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For a fixed linear circuit, self-inductance relates the magnetic flux linkage produced by its own current to that current: $\Lambda =LI$, with $\Lambda ={\sum }_{\mathrm{turns}}{\Phi }_B$. The self-induced electromotive force is $\mathcal{E}=-d\Lambda /dt=-LdI/dt$ when $L$ is constant. Its unit is the henry, $1\mathrm{H}=1Wb/A$.

mutual inductance relates the linkage in one circuit to current in another: ${\Lambda }_2=M_{21}{I}_1$ and ${\Lambda }_1=M_{12}{I}_2$. For stationary reciprocal linear media, $M_{12}=M_{21}=M$ with consistently oriented turns, and ${\mathcal{E}}_2=-Md{I}_1/dt$. Orientation fixes the sign. The coupled magnetic energy is $W=\frac{1}{2}{L}_1{I}_1^2+\frac{1}{2}{L}_2{I}_2^2+M{I}_1{I}_2$, so positivity requires $M^2\le L_1{L}_2$. Moving coils or nonlinear cores require differentiation of the full linkage rather than treating the inductance as a constant.

Faraday's law for a moving circuit is $\mathcal{E}=\oint (\mathbf{E}+\mathbf{v}\times \mathbf{B})\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$. The magnetic flux ${\Phi }_B$ is through a surface bounded by the instantaneous loop, with consistent orientation. For the stationary external field here, the motion supplies the motional electromotive force.

Let $x$ be downward penetration of the bottom segment, $v=\dot{x}$, and $K=B_{0}a$. While $0<x<b$, flux grows at rate $Kv$. Define positive current as the direction whose magnetic force is upward. With negligible self-inductance, $I=Kv/R$ andThe resistive magnetic entry of a falling loop givesThe initial acceleration is $g$ and the speed approaches $g\tau$ if the loop remains partly immersed long enough. Joule heating $I^{2}R$ removes gravitational energy. At the first $x=b$, both horizontal segments are inside the uniform field: the net flux becomes constant, the net magnetic force vanishes, and this zero-inductance model has $I=0$ thereafter. The subsequent speed increases with acceleration $g$.

For negligible electrical resistance, conservation of total linked flux instead gives $LI=Kx$ in the braking-current convention. Thereforeand the lossless limit of magnetic loop braking yieldsHere energy is stored reversibly in $L{I}^2/2$, not dissipated. If $b>2g/{\omega }^2$, the loop reverses before complete entry and executes the ideal oscillatory motion while straddling the boundary. If $b<2g/{\omega }^2$, it reaches complete entry on its first descent; thereafter the current is constant, the forces on its two horizontal segments cancel, and it falls with acceleration $g$. The exact grazing case needs care at the ideal discontinuous field boundary. An entry-only sinusoid must not be extrapolated through full immersion.

Self-inductance relates current to its own magnetic flux linkage: $\Lambda =LI$, so the induced electromotive force is $-L\dot{I}$ for fixed geometry. A long solenoid has $B={\mu }_{0}NI/l$ and area $\pi a^2$. Its $N$ turns giveThe homopolar generator uses a disc rotating through this axial field; integrate the motional Lorentz force from shaft to rim:For a homopolar disc dynamo, choose the wiring and rotation sense giving positive magnetic feedback. Kirchhoff's voltage law isfor prescribed constant $\omega$. ThusA nonzero seed grows when $\omega >{\omega }_c$, decays when $\omega <{\omega }_c$, and is marginal at equality. This explains the printed $\omega \ge {\omega }_c$ as the threshold including its boundary; equality does not itself give exponential growth. An exactly zero initial current remains zero in the ideal homogeneous equation, and reversing the feedback sense prevents the instability.

