itself dynamically — through the back-emf \(v = L\,dI/dt\) you will meet in Chapter 15.

Section 14-2

Mutual Inductance

Now place a second circuit nearby. Some of the flux produced by current \(I_1\) in circuit 1 links the \(N_2\) turns of circuit 2. The mutual inductance measures that shared linkage:

Mutual inductance and reciprocity
\[ M_{21} = \frac{N_2\,\Phi_{21}}{I_1}, \qquad M_{12} = M_{21} = M \]

The equality \(M_{12}=M_{21}\) is not obvious but always holds; it falls straight out of the symmetric Neumann formula \(M = \dfrac{\mu}{4\pi}\oint\!\oint \dfrac{d\vec{l}_1\cdot d\vec{l}_2}{R}\). How tightly the two coils share their flux is captured by the dimensionless coupling coefficient \(k\):

Coupling coefficient
\[ M = k\sqrt{L_1 L_2}, \qquad 0 \le k \le 1 \]

With \(k=1\) every flux line links both coils (a perfectly coupled, tightly wound transformer); with \(k\to 0\) the coils ignore each other. Mutual inductance is the entire basis of the transformer and of every coupled-resonator and wireless-power link.

Section 14-3

Computing Inductance

The recipe is always the same: assume a current \(I\), find \(\vec{B}\) (Chapters 11–12), integrate for the flux \(\Phi\), multiply by \(N\), and divide by \(I\). Applied to the standard geometries it gives a short table worth memorising:

Φ I N turns, length ℓ, area A
Solenoid: axial flux Φ links every one of the N turns → L = μN²A/ℓ
GeometryInductanceNotes
Long solenoid\(L = \dfrac{\mu N^2 A}{\ell}\)\(N\) turns, length \(\ell\), cross-section \(A\)
Toroid (mean radius \(\rho_0\))\(L = \dfrac{\mu N^2 A}{2\pi\rho_0}\)Flux confined to the core; little leakage
Coaxial cable (per metre)\(L' = \dfrac{\mu}{2\pi}\ln\dfrac{b}{a}\)External part; \(a\), \(b\) are inner/outer radii
Two-wire line (per metre)\(L' = \dfrac{\mu}{\pi}\ln\dfrac{d}{a}\)Separation \(d \gg\) wire radius \(a\)

Notice the family resemblance to the capacitance results of Chapter 8: the coax has \(C' = 2\pi\varepsilon/\ln(b/a)\) and \(L' = (\mu/2\pi)\ln(b/a)\), so their product \(L'C' = \mu\varepsilon\) is independent of the conductor sizes — a fact that returns with a vengeance when we reach transmission lines in Part 6.

Section 14-4

Internal & External Inductance

The flux that produces inductance lives in two places: outside the conductor (external inductance) and inside it, where the current itself threads its own field (internal inductance). For a long solid wire carrying a uniform current, integrating the field energy across the cross-section gives a result that is famously independent of the wire's radius:

Internal inductance of a round wire (per metre)
\[ L'_{\text{int}} = \frac{\mu}{8\pi} \quad\left(=\;5\times10^{-8}\ \text{H/m for }\mu=\mu_0\right) \]
a b external flux (a<ρ<b) internal
Coax cross-section: internal flux inside the core, external flux between a and b

The total is \(L = L_{\text{int}} + L_{\text{ext}}\). At low frequency both matter; but as frequency rises the skin effect crowds the current onto the conductor surface, hollowing out the interior. With no current deep inside there is no internal flux, so \(L_{\text{int}}\to 0\) and the inductance approaches its external value alone — one reason high-frequency line inductance is slightly lower than the DC figure.

