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An electric lamp whose current heats a filament until it emits visible thermal radiation. Joule heating converts electrical input into thermal energy and radiation. Its electrical resistance changes with temperature, so one fixed resistor value need not describe different operating voltages.

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.

The intrinsic planetary luminosity $L_{\mathrm{int}}$ is the power supplied by interior cooling, contraction, nuclear reactions if present, or internal heating, excluding reflected or passively reradiated stellar light. For fixed mass and no accretion or nuclear source, global conservation of energy gives $L_{\mathrm{int}}=Q_{\mathrm{deep}}-d(U+\Omega )/dt$, where $Q_{\mathrm{deep}}$ is any additional power deposited in the interior, such as tidal heating or Joule heating. For $Q_{\mathrm{deep}}=0$ and a gas-supported quasi-static sphere of fixed specific-heat ratio $\gamma$ and negligible surface pressure, the stellar virial theorem gives $\Omega =-3(\gamma -1)U$, henceFor a dynamically stable gas-supported model take $\gamma >4/3$. For $\gamma =5/3$ this is half the gravitational release. Degenerate giants require a different internal energy reservoir; the half-factor is not universal.

James Prescott Joule was an English physicist whose private experiments connected electricity, heat and mechanical work. His family ran a brewery in Salford; he developed an experimental laboratory while working in the family business.^[ref] The Royal Society's archive identifies his occupation as physicist and records his life as 1818–1889.^[ref]

His electrical heating experiments placed metal coils in water and measured the temperature change. Later, he used the brewery cellar to reduce ambient temperature fluctuations during friction experiments and took particular care over thermometer calibration.^[ref] These practical choices address a central difficulty of calorimetry: a small temperature rise can reflect both the intended input of energy and unwanted heat transfer to the surroundings. A quiet baseline and a known heat capacity are part of the measurement, not decoration around the formula.

The contrast between Joule heating and Joule's paddle-wheel experiment makes the connection especially clear. One supplies electrical work; the other supplies mechanical work. In each case, calorimetry tracks the resulting change of internal energy. Comparing quantitatively controlled inputs with thermal changes gives an experimental test of conservation of energy, rather than merely showing that stirring or an electric current can warm water. The paddle-wheel account describes the descending weights, vanes, thermal controls and corrections needed for that comparison.

Joule's free-expansion experiment asks a different question: how does a gas's temperature change during expansion without external work? The Joule-Thomson effect instead concerns flow through a restriction and a balance involving enthalpy. Keeping the two experiments distinct prevents a shared historical name from hiding different system boundaries and different conserved quantities.

Joule heating is the dissipative conversion of electrical work into internal energy in an electrical conductor. In a metal, the electric field supplies energy to charge carriers, which transfer it to the material. This makes an immersed resistor a useful calibrated heater and makes electrical resistance a source of losses in wiring. Wind-driven electric currents are a proposed source of heating of an inflated giant planet.

For a stationary, purely resistive Ohmic conductor, use the passive sign convention: $V$ is the terminal electric potential difference from the entry to the exit of positive electric current $I$. The absorbed electrical power and Ohm's law giveThe last two expressions assume $V=IR$ at the relevant material temperature. For constant electrical resistance and electric current over a duration $\Delta t$, the dissipated energy is $E_{\mathrm{diss}}=I^{2}R\Delta t$; otherwise use $E_{\mathrm{diss}}=\int V(t)I(t)dt$. In SI, $\mathrm{V}\mathrm{A}=\mathrm{W}=J/s$.^[ref]

Absorbed electrical power $VI$ is more general than Joule heating: a capacitor can store energy, a battery can change its chemical energy, and an electrical circuit can deliver mechanical work or emit light. Even a heated resistor need not show an indefinitely rising temperature: in a thermal steady state, its dissipated power leaves by heat transfer. With the heater inside a calorimeter boundary, electrical input crosses as work; the subsequent heat transfer to the room is a different term in the first law of thermodynamics.

No cyclic device can have as its sole effect the extraction of heat from one thermal reservoir and its complete conversion into net work. A putative device absorbing $Q_h>0$ and returning to its starting state with $W=Q_h$ and no other transfer would conserve energy, yet violate the Second law of thermodynamics. Rejecting heat to colder surroundings is an essential part of the ordinary two-thermal reservoir heat engine budget.^[ref]

The cyclic and sole-effect qualifications matter. A single isothermal expansion of an ideal gas can absorb heat and deliver equal work while leaving its volume changed; restoring the initial state must also be counted. Likewise, a device discharging stored internal energy or consuming fuel has another effect. Converting supplied electrical work completely to heat through Joule heating reverses the proposed conversion and is allowed. The statement also allows cyclic devices that exchange heat with one thermal reservoir while consuming net work; it forbids positive net work output with that single heat source as the only external change.