In the torque-controlled homopolar dynamo equilibrium, a radial current element exerts a braking torque $d{G}_B=IBrdr$, so $G_B=IB{a}^2/2=C{I}^2$. For steady rotation and nonzero steady current, $G=C{I}^2$ and $\omega ={\omega }_c$. Hence the field intensity magnitude isThe two current signs give opposite magnetic polarities. Finally,The steady magnetic and rotational energies do not change. This is a steady-state balance, not an assertion that every dynamical trajectory must approach that state.

For fixed, magnetically linear geometry, self-inductance relates a coil's own magnetic flux linkage to its electric current, ${\Lambda }_i=L_i{I}_i$. mutual inductance relates the linkage in one coil to the current in the other, ${\Lambda }_{21}=M{I}_1$. These coefficients enter the induced electromotive forces through ${\mathcal{E}}_i=-d{\Lambda }_i/dt$; reversing a chosen winding orientation reverses the sign assigned to $M$.

In the ideal confined-field toroid, a circular magnetostatic Ampère law contour encloses $N_{1}I$, soWrite $w=b-a$ for the height of the square cross-section. The inductance of a rectangular-section toroid follows by integrating the magnetic flux density over that cross-section; it cannot be replaced by its value at an arbitrary mean radius:Consequently, for winding normals chosen consistently,The equality means complete shared magnetic flux linkage; with leakage one instead has $|M|<\sqrt{L_1{L}_2}$.

For the crack, use the radial-path approximation for a cracked toroid, with an air length $d$ on each path and core length $2\pi r-d$. Continuity of the normal magnetic flux density at a crack face gives $H_g={\mu }_r{H}_c$. HenceIntegrating the radial paths yieldsHere $q=1.5915\mathrm{cm}$ and $M_d/M_0=0.47079$: about a 53% reduction. The small ratio $d/a$ does not justify expanding in ${\mu }_{r}d/a$, which is about ten. The crack's low magnetic permeability gives appreciable magnetic reluctance. This is an estimate within a confined-flux model; actual redistribution near the slit is outside its accuracy. Narrow-gap circuit models assume small gap width relative to transverse dimensions and approximately directed flux.^[ref]

For the intact ring, ideal poloidal winding currents and the no-leakage boundary condition admit $\mathbf{H}=N_{1}I\hat{\mathbf{\varphi }}/(2\pi r)$ with zero radial and vertical components. It is curl-free inside the current-free core, divergence-free for uniform magnetic permeability, tangent to the core boundary, and has the required circulation; uniqueness of the Neumann problem removes an additional homogeneous scalar-potential field. This explains zero non-azimuthal components in the stipulated intact-ring idealization. A literal slit breaks that rotational symmetry, so zero transverse components everywhere is not exact for the cracked ring: magnetic fringing and flux redistribution occur. In fact, extending the approximate $B(r)\propto 1/(r+q)$ unchanged around a current-free core would give $[\nabla \times \mathbf{H}{]}_z\ne 0$. Thus that radial-path calculation and a claim of exact purely azimuthal field in the cracked geometry cannot both be exact. The numerical answer above retains the intended circuit approximation while making its limitation explicit.

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.

The varying magnetic flux through the ring induces an electromotive force by Faraday's law of induction. Its direction follows Lenz's law. With solenoid current $I_s$, ring current $I_r$, ring self-inductance $L_r$, electrical resistance $R$, and mutual inductance $M(z)$,The ring sits in a nonuniform magnetic field: coupling decreases above the solenoid, so $M>0$ and $M^{\prime }<0$ with a common orientation for the two windings. For $I_s=I_0\cos \omega t$, the in-phase part of the induced ring current is negative:Thus alternating-current induction levitation can balance the ring's weight when $⟨F_z⟩=mg$. A falling mean lifting force with height supplies vertical stability. In the strongly inductive limit the two currents are almost opposite: the ring repels the solenoid during both half cycles. Finite self-inductance matters; an ideal purely resistive loop would carry a current exactly in quadrature with $I_s$ and give zero average force in this model. The source supplies the Joule heating in the real ring, and any mechanical work while it rises.

