Faraday's law of induction ($\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$)

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

Michael Faraday's induction ring separates two questions: can one electrical circuit affect another without an electrical connection, and does that effect continue under steady conditions? The surviving apparatus uses insulated copper windings around an iron ring; the Royal Institution dates the successful experiment to 29 August 1831.^[ref]

Faraday connected one winding group to a battery, the other to a galvanometer. Making battery contact deflected the needle; maintaining contact let it return to rest; breaking contact reversed the response. Reversing the battery reversed the deflections. His separate cylinder comparison enhanced induction with iron, whereas replacing it with copper gave no enhancement over the windings alone.^[ref]

The contrast between a switching response and a quiet steady state makes mutual induction tangible. A steady primary electric current can maintain magnetic flux without sustaining a secondary electromotive force. Changes in magnetic flux linkage matter, as expressed by Faraday's law of induction. This is why a driven electrical transformer uses repeated changes in magnetic flux to supply secondary power.

The galvanometer readout of an induction pulse also separates the measured motion from the inferred electric current. Its moment of inertia and damping mean that a deflection is not an instantaneous trace of electric current. The following coupled RL induction model calculates currents in a stipulated circuit, rather than fitting Faraday's needle motion or reconstructing his switching times.

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.

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.

Apply $V_0$ for $0\le t<T$ and then set the drive to zero while retaining the closed primary return. By superposition, each pulse-response electric current is its equal-loop RL induction step response at $t$ minus that response at $t-T$, with the step response defined as zero before its onset. All currents remain continuous at $T$; the drive changes their derivatives.

For the displayed pulse, the secondary responds negatively as the primary builds and positively as it decays. Between these transients, the primary is almost steady and the secondary is almost quiet. The energy balance of coupled RL circuits accounts for the entire process, including the energy remaining in the magnetic field when the drive changes.

The magnetic energy of this constant-inductance model isThe positive-definite matrix condition makes $W\ge 0$. Multiplying the two circuit equations by $i_1$ and $i_2$, then adding, gives the displayed conservation of energy statement: input power $V{i}_1$ supplies stored magnetic energy and Joule heating. With zero drive, $\dot{W}\le 0$. After the drive becomes zero, the stored magnetic energy supplies the remaining Joule heating.

A pickup coil links magnetic flux from a source and develops an electromotive force when that linked magnetic flux changes.

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.

If a loop of resistance $R$ has magnetic flux ${\Phi }_B=-kz$, then its induced current is $I=k\dot{z}/R$. When its magnetic force is $-kI$ in the vertical direction, its equation under gravity isso its downward terminal speed is $Rmg/k^2$.

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.