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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.

The differential form of Faraday's law of induction isBy Stokes theorem, for a stationary oriented circuit it becomes $\mathcal{E}=\oint \mathbf{E}\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$, where ${\Phi }_B=\int \mathbf{B}\cdot d\mathbf{S}$ is magnetic flux and $\mathcal{E}$ is electromotive force. Circuit orientation and surface normal obey the right-hand rule. Lenz's law is the physical meaning of the minus sign: the induced current's magnetic field opposes the change in linked magnetic flux. For a moving circuit, the force per unit electric charge is $\mathbf{E}+\mathbf{v}\times \mathbf{B}$, and its line integral is the corresponding electromotive force. Motion through a static magnetic field can therefore induce a voltage even when ${\partial }_t\mathbf{B}=0$.

Let $\theta$ be the wire's angle from the downward vertical, $r=L\sin \theta$, and choose $\mathbf{\omega }$ and $\mathbf{B}$ upward, as in the figure. The motional term along the wire has radial component $B\omega s$ at cylindrical radius $s$. In steady electrical equilibrium, integrating from the grounded axis to the bob givesThe Lorentz force on this positive electric charge is radially outward. Its ratio to the required centripetal term is $QB/(m\omega )\lesssim C{B}^2{L}^2/(2m)$, so the stated heavy-bob condition makes its effect on the zero-field conical pendulum geometry negligible. With wire tension $T_w$, $T_w\cos \theta =mg$ and $T_w\sin \theta =m{\omega }^{2}r$, giving $\cos \theta =g/({\omega }^{2}L)$. Hence the magnetically charged conical pendulum hasA nonzero-radius orbit requires ${\omega }^{2}L>g$. Reversing the sense of rotation or the magnetic field reverses the sign of $Q_0$.

The response to the added electric field requires an electrical convention. In the usual mechanical approximation, treat the pre-existing electric charge as $Q_0$, include its upward electric force $Q_{0}E$, and neglect all magnetic-force feedback. ThenThis predicts an increased radius for $Q_{0}E>0$. It is the result when the newly applied field is used only as a probe force on the original electric charge.

There is a first-order qualification if the bob remains electrically connected to the grounded pivot and the uniform added field's potential is included consistently. Put $s=\cos \theta$ and set the external electric potential to zero at the pivot. The external potential at the bob is then $ELs$, so the charging relation and full force balance becomeThe second equation follows by eliminating the wire tension between vertical balance and radial balance, including the outward magnetic Lorentz force. Expanding both equations to first order in $E$, and to leading order in $C{B}^2{L}^2/m$, with $s_0=g/({\omega }^{2}L)$, givesThe extra factor arises from the added field changing $Q$ by $-CEL{s}_0$ and therefore changing the radial magnetic force. That term is linear in $E$ too; the heavy-bob assumption alone does not justify retaining $Q_{0}E$ while dropping it. The two expressions refer to the two electrical approximations just specified. The expansion requires a small change relative to the initial radius and is nonuniform as that radius tends to zero.

For an electrical measurement, route the grounded connection through a sensitive electrometer operated near zero input voltage, and integrate the charging electric current when $B$ or $\omega$ is changed. Starting with the field off, the transferred electric charge is $Q_0$ in this ideal capacitance model. Reversing $B$ separates the induced signal from offsets. Alternatively, isolate the charged bob briefly and measure its electric potential with a high-input-impedance electrometer, accounting for the instrument's extra capacitance and for the fact that isolation changes the electrical boundary condition.

An inductor stores magnetic energy and opposes changes of current through electromagnetic induction. For a fixed linear device, the flux linkage is $\Psi =N\Phi =LI$, the self-induced electromotive force is $-LdI/dt$, and the passive terminal voltage is $V=LdI/dt$. HenceThe voltage leads a sinusoidal current by a quarter cycle. A long solenoid has $B\simeq \mu NI/l$ and $L\simeq \mu N^{2}A/l$; its energy agrees with the volume integral of $B^2/(2\mu )$.