Section 14-5

Magnetic Energy

Building a current from zero to \(I\) takes work, because the rising field induces a back-emf that opposes the change. Integrating that power against the linear \(\Lambda = LI\) relation gives the energy stored — the area of the shaded triangle below:

I Λ slope = L W = ½LI²
Energy = area under the Λ–I line = ½ΛI = ½LI²
Energy stored in an inductor
\[ W_m = \tfrac12 L I^2 = \tfrac12 \Lambda I = \frac{\Lambda^2}{2L} \]

But the energy does not really sit "in the coil" — it sits in the field, spread through all the space the field occupies. Rewriting the same energy as a volume integral gives the magnetic energy density, the exact twin of the electric \(\tfrac12\varepsilon E^2\) from Chapter 8:

Magnetic energy density and total energy
\[ w_m = \tfrac12\,\vec{B}\cdot\vec{H} = \frac{B^2}{2\mu} = \tfrac12\mu H^2, \qquad W_m = \int_V w_m\,dV \]
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Two views of the same energy
\[ \tfrac12 L I^2 \;=\; \int_V \frac{B^2}{2\mu}\,dV \]

Equating the circuit view (left) with the field view (right) is the energy method for finding inductance: compute the stored field energy, then read off \(L = 2W_m/I^2\). It often beats chasing flux linkages directly, especially for internal inductance.

Section 14-6

Coupled Coils

When two coils share flux, the total stored energy is no longer a simple sum — the cross term carries the sign of the coupling:

Energy of two coupled coils
\[ W_m = \tfrac12 L_1 I_1^2 + \tfrac12 L_2 I_2^2 \pm M I_1 I_2 \]

Connect the same two coils in series and the equivalent inductance depends on whether their fields add or oppose — set by the winding sense (the dot convention):

Series-connected coupled coils
\[ L_{\text{eq}} = L_1 + L_2 \pm 2M \qquad(+\text{ aiding},\ -\text{ opposing}) \]
One method, many devices. The same energy bookkeeping that gives \(L_{\text{eq}}\) also yields the force or torque on the magnetic parts — take the derivative of \(W_m\) with respect to position at constant flux, exactly as the dipole torque emerged in Chapter 13. Coupled-coil energy is what designers balance inside transformers, coupled resonators, energy-storage inductors, and the pulse-forming magnetics of high-voltage systems.
Section 14-7

Worked Examples

1 Solenoid inductance

Problem. An air-cored solenoid has \(N=500\) turns, cross-section \(A=4\ \text{cm}^2\), and length \(\ell=0.25\ \text{m}\). Find \(L\).

Solution. Apply \(L=\mu_0 N^2 A/\ell\):

Working
\[ L = \frac{(4\pi\times10^{-7})(500)^2(4\times10^{-4})}{0.25} \approx 0.50\ \text{mH} \]
2 Coaxial cable inductance

Problem. A coax has inner radius \(a=1\ \text{mm}\), outer \(b=4\ \text{mm}\), air dielectric. Find the external inductance per metre.

Solution. Use \(L'=(\mu_0/2\pi)\ln(b/a)\):

Working
\[ L' = \frac{4\pi\times10^{-7}}{2\pi}\ln 4 = (2\times10^{-7})(1.386) \approx 0.28\ \mu\text{H/m} \]
3 Energy in an inductor

Problem. A \(50\ \text{mH}\) inductor carries \(2\ \text{A}\). How much energy is stored?

Solution. Use \(W=\tfrac12 L I^2\):

Working
\[ W = \tfrac12 (0.05)(2)^2 = 0.10\ \text{J} \]
4 Magnetic energy density

Problem. Find the energy density in air where \(B=1.0\ \text{T}\).

Solution. Use \(w_m=B^2/2\mu_0\):

Working
\[ w_m = \frac{(1.0)^2}{2(4\pi\times10^{-7})} \approx 3.98\times10^{5}\ \text{J/m}^3 \]
5 Coupling coefficient

Problem. Two coils have \(L_1=10\ \text{mH}\), \(L_2=40\ \text{mH}\), and \(k=0.6\). Find the mutual inductance.

Solution. Use \(M=k\sqrt{L_1 L_2}\):

Working
\[ M = 0.6\sqrt{(10)(40)} = 0.6(20) = 12\ \text{mH} \]
6 Toroid inductance

Problem. A toroid (\(\mu_r=500\), mean radius \(\rho_0=5\ \text{cm}\), area \(A=3\ \text{cm}^2\)) is wound with \(N=200\) turns. Find \(L\).

