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A resistive loop with self-inductance $L_r$ coupled by mutual inductance $M(z)$ to an imposed alternating current obeys $L_r{\dot{I}}_r+R{I}_r=-M{\dot{I}}_s$. For $I_s=I_0\cos \omega t$, averaging the magnetic force $I_s{I}_r{M}^{\prime }(z)$ gives $-I_0^2{\omega }^{2}M{L}_r{M}^{\prime }/[2(R^2+{\omega }^2{L}_r^2)]$. It is repulsive where $M{M}^{\prime }<0$. The self-inductance supplies the current component opposite to the source current; a purely resistive, zero-inductance loop has zero average force in this model.

Coupled inductors are inductors whose magnetic flux linkages depend on each other's electric currents. A two-circuit linear model uses two self-inductances and a mutual inductance.

Take two electrical circuits containing coupled inductors. Let their constant self-inductances be $L_1,L_2>0$, mutual inductance $M$, and positive electrical resistances $R_1,R_2$. Choose current directions so that the off-diagonal entries of the inductance matrix are $+M$. Assume a linear, quasistatic model, with no capacitive effects or core hysteresis. The magnetic flux linkages are ${\lambda }_1=L_1{i}_1+M{i}_2$ and ${\lambda }_2=M{i}_1+L_2{i}_2$. Combining their time derivatives with resistive electric potential differences givesThe secondary is a closed, undriven electrical circuit. Require $M^2<L_1{L}_2$, making $\mathbf{L}$ a positive-definite matrix. The chosen current convention fixes the signs; reversing the secondary's reference direction changes the sign assigned to $M$ and $i_2$ without changing the physical circuit.

There is an essential switching assumption: the primary loop remains closed when $V$ becomes zero. An ideal voltage source then imposes zero electric potential difference, allowing the current to decay. Literally opening an inductive circuit requires additional physics to determine the interruption, including possible capacitance and an electric arc. This closed-return model explains transient induction while keeping that distinction from Faraday's historical making and breaking of contact explicit.

Electromotive force is the driving work per unit electric charge around an electrical circuit, measured in volts, despite its name containing "force". For a stationary loop its electromagnetic contribution is $\mathcal{E}={\oint }_C\mathbf{E}\cdot d\mathbf{l}$; for a moving electrical conductor, include the motional electromotive force. A nonzero closed-loop line integral of the electric field cannot be represented by a single-valued electric potential alone. An electromotive force specifies the drive; the resulting electric current also depends on the electrical resistance and self-inductance. ^[ref].

Two long round wires of radius $a_w\ll d$ carrying opposite electric currents form a loop with the displayed leading external self-inductance per unit length. Integrating their summed magnetic field across a connecting ribbon gives the logarithm; magnetic-energy integration gives the same leading value. The zero-radius limit diverges. Uniform direct current inside two nonmagnetic round wires contributes another ${\mu }_0/(4\pi )$, while an ideal surface current has no interior contribution.

Faraday's law of induction relates a changing magnetic field to circulation of the electric field. For a fixed oriented surface $S$ with boundary $C=\partial S$, choose the positive direction around $C$ by the right-hand rule: the fingers follow $C$ while the thumb follows the chosen unit normal. The fixed-loop electromagnetic electromotive force isThe final derivative of magnetic flux is an ordinary time derivative, with the surface held fixed. Reversing both the unit normal and the boundary direction reverses both signs. The loop may be an imagined path: the field law does not require a wire or an electric current. ^[ref].

For a computed example, a fixed single-turn loop has area $A=0.020{\mathrm{m}}^2$ and out-of-page unit normal. Prescribe an external uniform magnetic field in that direction with $B(t)=0.10\mathrm{T}+(0.30T/s)t$ for $0\le t\le 1.0\mathrm{s}$. ThenIn a quasistatic resistive model with electrical resistance $R=2.0\Omega$, neglect the loop's self-inductance and its own contribution to changing magnetic flux. Ohm's law givesThe magnetic flux rises from $0.0020\mathrm{Wb}$ to $0.0080\mathrm{Wb}$ while the induced electromotive force stays constant: it measures the rate of change, rather than the amount of magnetic flux. The negative sign means clockwise electric current; its magnetic field through the loop points into the page, as explained by Lenz's law. The toy model describes the interval of steady ramping, excluding switching transients at its ends.