Conversely, a hypothetical complete-conversion device could take $100\mathrm{J}$ from the hot thermal reservoir and supply $100\mathrm{J}$ of work to an ideal refrigerator that extracts $100\mathrm{J}$ from the cold thermal reservoir and delivers $200\mathrm{J}$ to the hot one. At $600\mathrm{K}$ and $300\mathrm{K}$, that refrigerator is itself admissible: its reservoir entropy changes cancel. The composite would consume no external work and move a net $100\mathrm{J}$ from cold to hot, violating the Clausius statement of the second law. Together with the construction there, this proves the equivalence of the two impossibility statements.^[ref]

For the two fixed positive-temperature thermal reservoir hypotheses, Carnot's theorem gives the sharper restriction $\eta \le 1-T_c/T_h<1$. A reversible thermodynamic process reaches that bound in the ideal Carnot cycle; an ordinary device's smaller thermal efficiency reflects its entropy production and measurement boundary. An inefficient observed engine is compatible with the Second law of thermodynamics, rather than evidence that the bound has been experimentally determined.

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.

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.

Low-temperature calorimetry measures a small sample's heat capacity by resolving a small temperature change after a known electrical energy input. The experimental challenge is that the support, heater and thermometer also warm up. Their calorimeter addenda must be removed before interpreting the sample's phonons or electronic heat capacity. The Debye T-cubed test turns the corrected response into a test of the low-temperature vibrational spectrum.

Manuel Brando's 2009 experiment used a sapphire platform suspended by four alloyed platinum wires from a copper ring fixed to a cryostat stage. Grease attached the sample; a chip heater and ruthenium-oxide resistance thermometer served platform PLS-2. A controller stabilized the bath; a lock-in amplifier supplied heating pulses and read the calibrated sensor. The steady rise calibrated thermal conductance as $G=P/\Delta T$. A separate $59.1\mathrm{mg}$, $99.99\%$ pure gold sample, compared with reference heat capacities, established the calorimeter addenda. Small pulses limited property variation; imperfect sample contact required a second relaxation time. The platform contribution rose below $1.5\mathrm{K}$, making simple extrapolation unsafe.^[ref]

For an original idealization, assume sample and platform share one temperature $T$, the bath remains at $T_b$, and both the total heat capacity and the weak-link thermal conductance $G$ are approximately constant across one small pulse. Ignore other heat transfer and any phase transition. The first law of thermodynamics becomesHere $C_s$ is sample heat capacity, $C_a$ is calorimeter addenda, $P$ is deposited electrical power from Joule heating, and $\tau$ is a time constant. This is the same balance as calorimeter heat-loss correction, used here with a deliberately weak cryogenic link and a separately subtracted platform. For a constant pulse $P_0$ beginning at bath temperature, then switched off at $t_h$,The cooling exponential decay determines $\tau$, while the heating plateau determines $G=P_0/(T_{\infty }-T_b)$. Together with calorimeter addenda, they give $C_s=G\tau -C_a$. Cooling alone determines a ratio, not an absolute heat capacity. With finite $t_h$, the observed end rise is smaller than the plateau; fit the full response or apply the displayed correction rather than identifying the two.

For the stipulated figure parameters, $G=5.00\times 10^{-8}\mathrm{W}{\mathrm{K}}^{-1}$, $\tau =20.0\mathrm{s}$, and $C_a=2.00\times 10^{-7}\mathrm{J}{\mathrm{K}}^{-1}$ give $C_s=8.00\times 10^{-7}\mathrm{J}{\mathrm{K}}^{-1}$. The asymptotic rise is $0.0400\mathrm{K}$, only $2\%$ of bath temperature. At $100\mathrm{s}$, the rise is $0.0397305\mathrm{K}$. Of $2.00\times 10^{-7}\mathrm{J}$ input, only $3.97305\times 10^{-8}\mathrm{J}$ remains stored: the other $1.60270\times 10^{-7}\mathrm{J}$ has flowed to the bath. Treating this long pulse as an isolated $Q/C$ experiment would overestimate total heat capacity by a factor of $5.034$.