For two coupled coils, ${\Psi }_1=L_1{I}_1+\mathcal{M}{I}_2$, ${\Psi }_2=\mathcal{M}{I}_1+L_2{I}_2$, with signs set by coil orientations. Reciprocity gives the same mutual inductance in both equations. For the energy stored in a nonlinear inductor, $U=\int Id\Psi$ replaces the simple quadratic formula, and hysteresis can dissipate energy. Real coils also have resistance, core loss and parasitic capacitance, so an ideal inductance model has a finite useful frequency range.^[ref]

Mutual inductance expresses how a changing electric current in one circuit induces an electromotive force in another. Choose winding orientations and define their linked magnetic fluxes byFor a reciprocal linear magnetic system, the same $M$ appears in both expressions. Faraday's law of induction gives ${\mathcal{E}}_2=-Md{I}_1/dt$ when the secondary electric current does not itself change. The sign depends on winding orientation and is consistent with Lenz's law.

The stored magnetic energy isIts nonnegativity for arbitrary electric currents requires $M^2\le L_1{L}_2$. Strong coupling makes $|M|$ approach $\sqrt{L_1{L}_2}$; flux leakage makes it smaller.

An ideal electrical transformer has two windings linked by the same time-varying core magnetic flux. The induced winding voltages are proportional to their turn counts. For appropriately chosen terminal polarities,The electric current convention in the power relation takes primary power into the device and secondary power out to the load. Consequently a load electrical impedance $Z_L$ appears at the primary as $Z_{\mathrm{in}}=(N_1/N_2{)}^2{Z}_L$. A larger secondary voltage therefore comes with a smaller available secondary electric current, rather than extra energy.

A steady direct electric current produces no sustained transformer action after its switching transient. Practical departures from the ideal include winding electrical resistance, leakage magnetic flux, finite magnetizing electric current, core saturation, eddy currents and magnetic hysteresis losses.

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.

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.

In the Lorentz law, $q$ is signed electric charge, $\mathbf{E}$ and $\mathbf{B}$ are the electric and magnetic fields at the bead, and the velocity in $q\mathbf{V}\times \mathbf{B}$ is its full laboratory velocity. Here $\mathbf{V}=v\hat{\mathbf{r}}+w\hat{\mathbf{\theta }}$. Since $\hat{r}\times \hat{z}=-\hat{\theta }$ and $\hat{\theta }\times \hat{z}=\hat{r}$,The tangential component integrated around an instantaneous circle of nearly fixed radius givesThis is the stated tangential circulation work expression in the $w\gg v$ limit. It is tangential circulation work. The tangential circulation work versus particle work distinction matters because the bead actually follows an expanding spiral. The radial displacement may be small, but its force is proportional to $w$, so its power is of the same order as the tangential power. In factThe cancellation expresses the fact that magnetic forces do no work. Ignoring radial work while retaining the first-order tangential effect would incorrectly conclude that a magnetic field supplies net particle energy.

The electric field giving the same tangential force would be $\boxed{{\mathbf{E}}_{\mathrm{eq}}=-vB\hat{\theta }}$. This is a force-equivalent field for the moving guide, not evidence that a real circulating electric field has appeared in the laboratory merely because the loop expands in a static magnetic field. An electromotive force is the tangential driving-force circulation per unit charge, with units of energy per charge. For a material circuit moving with local velocity ${\mathbf{v}}_{\mathrm{loop}}$,Orient the ring counterclockwise with positive normal $+\hat{z}$. Its flux is $\Phi =B\pi R^2$, so $\dot{\Phi }=2\pi BRv$. The tangential circulation therefore givesIt is the motional term that matters here. For a fixed loop, the circulation of the actual electric field instead follows the partial time derivative of the magnetic field. The distinction between moving-wire magnetic forces and induction in a stationary loop is essential to the flux rule.^[ref]

For the magnetic power balance on an expanding circular guide, the guide must supply an additional inward reaction $-qBw\hat{r}$ opposing the radial magnetic force. With radial guide speed $v$, this contribution to its power on the bead is $P_{\mathrm{guide}}^{(B)}=-qBvw$, equal to the tangential magnetic power. The radial magnetic power is positive by the same amount, so the total magnetic power still vanishes; the mechanical exchange is associated with the moving mechanical constraint.

If “the ring's work” is meant as its full work on a finite-mass bead, the source statement needs a further qualification. For a frictionless bead of mass $m_b$ with constant imposed radial speed, its radial acceleration is $-w^2/R$, so the complete radial guide reaction and power areThe first term is present even when $B=0$ and describes the familiar change of tangential kinetic energy on an expanding guide. The requested $-qBvw$ is exactly the additional magnetic-field-dependent term, not that full reaction power. No bead mass or extra agency fixing $w$ is supplied, so it would be unjustified to assign a numerical total mechanical power or silently discard its centripetal contribution.

Electromagnetic induction is the production of an electromotive force by changing magnetic flux or by moving a conductor through a magnetic field. With the loop and surface orientations related by the right-hand rule, Faraday's law of induction isHere $\mathbf{v}$ is the local wire velocity. For a stationary loop, the induced electric field can have a nonzero closed-path integral; its differential form is $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$. For a straight rod moving perpendicularly to itself and to a uniform magnetic field, the motional electromotive force has magnitude $B\ell v$.

The minus sign expresses Lenz's law: the induced electric current produces a magnetic field that opposes the change in linked magnetic flux. Pulling a conducting loop through a magnetic field therefore requires work against an electromagnetic resisting force; that work supplies the resulting heat or stored energy.

For an inductor with constant self-inductance, $\mathcal{E}=-LdI/dt$ and its stored magnetic energy is $L{I}^2/2$. A changing electric current in one coil can induce an electromotive force in another through mutual inductance, the operating principle of a electrical transformer. These examples connect the flux law, the direction of the response and conservation of energy.

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.

Choose a positive unit normal $\hat{\mathbf{n}}$ to the loop and choose positive circulation by the right-hand rule. Write the applied magnetic field as $\mathbf{B}(t)=B_0\cos (\omega t)\hat{\mathbf{b}}$, where $\hat{\mathbf{b}}\cdot \hat{\mathbf{n}}=\cos \theta$. Because the field is uniform, the external magnetic flux isFaraday's law of induction gives the induced electromotive force, and Ohm's law gives the current through the loop's electrical resistance:This is the induced current in a tilted resistive loop with the self-generated magnetic-flux contribution neglected as requested. The factor has units $\mathrm{Wb}{\mathrm{s}}^{-1}{\mathrm{\Omega }}^{-1}=\mathrm{A}$. A field parallel to the loop plane has $\cos \theta =0$ and induces no current in this uniform-field model.

For $\cos \theta >0$, the initial normal flux is positive. At $\omega t=\pi /2$, the external magnetic field is instantaneously zero but its normal component is decreasing, so $d{\Phi }_B/dt<0$ and $I>0$. The current is counterclockwise when viewed from the positive-normal side. Its own magnetic field points along $+\hat{\mathbf{n}}$ through the loop, preserving the initial normal flux in accordance with Lenz's law.

For $\cos \theta <0$, both circulation and the induced normal-field direction reverse. The current follows the rate of change of magnetic flux; the quarter-cycle current is nonzero even though the external field itself vanishes then.

For causal reachability on a spatial circle, put $P=2\pi R$ and use a representative $\bar{x}\in [-P/2,P/2)$ of the spatial circle. The lower-left panel of part c is a fundamental strip with its vertical edges identified. On the covering space the charged world line is still the solution of part b; on the quotient it isA jump between opposite strip edges is a coordinate seam, not a discontinuity of the physical world line. At its $n$th complete lap, $x=nP$, soFor $a=q{E}_x/m>0$, the speed increases on each return while remaining below $c$. Its kinetic energy increases byThere is no single-valued electrostatic potential energy $-q{E}_x\bar{x}$ on a circle: the nonzero electromotive force is $\oint \mathbf{E}\cdot d\mathbf{l}=E_{x}P$. A globally defined magnetic vector potential one-form $A_x=-E_{x}t$ instead makes the driving explicit. In an embedded circuit, Faraday's law realizes this electromotive force through changing magnetic flux. More generally an external apparatus maintaining the field supplies the increasing particle energy. In an isolated dynamical field-plus-particle system, backreaction would change the field. Thus this is not a cyclic machine extracting work from a conservative potential, and it does not violate conservation of energy.

