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An eddy current is a circulating electric current induced within an electrical conductor by changing magnetic flux, as described by Faraday's law of induction. Its magnetic field opposes the inducing change according to Lenz's law; electrical resistance converts electrical energy into heat.

A conducting rod sliding on rails separated by $\ell$ in a transverse magnetic field has induced current magnitude $B\ell v/R$ when self-inductance is neglected and resistance is $R$. The Lorentz force opposes motion with magnitude $B^2{\ell }^{2}v/R$, so free motion decays exponentially on the displayed time constant. Joule heating integrates to the rod's lost kinetic energy. Current direction follows the Lenz's law.

For an exactly soluble example, set $L_1=L_2=L$, $R_1=R_2=R$, and $0<M<L$. Start with zero electric currents, and apply a constant electric potential difference $V_0>0$ at $t=0$. The sums $s=i_1+i_2$ and $d=i_1-i_2$ obey separate first-order linear differential equations:Their time constants are ${\tau }_{+}=(L+M)/R$ and ${\tau }_{-}=(L-M)/R$. Thus, for $t\ge 0$,Since ${\tau }_{-}<{\tau }_{+}$, $i_2(t)<0$ for $t>0$, but $i_2(t)\to 0$ as $t\to \infty$. The primary approaches $V_0/R$. The secondary response opposes the rise of linked magnetic flux, in accordance with Lenz's law. Removing mutual inductance by setting $M=0$ makes the two exponentials equal and gives zero secondary electric current.

For stipulated $L=0.1\mathrm{H}$, $M=0.06\mathrm{H}$, $R=10\Omega$ and $V_0=1\mathrm{V}$, the two time constants are $16\mathrm{ms}$ and $4\mathrm{ms}$. Setting ${\dot{i}}_2=0$ gives its largest negative magnitude atThe continuing primary electric current eventually reaches $100\mathrm{mA}$ while the secondary returns to zero. These numbers belong to the stated model, not the 1831 apparatus.

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].

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 $L\dot{I}+RI=-{\dot{\Phi }}_{\mathrm{ext}}$.

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.

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.

Let ${\rho }_q=3q/(4\pi a^3)$ be the uniform charge density and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. Rigid rotation produces current density $\mathbf{J}={\rho }_q\mathbf{\omega }\times \mathbf{r}$. The magnetic dipole moment isSpherical symmetry gives $\int r_i{r}_j{d}^{3}r={\delta }_{ij}\int r^2{d}^{3}r/3$, while $\int r^2{d}^{3}r=4\pi a^5/5$. ThereforeIt is parallel to the angular velocity for positive $q$ and antiparallel for negative $q$. The fully signed magnitude statement is $|{\mathbf{m}}_s|=|q||\omega |a^2/5$; the printed magnitude assumes positive $q$ and positive angular speed.

Axisymmetry permits a purely azimuthal magnetic vector potential, $\mathbf{A}=f(r)\sin \theta \hat{\mathbf{\phi }}$, with $A_r=A_{\theta }=0$. For the source's associated Legendre function convention, $P_1^1(\cos \theta )=+\sin \theta$, since its definition omits the Condon–Shortley phase. The given differential operator reduces toInside the sphere, $J_{\phi }={\rho }_q\omega r\sin \theta$. Since the radial operator maps $r^3$ to $10r$, the particular coefficient is ${\alpha }_0=-{\mu }_0{\rho }_q\omega /10$. Regularity at the origin and decay at infinity giveThere is no surface sheet electric current, so both normal and tangential magnetic field components are continuous at $r=a$. Usingthis is equivalent to continuity of $f$ and $f^{\prime }$. The two equations give $A={\mu }_0{\rho }_q\omega a^2/6$ and $C={\mu }_0{\rho }_q\omega a^5/15$. ThusIn the homogeneous expansion, ${\alpha }_1=A$, ${\beta }_1=C$ in their respective interior and exterior regions; all higher angular orders vanish. Indeed, matching the regular $r^{\ell }$ and decaying $r^{-\ell -1}$ terms with zero source at $\ell >1$ forces both coefficients to zero. The outside magnetic field is exactly the magnetic dipole field of ${\mathbf{m}}_s$, not merely its leading far-field approximation. For a positive moment along $z$, $A_{\phi }={\mu }_0{m}_s\sin \theta /(4\pi r^2)$, which is the printed cross-product magnitude expression.

For the flux coupling of a rotating charged sphere to a circular loop, take a loop of radius $b>a$ in the equatorial plane and use Stokes theorem on the regular full vector potential, evaluated around the rim. Its linked magnetic flux isWith $m_s=\pi a^2{I}_0$, the mutual inductance defined using this effective electric current isIntegrating an ideal magnetic dipole field through the entire disk would incorrectly use its singular expression inside the actual sphere; the rim calculation avoids that error.