For a moving conducting loop $C(t)$, the electromagnetic electromotive force instead includes the Lorentz force contribution from the local velocity $\mathbf{v}$ of each conductor segment:This form assumes a thin-wire model with a well-defined closed contour and spanning oriented surface that move compatibly, a magnetic field satisfying Gauss's law for magnetism, and conductor speeds small compared with the speed of light. Ideal sliding contacts such as those in motional electromotive force can be treated piecewise. Arbitrary changes of electrical connections require checking the actual contour and conductor velocities; differentiating the flux through an arbitrary chosen moving path is insufficient. The velocity here describes conductor motion, not the charge carriers' drift along the wire. ^[ref].

A short derivation of the fixed-loop integral statement applies Stokes theorem to the local curl equation and differentiates under the fixed surface integral. For the moving loop, the magnetic flux transport for a moving circuit adds the swept-area term $-{\oint }_C(\mathbf{v}\times \mathbf{B})\cdot d\mathbf{l}$ to $d{\Phi }_B/dt$. Combining these two identities gives the moving-loop statement. This derives the integral forms from the local field law; it is not a proof of that empirical law from mathematics alone.

The useful unity is geometric: changing the magnetic field, the enclosed area, or the orientation can change magnetic flux and hence produce an electromotive force. A coil's multiple turns are accounted for by magnetic flux linkage, rather than by multiplying a single-turn flux without checking which flux each turn receives. ^[ref].

Faraday's report read on 24 November 1831 describes opposite momentary responses when a neighboring circuit is connected and disconnected, with no continuing response while its steady electric current is maintained. These are concrete experimental roots of the changing-flux idea. The report predates the modern Maxwell equations notation used above. ^[ref].

In a fixed square coil beside a long wire, neglect self-inductance compared with resistance. A changing wire current induces coil current proportional to $dI/dt$. Forces on the two perpendicular sides cancel, and the unequal parallel-side fields give magnitude ${\mu }_{0}Na|Ii|[1/s-1/(s+a)]/(2\pi )$. When $I=I_0\sin \omega t$, the signed force alternates at twice the drive frequency and has zero cycle mean, while Joule dissipation is positive with mean ${\mathcal{M}}^2{\omega }^2{I}_0^2/(2R)$. A fixed coil has no mechanical work despite its time-dependent force.

The henry is the SI unit of inductance:A coil with self-inductance $1\mathrm{H}$ has magnetic flux linkage $1\mathrm{Wb}$ at electric current $1\mathrm{A}$. A rate of change of electric current $1A/s$ produces induced electromotive force of magnitude $1\mathrm{V}$. The weber measures magnetic flux and magnetic flux linkage; the henry measures their ratio to electric current. ^[ref].

Choose the positive loop normal and its positive circulation by the right-hand rule. A spatially uniform external magnetic field of fixed direction and signed magnitude $B_0\cos \omega t$, at angle $\theta$ to that normal, gives magnetic flux $\Phi =\pi r^2{B}_0\cos \theta \cos \omega t$. Neglecting the loop's self-inductance, Faraday's law of induction and Ohm's law give the displayed electric current. At a quarter-cycle the external field is zero but its normal component is decreasing most rapidly, so the current produces a field in the direction of the initial normal flux, as required by Lenz's law. With appreciable self-inductance the circuit instead obeys $L\dot{I}+RI=-{\dot{\Phi }}_{\mathrm{ext}}$.

Inductance relates electric currents to the magnetic flux linkage they produce. For a single linear coil with fixed geometry and magnetic response, its self-inductance $L$ is the constant inHere $\Lambda$ is magnetic flux linkage from the coil's own electric current $I$. The SI unit of inductance is the henry. Knowing inductance lets us predict the induced electromotive force when electric current changes, and the magnetic energy stored by an inductor. ^[ref].

For example, a coil of self-inductance $0.5\mathrm{H}$ carrying electric current $3\mathrm{A}$ has magnetic flux linkage $1.5\mathrm{Wb}$. If its $100$ turns link the same magnetic flux, each turn has magnetic flux $0.015\mathrm{Wb}$. Keeping the geometry and magnetic response fixed while doubling electric current doubles magnetic flux linkage and quadruples the inductor's magnetic energy.

The comparison with capacitance is useful: a linear capacitor has $Q=CV$ and stores capacitor energy $C{V}^2/2$ in its electric field, whereas a linear inductor has $\Lambda =LI$ and stores magnetic energy $L{I}^2/2$ in its magnetic field. The RLC circuit brings both forms of energy storage together.