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.

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

For steady axisymmetric ideal flow with only poloidal components, induction implies $\mathbf{u}\times \mathbf{B}=C{\mathbf{e}}_{\varphi }/R$. Axis regularity, or zero toroidal electromotive force, sets $C=0$. Continuity then gives $\rho \mathbf{u}=k(\psi )\mathbf{B}$, locally on connected magnetic surfaces of the poloidal magnetic flux function. Without the zero-circulation condition, annular domains can support steady cross-field flow.

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].

The Rüchardt experiment infers heat capacity ratio from the period of an oscillation of a piston supported by trapped gas. Use a vertical precision cylinder with a close-fitting movable piston, known moving mass $M$, face area $A$, trapped equilibrium volume $V_0$, and local atmospheric pressure $p_{\mathrm{atm}}$. With negligible static friction, force balance gives the absolute gas pressure $p_0=p_{\mathrm{atm}}+Mg/A$, where $g$ is gravitational acceleration. Displace the piston slightly and release it: compression raises pressure, and expansion lowers it, producing a restoring force. The oscillation is a mechanical probe of thermal response, without separately measuring two heat inputs. ^[ref]

Let $x$ be upward displacement from equilibrium, so $V=V_0+Ax$. For constant heat capacity ratio and a reversible adiabatic process, the net upward force, including the ambient load and weight, isThe linear approximation requires $|Ax|/V_0\ll 1$. Applying Newton's second law gives simple harmonic motion withHere ${\omega }_0$ is the undamped angular frequency and $\mathcal{T}$ the undamped oscillation period, rather than temperature. Record repeated cycles, measure $M,A,V_0,p_{\mathrm{atm}}$ independently, and include all connected trapped volume. An illustrative model with $M=0.020\mathrm{kg}$, $A=2.0\times 10^{-4}{\mathrm{m}}^2$, $V_0=0.010{\mathrm{m}}^3$, $p_{\mathrm{atm}}=100000\mathrm{Pa}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\gamma =1.4$ gives $p_0=100981\mathrm{Pa}$ and $\mathcal{T}\simeq 1.182\mathrm{s}$. These stipulated values are not measurements.

In Shearwood and Sloan's implementation, piston-mounted magnets induce an electromotive force in a stationary coil through Faraday's law of induction; an oscilloscope records the signal. They observed faster oscillations at smaller trapped volume values and stronger damping at small cylinder volume values. The signal tracks velocity locally, so its amplitude is not directly piston displacement. Calibration also identified inlet dead space beyond the graduated cylinder volume. ^[ref]

Repeat at smaller initial amplitudes to test the linear approximation, and compare several $V_0$ values. Heat transfer weakens the restoring force toward the isothermal process limit, biasing the inference toward $1$. Check leakage through resting-piston drift and before-and-after trapped volume readings. For viscous damping with envelope $e^{-t/\tau }$, the measured angular frequency obeys ${\omega }_0^2={\omega }_d^2+{\tau }^{-2}$; this correction does not remove thermal leakage. Internal equilibration must be faster than an oscillation period, with heat transfer and leakage slow over a run. Large amplitude or sticking friction require a richer model. ^[ref]

The Seebeck effect is the generation of a thermoelectric electromotive force by a temperature difference. In a thermocouple, two different homogeneous electrical conductors A and B connect a measuring junction at $T_h$ to two reference connections at a common $T_c$. With negligible electric current, the measured terminal voltage is determined by the difference of their Seebeck coefficients. A hot junction alone does not specify the reading: the reference temperature, the second material, and the terminal order matter.

For a motivating experiment, join one end of A to one end of B, insulating the wires elsewhere. Connect their free ends to identical meter leads at reference temperature $T_c$, for example in an ice-water heat bath. A meter with input electrical resistance much larger than the thermocouple's electrical resistance approximates an open circuit, meaning $I\approx 0$. Keep the measuring junction in a separately controlled heat bath at $T_h$ and wait for stable junction temperatures. Both reference connections must share $T_c$; identical leads with identical endpoint temperatures have cancelling thermoelectric contributions, allowing the meter to be elsewhere. ^[ref].