Solution. Use \(L=\mu_0\mu_r N^2 A/2\pi\rho_0\):

Working
\[ L = \frac{(4\pi\times10^{-7})(500)(200)^2(3\times10^{-4})}{2\pi(0.05)} \approx 24\ \text{mH} \]
Review

Chapter Summary

Self-inductance

\(L=N\Phi/I\); pure geometry and \(\mu\); unit henry; the magnetic twin of \(C\).

Mutual inductance

\(M=N_2\Phi_{21}/I_1\); \(M_{12}=M_{21}\); \(M=k\sqrt{L_1L_2}\), \(0\le k\le1\).

Standard formulas

Solenoid \(\mu N^2A/\ell\); coax \((\mu/2\pi)\ln(b/a)\) per metre; \(L'C'=\mu\varepsilon\).

Internal & external

\(L=L_{\text{int}}+L_{\text{ext}}\); wire \(L'_{\text{int}}=\mu/8\pi\); skin effect kills \(L_{\text{int}}\).

Magnetic energy

\(W=\tfrac12 LI^2\); density \(w_m=B^2/2\mu\); energy method \(L=2W/I^2\).

Coupled coils

\(W=\tfrac12 L_1I_1^2+\tfrac12 L_2I_2^2\pm MI_1I_2\); series \(L_1+L_2\pm2M\).

Practice

Problems

For each item, decide first whether it is a flux-linkage, a standard-formula, an energy, or a coupling problem, then apply the matching relation. Difficulty rises down the list.

  1. A coil of \(200\) turns links \(0.4\ \text{mWb}\) of flux when it carries \(2\ \text{A}\). Find its self-inductance.
  2. An air-cored solenoid is \(0.3\ \text{m}\) long with \(800\) turns and cross-section \(5\ \text{cm}^2\). Find \(L\).
  3. A coaxial cable has \(a=2\ \text{mm}\), \(b=6\ \text{mm}\), air dielectric. Find the external inductance per metre.
  4. An inductor stores \(0.45\ \text{J}\) when carrying \(3\ \text{A}\). Find its inductance.
  5. Find the magnetic energy density in air where \(H=2000\ \text{A/m}\).
  6. Two coils have \(L_1=20\ \text{mH}\), \(L_2=45\ \text{mH}\), \(M=24\ \text{mH}\). Find the coupling coefficient.
  7. A toroidal core (\(\mu_r=1000\), mean radius \(4\ \text{cm}\), area \(2\ \text{cm}^2\)) carries \(300\) turns. Find \(L\).
  8. The two coils of Problem 6 are connected in series. Find \(L_{\text{eq}}\) for both aiding and opposing connections.
  9. A long solid wire of radius \(1\ \text{mm}\) carries \(10\ \text{A}\). Find its internal inductance per metre and the internal stored energy per metre.
  10. For a coax with \(a=1\ \text{mm}\), \(b=5\ \text{mm}\), filled with a dielectric of \(\varepsilon_r=2.25\) and \(\mu_r=1\), verify that \(L'C'=\mu\varepsilon\).
  11. Starting from the back-emf \(v=L\,dI/dt\), derive \(W=\tfrac12 LI^2\) by integrating the input power.
  12. Explain physically why the internal inductance of a round wire is independent of its radius, while the external inductance is not.
Tip: almost every inductance problem reduces to one of two moves — flux linkage (\(L=N\Phi/I\): assume \(I\), find \(B\), integrate, divide) or energy (\(L=2W_m/I^2\): integrate \(B^2/2\mu\)). Reach for the energy method whenever the field is messy or when internal inductance is involved. For mutual inductance, remember it is symmetric (\(M_{12}=M_{21}\)) and bounded (\(k\le1\)), and that everything here gains its dynamical meaning only once Faraday's law in Chapter 15 turns \(L\) into a voltage \(v=L\,dI/dt\).