Faraday's law for the moving loop is $\mathcal{E}=-d{\Phi }_B/dt$, equivalently $\mathcal{E}=\oint (\mathbf{E}+\mathbf{v}\times \mathbf{B})\cdot d\mathbf{\ell }$ for its moving circuit. Lenz's law fixes its orientation: the induced current opposes the change in linked magnetic flux. The minus sign represents this opposition, not opposition to the existing field in every phase of motion.

Unroll the small loop's motion locally into translation to the right at speed $v\simeq b\omega$, where $b$ is the radius marked in the source figure. Take the imposed field into the page. On entry the linked into-page flux increases, so the current is counterclockwise. The leading vertical side carries current upward and feels a leftward magnetic force $iBa$. The in-field parts of the horizontal sides feel equal opposite vertical forces; the trailing side is outside the field. On exit the current is clockwise. The trailing vertical side is now the one in the field and again carries upward current, producing leftward drag; the horizontal forces reverse and still cancel. When the entire loop is within the uniform region, its flux is constant, so $i=0$ and every side force is zero under the stated negligible-self-inductance approximation.

For a loop crossing a magnetic field boundary, overlap changes at rate $av$, henceFor approximately constant speed this gives two negative rectangular force pulses of duration $a/v$, separated by a short zero-force interval while the loop is fully inside. Their time integral isThe quoted impulse is its magnitude; this impulse of magnetic braking opposes the motion. In fact its independence of speed remains exact within this locally translational sharp-boundary model even if speed changes: each crossing contributes $-(B^2{a}^2/R)\int vdt=-(B^2{a}^2/R)\int dx$ over distance $a$.

Write the system's moment of inertia as $\mathcal{I}$. Since $d{K}_{\mathrm{rot}}/dt=\tau \omega =F_{x}b\omega =F_{x}v$, the local energy balance during overlap isThus rotational energy lost is exactly resistive heat in this model. At nearly constant $v$, each pass givesFor a finite speed reduction the exact relation is the integral of $i(t{)}^{2}R$, rather than initial speed times impulse; the equality of work and heating still holds. Fringing fields, loop curvature and finite self-inductance smooth the ideal pulses. A practical user control is to vary the electromagnet current, hence $B$: braking scales as $B^2$.

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 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.

The regular hexagon has area $A=6(\sqrt{3}{a}^2/4)=3\sqrt{3}{a}^2/2$. Choose $t=0$ with its positive normal parallel to the imposed field. The external magnetic flux is $\Phi =B_{0}A\cos \omega t$, so Faraday's law of induction givesTaking the stated resistance to determine the loop response,A different initial orientation adds a phase. The sign is with respect to the chosen loop orientation and obeys Lenz's law; reversal of that orientation reverses the signed current. This expression neglects self-inductance, as in the resistive-loop model.

A Josephson junction weakly couples two superconducting condensates. Its coupling energy is $-E_J\cos \phi$, with a gauge-invariant phase difference across the junction. Choose current and voltage orientations consistently so thatThe DC Josephson effect permits a stationary zero-voltage supercurrent up to $I_c$. A constant nonzero voltage produces the AC Josephson effect, with frequency $f=2e|V|/h$. Resistive and capacitive currents are additional circuit contributions, not part of the ideal sinusoidal supercurrent alone.

In a symmetric DC SQUID with negligible loop self-inductance, the two junction phases differ by $2\pi \Phi /{\Phi }_0$. Adding their currents and maximizing over the common phase givesFlux changes therefore modulate the critical current or a suitably biased output voltage. This converts phase interference into sensitive flux measurement. Finite loop inductance, unequal junctions, noise and readout conditions alter the simple modulation; the loop flux then includes screening currents. A SQUID is useful in magnetometry, but achieving sensitivity requires both a suitable circuit and controlled environmental noise.

A superconducting persistent current is an electric current circulating in a closed path with superconductivity after the external driving electromotive force has ceased. Its associated magnetic field makes the circulation observable without inserting a measuring instrument into the path. This article concerns macroscopic circulation in superconductivity. A finite experiment establishes persistence over its observation window with a stated measurement error, rather than directly observing an infinite lifetime.