Suppose independent standard uncertainties are $u(G)=0.05\times 10^{-8}\mathrm{W}{\mathrm{K}}^{-1}$, $u(\tau )=0.2\mathrm{s}$ and $u(C_a)=0.10\times 10^{-7}\mathrm{J}{\mathrm{K}}^{-1}$. First-order propagation of uncertainty givesThe uncertainty is one standard deviation under those assumptions, about $2.2\%$ of sample heat capacity. Shared thermometer calibration errors can correlate fitted $G$ and $\tau$; then their covariance contributes too. Smaller samples amplify the relative influence of calorimeter addenda. Proposed checks include varying pulse amplitude, repeating with a reference sample, checking sensor self-heating, and comparing single- and two-time-constant fits. A drifting bath, slow internal equilibration or a transition invalidates this simple reduction even if a cooling curve looks smooth.

For constant electrical resistivity $\eta$ and magnetic permeability ${\mu }_0$, let $E_B={\int }_V{B}^2/(2{\mu }_0)dV$ and $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$. In a fixed volume with zero boundary energy flux, multiply the resistive induction equation by $\mathbf{B}/{\mu }_0$ and integrate by parts. This gives the displayed balance. The first term transfers energy between the magnetic field and the fluid; the second is nonpositive and represents Joule heating. The latter does not imply that the total magnetic energy must decrease if fluid motion continually supplies it.

Ohm's law states that an Ohmic conductor has electric current proportional to its applied electric potential difference, with constant electrical resistance under fixed material conditions:Here $V=\varphi (A)-\varphi (B)$ is the terminal electric potential difference, and positive $I$ enters at $A$ and exits at $B$. This sign convention makes $VI$ the electrical power absorbed. The straight-line relation through the origin lets one constant predict the response at many applied electric potential differences. Merely defining $R=V/I$ at one nonzero operating point does not establish Ohm's law: a component may have a ratio that varies with applied electric potential difference. Material and temperature must be specified and held fixed; many components have a nonlinear response. ^[ref].

For the ideal resistor model $R=2\Omega$ at electric potential difference $V=5\mathrm{V}$,This power is converted to internal energy by Joule heating; a watt is a joule per second. If the model holds for time $10\mathrm{s}$, the transferred energy is $125\mathrm{J}$. Doubling electric potential difference doubles electric current and quadruples power for the same fixed $R$. The model keeps material temperature fixed even while calculating Joule heating; appreciable changes in temperature may require a changing electrical resistance. ^[ref].

In 1827, Georg Ohm published his mathematical account of electrical transport in Die galvanische Kette, mathematisch bearbeitet. On printed page 36, his relation $S=A/L$ expresses electric current in terms of total driving tension and a reduced length incorporating electrical conductivity and geometry. This is historical notation: his $A$ is not the wire cross-sectional area used here, and his $L$ is not simply the wire's physical length. The striking idea is that material response and geometry can be combined into one quantity governing an electrical circuit. ^[ref].

For the damped harmonic oscillator, the mechanical energy is the sum of kinetic energy and spring potential energy:Thus $b{\dot{x}}^2$ is the power transferred out of the oscillator's mechanical energy. For $b>0$, the rate vanishes instantaneously at a turning point, despite nonzero stored potential energy; it is not a constant loss per unit time. In eddy-current damping the loss becomes Joule heating. ^[ref]

For an underdamped response, advancing one oscillation period multiplies both displacement and velocity by $e^{-\zeta {\omega }_0{T}_{\mathrm{d}}}$, so $E(t+T_{\mathrm{d}})=e^{-2\zeta {\omega }_0{T}_{\mathrm{d}}}E(t)$. This exact cycle relation does not imply that instantaneous energy is a pure exponential function throughout each cycle.

Consider an original electrical circuit model between two equipotential cross-sections. A normal path has electrical resistance $R_n>0$. Let $b\in \{0,1\}$ indicate whether an additional path connects those same cross-sections; when connected it has electrical resistance $r\ge 0$. Ideal voltage contacts measure those cross-sections. This collapses a spatial network to two parallel branches; it is not a fit to the Bednorz-Müller experiment.

For $r>0$, both branches have the same electric potential difference $V$. Ohm's law and the Kirchhoff node law give $I=V/R_n+bV/r$, henceThe last case follows directly from the ideal short's constraint $V=0$, provided it can carry the imposed electric current below its superconducting critical current. It does not follow by substituting $r=0$ into $V/r$. A normal bridge in series with otherwise superconducting pieces belongs in $r$; superconducting grains alone need not short the measured interval.

Choose $R_n=1\Omega$. A connected bridge with $r=0.01\Omega$ gives $R_{\mathrm{eff}}=1/101\Omega \simeq 9.901\mathrm{m\Omega }$, almost a hundredfold drop despite finite electrical resistance. An ideal connected superconducting filament instead gives zero in this model even when its volume is a small part of the specimen. