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A current in one circuit links magnetic flux through another. It produces magnetic flux linkage through another. With fixed geometry and a linear reciprocal medium, choose consistent winding orientations and writeThe shared coefficient is mutual inductance; reversal of one winding reverses its sign. Faraday's law of induction gives the mutual induced emf ${\mathcal{E}}_2=-M{\dot{I}}_1$ for fixed $M$. The minus sign expresses Lenz's law.

The magnetic energy isNonnegative energy for every current pair requires $|M|\le \sqrt{L_1{L}_2}$, motivating the coupling coefficient $M/\sqrt{L_1{L}_2}$. Transformers use strong common flux to exchange energy between circuits; leakage flux reduces coupling. A shared core increases linkage, but saturation, hysteresis and eddy-current losses limit the linear description. A moving geometry also contributes $-I_1\dot{M}$ to the mutual emf.

Electromagnetic induction produces an electromotive force when magnetic flux linkage changes. For an oriented moving loop,For a stationary loop this reduces to Faraday's law of induction, $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$. Motion in a static field can instead supply the motional term. Lenz's law gives the minus sign: the induced electric current opposes the flux change, with the required electrical energy supplied by mechanical work or the field source.

For linear fixed-geometry circuits, flux linkages satisfy ${\Lambda }_1=L_1{I}_1+M{I}_2$ and ${\Lambda }_2=M{I}_1+L_2{I}_2$. Self-inductance gives ${\mathcal{E}}_1=-L_1{\dot{I}}_1$; mutual inductance gives the additional term $-M{\dot{I}}_2$. Reciprocity gives the same mutual coefficient for both circuits, with its sign determined by winding orientation. The stored field energy isThe bound follows from nonnegative energy for arbitrary electric currents. In a linear medium the equivalent field expression is $W_B=\frac{1}{2}\int \mathbf{B}\cdot \mathbf{H}dV$.

An ideal electrical transformer links both windings with one changing core flux, giving $V_1/V_2=N_1/N_2$ for corresponding voltage magnitudes. A loaded secondary drives an opposing magnetomotive force and increases primary electric current; ideally $I_1/I_2=N_2/N_1$ and input and output powers agree. A electrical transformer exchanges voltage for electric current, not for extra power. Real devices have winding resistance, leakage flux, eddy-current and hysteresis losses, and finite magnetizing electric current. A steady DC input gives no continuing secondary voltage after the switching transient and may cause excessive primary electric current or core saturation.

Inductance and interface conditions. For a linear coil, self-inductance is the ratio $L=\Lambda /I$ of its own magnetic flux linkage $\Lambda =N\Phi$ to current. Its induced electromotive force is $-LdI/dt$ for fixed $L$. Across an interface with unit normal $\mathbf{n}$ from medium 1 to 2,These electromagnetic boundary conditions follow from Gauss's law for magnetism and Ampère's law. Here ${\mathbf{K}}_f$ is free surface current. At the iron–air faces there is no free surface current, so normal magnetic flux density and tangential magnetic field intensity are continuous. In each isotropic linear medium, $\mathbf{B}={\mu }_0{\mu }_r\mathbf{H}$.

Complete core. A slender toroid has approximately uniform $H$ along mean path length $\ell$. Ampère's law gives $H\ell =NI$, so $B=\mu {\mu }_{0}NI/\ell$. The flux per turn is $\Phi =BA$, yielding the inductance of a toroidal winding,This uses constant relative magnetic permeability $\mu$ and neglects leakage, fringing and saturation.

Why the gap lowers inductance. The gap replaces an easy magnetic path in iron with air. Maintaining the same flux now requires a much larger magnetic field intensity in part of the path and therefore more ampere-turns. Equivalently, magnetic reluctance increases, so the flux linkage for a fixed current and hence self-inductance decrease. A narrow gap can matter greatly because its permeability is smaller by the factor $\mu$.

Oblique-gap calculation. In the original diagram $t$ is the perpendicular separation of the cut faces. Let $\mathbf{s}$ point along the core, and let $\mathbf{n},\mathbf{q}$ be normal and tangential to the gap faces. With the face at angle $\theta$ to the core direction, $\mathbf{s}=\sin \theta \mathbf{n}+\cos \theta \mathbf{q}$. Write the core flux density as $B\mathbf{s}$. The boundary conditions givesoThe gap area normal to its nearly normal field is $A/\sin \theta$, not $A$. Its leading reluctance is consequently $t\sin \theta /({\mu }_{0}A)$. Adding the core reluctance gives the inductance of a toroid with an oblique narrow air gap,Thus at $\theta =45^{\circ }$,The approximation retains the possibly order-one ratio $\mu t/\ell$ while neglecting $t/\ell$, fringing and the small tangential gap contribution. More precisely, a local uniform-field path along the core, with missing iron length $t/\sin \theta$ subtracted, gives denominator $\ell /\mu +t\sin \theta (1-1/\mu )$. Its correction is beyond the stipulated large-$\mu$, narrow-gap accuracy; simply neglecting $\mu t$ would not be justified.

Field in the slot. The component expression above givesFor the $45^{\circ }$ gap and $\mu \gg 1$,Its units are $\mathrm{A}{\mathrm{m}}^{-1}$. The field is almost perpendicular to the cut faces, rather than following the iron direction straight across the cut. Its small angle away from the face normal satisfies $\tan \alpha =1/\mu$ at $45^{\circ }$; the sense reverses if the current reverses.

In magnetostatics Ampère's circuital law is $\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_{\mathrm{enc}}$. Its general vacuum version adds ${\mu }_0{ϵ}_{0}d{\Phi }_E/dt$ on the right. Faraday's law is $\mathcal{E}=\oint \mathbf{E}\cdot d\mathbf{l}=-d{\Phi }_B/dt$ for a fixed loop. A circular Amperian path around a long straight wire giveswith direction given by the right-hand rule. The force on a wire element is $d\mathbf{F}=Id\mathbf{l}\times \mathbf{B}$; its magnitude is $IBdl\sin \chi$ and its direction follows this cross product.

Choose the straight-wire current rightward and the square below it, as printed. Its positive magnetic field there is into the page. With an into-page loop normal, the magnetic flux linkage isHere $\mathcal{M}$ denotes mutual inductance, not a mass. Choose positive coil current to produce an out-of-page field. Then its induced current in the resistance-dominated approximation isThis reproduces the specified sign with the declared orientation. It neglects coil self-inductance, requiring $\omega L_{\mathrm{coil}}\ll R$, as implicit in the requested expression.

The nearest side contains N parallel current segments and is at a uniform distance $s$, so its force magnitude isThe two sides perpendicular to the wire have opposite currents and equal magnetic fields at corresponding distances; their forces cancel pairwise. The far side opposes the near-side force and has the smaller field. ThusWhen $I$ and $i$ have the same sign under the chosen conventions, the near-side coil current opposes the straight-wire current and the total force is away from the wire; otherwise it is toward it. In particular the signed force alternates with $\sin (\omega t)\cos (\omega t)$. The coil is fixed, so these forces do no mechanical work on it. The mean Joule heating is

Faraday's law relates induced electromotive force to the rate of change of magnetic flux linkage. For a coil with $N$ turns and a consistent right-hand orientation,Flux can change through field variation or circuit motion. Lenz's law is the minus sign: the induced current's field opposes the flux change, consistent with energy conservation. It opposes the change rather than necessarily opposing the original field.

For a linear fixed-geometry inductor, $N{\Phi }_B=LI$ and the self-induced electromotive force is $-LdI/dt$. The passive terminal voltage drop is instead $V=LdI/dt$. The stored magnetic energy isInductance is measured in henries. Transformers use mutual inductance: changing one winding's current induces an electromotive force in another.

Faraday's law of induction relates an induced electromotive force to the rate of change of magnetic flux. Choose a positive orientation around a loop and its associated surface normal by the right-hand rule. For a stationary conducting loop,Unlike an electrostatic field, the induced electric field generally has nonzero curl; the differential form is $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$. For a moving circuit, magnetic flux transport for a moving circuit includes the magnetic Lorentz force on the charges:where $\mathbf{u}$ is the local wire velocity. Thus changing the field or moving a conductor can produce induction.

The minus sign is Lenz's law: an induced electric current, when a closed conducting path permits it, produces a magnetic field opposing the change of linked flux. It need not oppose the original field itself. For example, increasing flux through a loop produces a current tending to reduce that increase; decreasing the same flux reverses the current. The opposition is consistent with conservation of energy: external mechanical or electrical work is needed to maintain a change against its inductive reaction.

For a coil of $N$ turns, define the magnetic flux linkage $\Lambda ={\sum }_{j=1}^N{\Phi }_j$, reducing to $N{\Phi }_B$ when every turn links the same flux. For fixed geometry and a linear magnetic response, self-inductance $L$ is defined by $\Lambda =LI$, and the coil's induced electromotive force isThe corresponding terminal voltage drop, with the passive sign convention, is $V_L=LdI/dt$. It opposes changes in electric current; it does not require the current itself to vanish.

Building the current from zero requires power $P=I{V}_L=LIdI/dt$. Integrating gives the energy stored by an ideal inductor:This energy is stored in the magnetic field and can be returned when the current falls. In a series RL circuit driven by a constant voltage $V_0$, Kirchhoff's voltage law gives $L\dot{I}+RI=V_0$. Starting with zero current, $I=(V_0/R)(1-e^{-tR/L})$, with time constant $L/R$. This makes the resistance to rapid current changes directly observable. Two coils also have mutual inductance: changing primary current induces secondary electromotive force $-Md{I}_1/dt$ with a polarity determined by the chosen winding orientations, the principle behind a transformer.

Self-inductance is inductance associated with a coil's own electric current, rather than magnetic flux supplied by another coil. In the linear model, fixed geometry and magnetic response give constant $L$ in $\Lambda =LI$. With the turn orientations defined in magnetic flux linkage, Faraday's law of induction gives the self-induced electromotive forceThe final equality requires constant $L$; if the geometry changes, use $-d\Lambda /dt$. The minus sign expresses Lenz's law: the induced electromotive force opposes the change in electric current. It does not always oppose the electric current itself. When positive electric current decreases, the induced electromotive force acts in its direction. ^[ref].

For terminal sign bookkeeping, label the ideal inductor's terminals $a,b$, choose positive electric current entering $a$ and leaving $b$, and define $v_L$ as the terminal electric potential difference drop from $a$ to $b$ in the lumped electrical circuit model. This passive convention givesThus a coil with self-inductance $0.5\mathrm{H}$ and increasing electric current $dI/dt=12A/s$ has terminal drop $v_L=6\mathrm{V}$ and induced electromotive force ${\mathcal{E}}_{\mathrm{ind}}=-6\mathrm{V}$. These signs describe the same process with different reference conventions. The RL circuit uses the terminal-drop convention throughout.

A finite terminal electric potential difference cannot make the electric current of an ideal inductor with constant $L>0$ jump. Integrating its differential equation givesIf $v_L$ stays bounded, the integral tends to zero as the interval shrinks, ensuring continuity of electric current. A step in terminal electric potential difference changes $dI/dt$ immediately, while electric current remains continuous. At constant electric current, the self-induced electromotive force is zero, although the inductor's magnetic energy can remain nonzero.

For a long vacuum-core or air-core solenoid with fixed geometry and linear magnetic response, neglecting end effects gives self-inductanceHere $N$ is the total number of turns, $n=N/\ell$ is the solenoid turn density, and $A$ is the interior cross-sectional area. The reason for the square of $N$ is physical: more turns produce more magnetic field, and more turns link the resulting magnetic flux. Approximately equal magnetic flux $\Phi \approx BA$ through each turn gives magnetic flux linkage $\Lambda \approx N\Phi \approx {\mu }_0{N}^{2}AI/\ell$; division by the electric current yields $L$. ^[ref].

For the same $N=1000$, $\ell =0.50\mathrm{m}$, and $a=0.025\mathrm{m}$, the area is $A=\pi (0.025{)}^2{\mathrm{m}}^2\approx 1.9635\times 10^{-3}{\mathrm{m}}^2$. The long-coil estimate isAt $I=0.50\mathrm{A}$, each turn links approximately $2.4674\times 10^{-6}\mathrm{Wb}$ and the total magnetic flux linkage is approximately $2.4674\times 10^{-3}\mathrm{Wb}$. Their ratio to $I$ is the stated self-inductance. A real finite coil need not give precisely these values: the axial centre correction above is not by itself a correction formula for inductance, which depends on magnetic flux across all the turns.

At fixed $A$ and $\ell$, doubling $N$ quadruples $L$. At fixed solenoid turn density $n$ and area $A$, doubling $\ell$ doubles both $N$ and $L$. These are different changes to the geometry, so there is no contradiction between $L\propto 1/\ell$ at fixed $N$ and $L\propto \ell$ at fixed $n$. A magnetic core requires its own magnetic permeability model; nonlinear magnetization can make a constant-$L$ description inadequate.