For constant self-inductance L, the steady induced current satisfies $L\dot{I}+RI=BA\omega \sin \omega t$, giving amplitude $I_0=BA\omega /\sqrt{R^2+{\omega }^2{L}^2}$ and phase lag. Multiplication by the rotating loop normal again gives a static moment plus a twice-frequency rotating moment of magnitude $A{I}_0/2$, independently of lag. Therefore replace the resistance-only denominator $R^2$ in the radiation power by $R^2+{\omega }^2{L}^2$. A wire resistance alone is not sufficient for an exact prediction if inductance is appreciable.

For a uniformly permeable toroid with inner radius $a$, outer radius $b$ and axial height $h$, an ideal poloidal winding gives $H_{\varphi }=NI/(2\pi r)$. Integrating the magnetic flux density across its section gives $\Phi =\mu NIh\ln (b/a)/(2\pi )$ and hence the displayed self-inductance. It retains radial field variation when $b-a$ is not small. For two fully linked windings, replace $N^2$ by $N_1{N}_2$ to obtain their mutual inductance. Discrete turns, leakage and nonlinear core response limit the ideal formula.

Let a slender toroid have core cross-section $A$, mean length $\ell$ and large relative magnetic permeability $\mu$. Cut a narrow gap of perpendicular width $t$ whose faces make angle $\theta$ with the local core direction. Continuity of normal magnetic flux density gives $B_{g,n}=B\sin \theta$; continuity of tangential magnetic field intensity gives $H_{g,t}=B\cos \theta /(\mu {\mu }_0)$. The gap field is nearly normal to the faces for large $\mu$. Its effective flux area is $A/\sin \theta$, so its leading magnetic reluctance is $t\sin \theta /({\mu }_{0}A)$. Adding core reluctance gives the displayed self-inductance. At $45^{\circ }$, $L\simeq {\mu }_0{N}^{2}A/(\ell /\mu +t/\sqrt{2})$ and $H_g\simeq NI/(t+\sqrt{2}\ell /\mu )$. This retains $\mu t/\ell$ while neglecting fringing and corrections of relative order $t/\ell$ or $1/\mu$.

Following a change in applied electromotive force, a series inductor and resistor obey $L\dot{I}+RI=\mathcal{E}(t)$. For a constant applied electromotive force, the difference from the final electric current decays as $e^{-tR/L}$. Neglecting self-inductance is justified only when the forcing varies slowly compared with the time constant $L/R$, or when considering a steady state.

An inductor is an electrical circuit component designed to provide self-inductance, often a coil such as a solenoid. An ideal linear inductor has constant $L>0$ and zero electrical resistance. Its stored magnetic energy, with zero chosen at zero electric current, isFor self-inductance $0.5\mathrm{H}$ and electric current $3\mathrm{A}$, this gives $U=2.25\mathrm{J}$. Unlike Joule heating, ideal storage is recoverable: the inductor can return energy to an electrical circuit when its electric current decreases. ^[ref].

The short derivation uses the terminal convention in self-inductance. The power absorbed by an ideal inductor isStarting from zero electric current, the integral of this power givesPositive $P_L$ increases stored magnetic energy; negative $P_L$ returns energy. At constant electric current $3\mathrm{A}$ in the example, $P_L=0$ and induced electromotive force is zero, but $U=2.25\mathrm{J}$ still resides in the magnetic field. The battery in the steady RL circuit supplies Joule heating in the electrical resistance, rather than further increasing the inductor's magnetic energy.

During the closed-loop release in RL circuit, stored magnetic energy plus accumulated Joule heating stays $2.25\mathrm{J}$. A real coil also has electrical resistance, so its Joule heating must be included separately from ideal inductor storage. The constant-$L$ formulas assume a fixed linear magnetic response.

Lenz's law expresses the minus sign in Faraday's law of induction: the induced electromotive force drives a passive resistive loop's electric current in the direction whose own magnetic field opposes the imposed change in magnetic flux. It opposes the change of flux, not invariably the external magnetic field. Choose an oriented surface and its positive boundary direction first, find the sign of $d{\Phi }_{\mathrm{ext}}/dt$, and then use the right-hand rule to translate the signed electric current into a physical direction. ^[ref].

In the fixed-loop example in Faraday's law of induction, out-of-page magnetic flux increases at $0.0060Wb/s$. With counterclockwise positive, $I=-0.0030\mathrm{A}$ is clockwise, producing an in-page magnetic field through the loop. Reverse the ramp to $B(t)=0.40\mathrm{T}-(0.30T/s)t$ on the same one-second interval: now $\mathcal{E}=+0.0060\mathrm{V}$ and $I=+0.0030\mathrm{A}$. The counterclockwise electric current produces an out-of-page magnetic field, supporting the decreasing external magnetic flux. A steady out-of-page magnetic field with fixed loop geometry gives no induced electromotive force in either direction.

For the increasing-area example in motional electromotive force, clockwise electric current runs down the right-hand rod. Its magnetic force on a current-carrying wire points left, opposing the imposed motion to the right. Writing $I_{\mathrm{mag}}=B\ell v/R$ for the current magnitude, the power required to maintain constant speed isThus the sign agrees with conservation of energy: the external drive supplies the Joule heating. The magnetic forces do no work on individual charges; mechanical work by the drive supplies the electrical energy. This also explains the usefulness of electromagnetic braking of a sliding rod and the terminal speed of a falling conducting loop. ^[ref].

The sign of induced electromotive force alone does not determine an arbitrary circuit's instantaneous electric current. Retaining constant self-inductance $L$ gives $LdI/dt+RI=-d{\Phi }_{\mathrm{ext}}/dt$, so a pre-existing electric current can persist while the induced drive opposes its change. The distinction matters in the lossless limit of magnetic loop braking. The resistive toy examples use $L=0$ as an approximation, not a general consequence of Lenz's law.

A square resistive loop of side $a$, moving normally across a sharp field edge at speed $v$, has induced current magnitude $Bav/R$ and drag magnitude $B^2{a}^{2}v/R$ while partially overlapping the field. Completely enclosed in a uniform static field it has no changing flux and, with neglected self-inductance, no current.

For an external loop flux $Kz$, positive current orientation and upward Lorentz force $KI$, including self-inductance gives $\mathcal{L}\dot{I}+RI=-K\dot{z}$ and $m\ddot{z}=-mg+KI$. At zero resistance total linked flux is conserved. If initially $I=0$ at $z=z_0$, then $I=-K(z-z_0)/\mathcal{L}$ and the mechanical motion is an undamped oscillator about $z_0-mg\mathcal{L}/K^2$. The finite-resistance terminal speed tends to zero as $R\to 0$, but neglecting self-inductance is a singular approximation in that limit. A lossless loop does not dispose of energy through Joule heating.

Motional electromotive force is the contribution ${\oint }_C(\mathbf{v}\times \mathbf{B})\cdot d\mathbf{l}$ from an electrical conductor moving with local velocity $\mathbf{v}$ through a magnetic field. The cross product comes from the magnetic part of the Lorentz force. For a straight rod of length $\ell$, perpendicular to both its velocity and a uniform magnetic field, its magnitude is $B\ell v$. The sign depends on the direction of integration along the rod. ^[ref].

For a computed example, take $\ell =0.50\mathrm{m}$, velocity $\mathbf{v}=2.0\hat{\mathbf{x}}m/s$ to the right, and magnetic field $\mathbf{B}=0.40\hat{\mathbf{z}}\mathrm{T}$ out of the page. The rod lies along $y$. ThenPositive electric charge is driven downward. For an isolated rod after charge separation has equilibrated, $\mathbf{E}=-\mathbf{v}\times \mathbf{B}$ inside it, and the electric potential difference is $V_{\mathrm{bottom}}-V_{\mathrm{top}}=+0.40\mathrm{V}$; the bottom is positive. This open-circuit statement does not imply a sustained electric current.

Close the rod's path with stationary U-shaped rails and an electrical resistance $R>0$ at the left. With out-of-page unit normal, the positive loop direction is counterclockwise, hence upward along the rod. If the rod's horizontal position is $x(t)$, the enclosed area is $A=\ell x$ and the external magnetic flux is ${\Phi }_{\mathrm{ext}}=B\ell x$. ThusHere the resistive approximation neglects self-inductance. The negative electric current is clockwise, downward along the rod, consistent with the charge-separation calculation and Lenz's law.

The closed-loop result illustrates why the moving-circuit form of Faraday's law of induction includes motional electromotive force: the magnetic field is constant in time, so the electric field circulation alone would miss this drive. Its mechanical consequence is electromagnetic braking of a sliding rod.

For two fixed linear coils, mutual inductance $M$ relates magnetic flux linkage in one coil to electric current in the other: ${\Lambda }_{21}=M{I}_1$. The induced electromotive force is ${\mathcal{E}}_2=-Md{I}_1/dt$ with consistent winding orientations. For reciprocal coils, $|M|=k\sqrt{L_1{L}_2}$, where $L_1,L_2$ are self-inductances and $0\le k\le 1$ measures magnetic coupling. Its SI unit is the henry.