The ideal circuit connection is flux conservation in a superconducting loop. In a fixed loop with self-inductance $L$, a single circulating electric current $I$, and externally imposed linked magnetic flux ${\Phi }_{\mathrm{ext}}$, the lumped circuit approximation givesWhen the electrical resistance $R$ vanishes, integrating gives $LI+{\Phi }_{\mathrm{ext}}=\text{constant}$. Removing the external linked magnetic flux therefore changes the circulating electric current; holding it fixed afterwards preserves that electric current. This is a circuit approximation with a specified self-inductance, not a derivation of the microscopic state of superconductivity. For thin paths where the carrier contribution matters, fluxoid and flux quantization in a superconductor supplies the more careful description.

For constant $L>0$, constant effective electrical resistance $R\ge 0$, and no changing external linked magnetic flux, the RL circuit instead predictsThis exponential decay follows by separating $dI/I=-(R/L)dt$. A long relaxation time tests extremely small electrical resistance without trying to resolve an equally tiny steady electric potential difference. The stored magnetic energy is $U=L{I}^2/2$, so the same model gives $dU/dt=-R{I}^2$, exactly the Joule heating loss. The finite energy stored in the loop supplies the magnetic readout; the circulating electric current is not a continuously powered source of useful work.

Suppose a calibrated measurement bounds the fractional loss of a nonzero superconducting persistent current by $ϵ$, with $0<ϵ<1$, over a duration $T>0$. Take $I_0>0$ by choosing the circulation direction. Under the constant-$L$, constant-$R$, undriven RL circuit model,The inequality follows because the logarithm is increasing. The final approximation requires $ϵ\ll 1$ and slightly understates the exact upper bound. If $L$ is uncertain, an independently established upper bound on $L$ gives the conservative electrical resistance bound. The uncertainty in the initial reading, the final reading, background subtraction and drift must all be included in $ϵ$; a display resolution alone does not provide it.

For an original illustrative example, let $L=1\mu \mathrm{H}$ and suppose the corrected electric current ratio after $T=3\mathrm{h}=10800\mathrm{s}$ is at least $0.999$. ThenThese are conditional bounds for stipulated parameters, not values attributed to the 1914 apparatus. With the same $L$ but $R=10^{-11}\Omega$, the model instead gives $\tau =10^5\mathrm{s}$ and $I(24\mathrm{h})/I_0=e^{-0.864}\simeq 0.42147$: a readily visible loss on the chosen scale.

A resolved decay can be checked using its logarithm:A straight line supports this constant-parameter model; curvature or jumps demand a different account. If an effective electrical resistance varies with time while $L$ remains fixed and the loop remains undriven, only its time average is constrained:This does not bound every instantaneous value of $R(t)$. Changing self-inductance or external linked magnetic flux adds terms to the circuit equation, so an uncorrected magnetic signal cannot simply be substituted for $I(t)$. Nonobservation of decay leaves arbitrarily small positive constant electrical resistance compatible with a finite observation: it is not a logical proof that $R=0$.

A rectangular loop entering a uniform magnetic field while only one horizontal segment is immersed has motional electromotive force $Bav$. Neglecting self-inductance, its electric current is $Bav/R$ and its upward magnetic force is $(Ba{)}^{2}v/R$. The equation $m\dot{v}=mg-(Ba{)}^{2}v/R$ gives exponential relaxation toward $mgR/(Ba{)}^2$. This entry model ends when both horizontal segments enter the field; then their net magnetic force and the rate of change of external magnetic flux vanish.

An RL circuit combines electrical resistance $R$ and self-inductance $L$. In the series model, the same electric current $I$ flows through both components. An ideal battery maintains terminal electric potential difference $V$; the drops in the direction of positive electric current obey Ohm's law and the inductor relation:This is Kirchhoff's voltage law with the inductor's terminal drop, rather than its oppositely signed induced electromotive force. For constant $R>0$, $L>0$, and $V$, the first-order linear differential equation has solutionThe time constant $\tau$ measures exponential decay of the difference from the final electric current. ^[ref].