For the prescribed time dependence, the quasistatic Faraday's law of induction result and the purely resistive loop approximation givePositive loop electric current is defined to produce a positive $z$ magnetic dipole moment, so the minus sign is Lenz's law. This assumes negligible retardation and $\nu L_b\ll R$, where $L_b$ is the loop's self-inductance. If that is not negligible, the governing equation is instead $L_b{\dot{I}}_b+R{I}_b=-{\dot{\Phi }}_b$; $R$ alone is insufficient for the exact dynamic response.

Choose positive circulation anticlockwise as viewed from positive $z$. The linked magnetic flux is $\Phi =2\ell XB$, so Faraday law, including the motional contribution, givesThe conducting circuit consists of two arms of length $X$ and two transverse pieces of length $2\ell$. Its resistance is $\mathcal{R}=2R(X+2\ell )$, where $R$ is resistance per unit length. Neglecting self-inductance as in the quasistatic circuit model, the current isOn the moving rod this positive current points along $+y$, so the Lorentz force is $I(2\ell {\mathbf{e}}_y)\times (B{\mathbf{e}}_z)=2\ell IB{\mathbf{e}}_x$. Its mass is $2\ell M$, not the mass of the entire circuit. Newton's equation consequently givesFor constant $B$, the force opposes $\dot{X}$, as expected from Lenz's law. The prescribed time dependence of the external field can also supply energy to the rod.

The bar's motion increases positive $z$-directed magnetic flux at rate $B\ell v$. By Lenz's law, induced current circulates clockwise as viewed from positive $z$, producing a field into the page. Equivalently $\mathbf{v}\times \mathbf{B}=-vB\hat{\mathbf{y}}$ drives positive charge down the moving bar. The motional electromotive force has magnitude $B\ell v$, soThe current flows downward in the bar, left along the lower rail, upward through the resistor, and right along the upper rail. The ideal circuit approximation neglects self-inductance and the bar's resistance.

For motion to the right in $B_0{\mathbf{e}}_z$, $B_0>0$, the induced current is clockwise. Its own magnetic field points into the page inside the circuit, opposing the increase in external flux, as required by Lenz's law. As that current changes, its changing flux also produces an electromotive force opposing the change in current: the circuit has inductance.

Consequently the algebraic current law used in parts (i)-(iv) must be replaced by a dynamical circuit equation that includes the derivative of the self-flux. The induced current cannot in general adjust instantaneously to the bar velocity. Energy is temporarily stored in the self-field and can later return to the motion or be dissipated; the simple exponential braking law can change. Self-field forces and changes of inductance with circuit geometry may also matter. The magnetic self-field changes the transient dynamics but does not create energy: mechanical energy, magnetic-field energy and Joule heat must be included in the conservation balance. A quantitative answer would require the circuit's inductance and its dependence on $s$.

For a moving circuit, Faraday's law states that the electromotive forceequals $-d{\Phi }_B/dt$, where ${\Phi }_B={\int }_S\mathbf{B}\cdot d\mathbf{S}$ is the magnetic flux through any spanning surface. Choose the upward normal. The horizontal square hasso Ohm's law gives the induced currentwith positive current taken counterclockwise from above.

On the edge $x=a$, $d\mathbf{\ell }=dy{\mathbf{e}}_y$. The magnetic force on a current-carrying wire is $d\mathbf{F}=I,d\mathbf{\ell }\times \mathbf{B}$, henceThe ratio of the horizontal to vertical magnitudes is $2|z|/a$, so the directed force makes angle ${\tan }^{-1}(2z/a)$ with the vertical, with the sign fixing the horizontal side.

The horizontal forces on opposite edges cancel, while all four vertical contributions add:This is an electromagnetic drag force opposing the motion, as required by Lenz's law. Including gravity givesAt the terminal velocity, $\ddot{z}=0$, and the constant downward speed has magnitude

At time $t$, let the centre of the loop have coordinate $Y=d+vt$. With normal $+{\mathbf{e}}_x$, the magnetic flux isFaraday's law and Ohm's law therefore give a current of magnitudeThe original flux points into the plane and decreases in magnitude as the loop recedes. By Lenz's law, the induced current reinforces the into-plane field: it is clockwise when viewed from the $+x$ side.

Faraday's law isChoose the positive circulation to be anticlockwise as viewed from $+z$. Since ${\Phi }_B=\pi B{r}^2$,

If $\dot{r}>0$, the current is clockwise and its field points in the $-z$ direction, reducing the field inside the loop. Since $d\mathbf{F}=Id\mathbf{l}\times \mathbf{B}$, the force due to the imposed field is radially inward. If $\dot{r}<0$, the current is anticlockwise, its field points in the $+z$ direction and increases the field inside, while the force is radially outward. In both cases the magnetic force opposes the radial motion, in accordance with Lenz's law.

Faraday's law is $\mathcal{E}={\oint }_C(E+v\times B)\cdot dl=-d{\Phi }_B/dt$. Here $I=-L(B\dot{x}+x\dot{B})/R$ (sign relative to the chosen circuit orientation), and the bar force is $ILB$. For $B=B_0/(1+t)$ this gives the displayed equation with $\alpha =B_0^2{L}^2/(mR)$. At $\alpha =1$, the independent solutions are $u$ and $u{e}^{1/u}$, $u=1+t$; the data give $x=x_{0}u{e}^{1/u-1}$. Thus $\dot{x}\to x_0/e$ and total mechanical work is $m{x}_0^2/(2e^2)$. The current circulates to reinforce the decaying $+z$ flux, and its force drives increasing area, both opposing the flux decrease as Lenz's law requires.

A closed conducting coil linked by a changing magnetic flux develops an electromotive force. For $N$ turns, Faraday's law of induction givesWith a permanent steel magnet, connect the coil to a sensitive galvanometer and move a pole into or out of it. Entry and withdrawal give opposite deflections; exchanging the pole reverses the signs. Holding both magnet and coil stationary gives no sustained induction merely because a magnetic field is present. Moving the coil instead, or changing its orientation, can produce the same necessary flux change.

Earth's magnetic field can serve as the field source. Turn a many-turn coil so that its area normal changes direction relative to the terrestrial field. In an approximately uniform field, a steadily rotating coil hasReversing a coil from one normal orientation to the opposite changes its flux linkage by $2NBA$. The Earth's weak field calls for an appropriate coil area, number of turns and detector sensitivity; it does not require a chemically supplied current in the detecting circuit. Faraday described terrestrial magneto-electric induction in his Second Series.^[ref]

The coil must provide a conducting return path for a circulating current. Its electrical resistance and self-inductance affect the response. Lenz's law fixes the sign: the induced effect opposes the change responsible for it. The moving system supplies mechanical work when current delivers energy or dissipates heat; a permanent magnet is not thereby an inexhaustible fuel.

The coils form a mutual induction apparatus. Changing primary electric current changes secondary magnetic flux linkage ${\Lambda }_2$. Faraday's law of induction givesOn closing the primary circuit, increasing electric current produces a transient secondary electric current and galvanometer deflection. By Lenz's law its direction opposes the increasing magnetic flux. Once primary electric current and core magnetization are steady, the derivative vanishes: the galvanometer returns to zero despite the continuing magnetic field.

On opening the circuit, the decreasing magnetic flux induces a secondary electric current tending to sustain that flux. The galvanometer deflects oppositely, then returns to zero. Making and breaking produce opposite transient deflections; steady primary current produces no sustained induced current.

For fixed linear coils, ${\Lambda }_2=M{I}_1$ and ${\mathcal{E}}_2=-Md{I}_1/dt$, with mutual inductance $M$. Opposite equal changes of primary electric current give opposite equal areas under the induced-electromotive force curves in that model. Peak sizes can differ with switching times. Real iron-core magnetic hysteresis and remanent magnetization can also change the flux excursion, so equality of the actual peaks is not required.

Faraday's law statesfor a stationary loop, with ${\Phi }_B=\int \mathbf{B}\cdot d\mathbf{S}$ and consistent contour and area orientations. The minus sign expresses Lenz's law: the induced response opposes the change of magnetic flux. An emf can exist across an open loop even when no conduction current flows.

Faraday's law states that the induced electromotive force equals minus the rate of change of linked magnetic flux:The minus sign expresses Lenz's law: the induced electric current acts to oppose the change in magnetic flux. For a fixed contour this is equivalent to $\oint \mathbf{E}\cdot d\mathbf{\ell }=-d\int \mathbf{B}\cdot d\mathbf{S}/dt$.

Faraday's law of induction states that the induced electromotive force around a material electrical circuit equals the negative rate of change of its linked magnetic flux:The surface orientation and direction around its boundary are related by the right-hand rule. Lenz's law supplies the physical meaning of the minus sign: the induced electric current produces a magnetic field opposing the change in linked magnetic flux. For a stationary contour the local form is $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$; for this moving wire, the magnetic Lorentz force term is also needed in the electromotive force.