Define the reading as $V_{AB}=V_A-V_B$, with the positive input on the A reference terminal and the negative input on B. Use the Seebeck coefficient convention in which the voltage change along a homogeneous wire at zero electric current is $dV=-S(T)dT$. Following the path from A at $T_c$ to the common junction at $T_h$, then back along B to $T_c$, givesThis formula assumes homogeneous materials whose Seebeck coefficients depend only on the local temperature, and no additional uncompensated lead contributions. Here voltage is the electrical transport voltage associated with electrochemical potential, rather than a direct measurement of an isolated equilibrium contact electric potential.

Start with $T_h=T_c$, then increase $T_h$ while holding $T_c$ fixed and record $V_{AB}$ at each stable setting. In the ideal model the equal-temperature reading is zero. If $S_A-S_B>0$ throughout the interval, A reads positive when $T_h>T_c$. Exchanging meter inputs reverses the reading; exchanging the two junction temperatures reverses the oriented integral. Repeating the run with identical homogeneous materials gives zero because $S_A=S_B$. These controls test the stated model; a nonzero experimental baseline should be measured rather than silently removed by assuming it is zero.

For an explicitly toy linear example, stipulate constant Seebeck coefficients $S_A=25\mu V/K$ and $S_B=-15\mu V/K$. Choose reference temperature $T_c=20{}^{\circ }\mathrm{C}$ and measuring-junction temperature $T_h=70{}^{\circ }\mathrm{C}$. A temperature difference has the same numerical size in kelvins and on the degree Celsius scale, soSwapping the inputs gives $-2.00\mathrm{mV}$. Bringing both junction regions to the same temperature gives zero. With temperature-dependent Seebeck coefficients, shifting both junction temperatures equally while keeping their difference fixed can change the reading. No actual material pair is claimed to have the stipulated coefficients over this interval.

Real thermocouples need a calibrated voltage-versus-temperature relation: the constant-Seebeck coefficient approximation is useful only where the coefficient difference changes little. Wire inhomogeneity can make the result depend on where the temperature gradient lies, and reference connections at unequal or unknown temperatures introduce errors. An electronic reference correction uses a measured reference temperature in place of an ice bath. ^[ref]. Burns and colleagues provide the standardized reference functions for common letter-designated thermocouple types; the toy line above is not a replacement for them. ^[ref].

The historical experiment used a different detector. In his 1821–1823 investigations, Seebeck found that a closed electrical circuit of two dissimilar materials with junctions at different temperatures deflected a compass needle. The modern explanation is that thermoelectric electromotive force drives an electric current, whose magnetic field deflects the needle. The open-circuit experiment above detects voltage while drawing negligible electric current. The striking connection is that a thermal imbalance can become an electrical signal and then a magnetic response. ^[ref].

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.

A series circuit connects components along one unbranched path, so any electric current through the connected components is the same. Oriented terminal electric potential differences add along that path. In an open-circuit measurement, negligible electric current permits the electromotive forces of serially connected sources to add without appreciable internal electrical resistance drops.

A thermocouple uses two different electrical conductors and the Seebeck effect to infer a measuring-junction temperature from a voltage and a known reference temperature. The reference material and the reference temperature are distinct choices: B specifies the comparison Seebeck coefficient, while $T_c$ specifies the lower limit of the integral. The two-material experiment in Seebeck effect gives the geometry, sign convention, and calibration limitations.

The useful measurement is a pair: the raw terminal electric potential difference and the thermocouple reference junction temperature. For the A-positive terminal order and homogeneous-wire assumptions of Seebeck effect, write the open-circuit electromotive force asHere $T_h$ is the thermocouple measuring junction temperature and $T_c$ is the common temperature of the reference connections. The subscript $h$ names the measuring end; it need not be hotter. A voltmeter senses the complete connected thermoelectric path through the wires and its leads, in the electrochemical potential convention already described under Seebeck effect. It does not measure an absolute temperature, the individual Seebeck coefficient of A, or an isolated contact electric potential.