For a stipulated example, take electrical resistance $R=2\Omega$, self-inductance $L=0.5\mathrm{H}$, terminal electric potential difference $V=6\mathrm{V}$, and initial electric current $I(0)=0$. ThenWith $t$ measured numerically in seconds, this is $I(t)=3(1-e^{-4t})$ in amperes. At one time constant,and the terminal electric potential difference drops areTheir sum is exactly $6\mathrm{V}$. At $t=0^{+}$, the inductor takes the entire terminal electric potential difference while the electric current is still zero. At steady electric current, the inductor's terminal drop is zero and the entire drop is across the electrical resistance.

The final magnetic energy stored by the inductor is $U_{\infty }=L{I}_{\infty }^2/2=2.25\mathrm{J}$. At one time constant, the electric current has reached about $63.21\%$ of its final value, but its magnetic energy has reached only about $39.96\%$:This difference follows from the square of electric current in the inductor's magnetic energy.

For release, take $t=0$ at the start of a separate experiment with initial electric current $3\mathrm{A}$. Remove the battery from the model while maintaining a closed path through the same electrical resistance and inductor. ThenNow $v_L=-v_R$: the induced electromotive force supports the continuing positive electric current as its magnitude decreases. This closed-loop model describes exponential decay, not opening the sole path and imposing an instantaneous jump to zero electric current. The inductor's magnetic energy and the power of Joule heating areTheir time constant is $\tau /2=0.125\mathrm{s}$. The total energy dissipated by Joule heating is exactly the initial magnetic energy:The second diagram in inductor shows this energy transfer. A finite closed path lets us predict both how long the electric current persists and how much energy becomes heat. ^[ref].

For negligible self-inductance, a loop with electrical resistance $R$ obeys $I=-{\dot{\Phi }}_{\mathrm{ext}}/R$. With ${\Phi }_{\mathrm{ext}}=AB(t)\cos \theta$, this gives $I=-(A/R)\dot{B}\cos \theta +(A/R)B\sin \theta \dot{\theta }$. Its magnetic torque isFor $B=B_0\sin \omega t$ and slow mechanical motion, the first term averages to zero while the second gives the displayed opposing torque. A stationary loop nevertheless experiences an alternating torque at $2\omega$ and generally executes small forced motion. The zero leading mean aligning torque is not a claim of exactly zero instantaneous motion or of no higher-order effect.

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.

The magnetic energy stored in a linear solenoid of constant self-inductance $L$ isin the long-solenoid approximation. The first formula is an inductor relation; the second identifies its storage with magnetic energy in the interior magnetic field, under the vacuum-core approximation. This agreement joins the electrical circuit description to the spatial magnetic field description. The expression $B^2/(2{\mu }_0)$ is the vacuum magnetic energy density. ^[ref].

A short derivation starts from the terminal electric potential difference $v_L=LdI/dt$, with positive electric current entering the terminal at higher potential. The absorbed power is $v_{L}I$, and integrating from zero electric current givesThe result describes recoverable storage in the inductor; Joule heating in the wire is a separate expenditure of energy. ^[ref].

For the long-coil example, $L\approx 4.9348\mathrm{mH}$ and $I=0.50\mathrm{A}$ give $U\approx 6.1685\times 10^{-4}\mathrm{J}=0.61685\mathrm{mJ}$. Substituting the ideal $B\approx 1.256637\mathrm{mT}$ and volume $A\ell \approx 9.8175\times 10^{-4}{\mathrm{m}}^3$ into $B^{2}A\ell /(2{\mu }_0)$ gives the same value. Doubling electric current quadruples stored magnetic energy. In a real coil, reducing electric current can return this energy to an electrical circuit, or release it as Joule heating in a closed RL circuit.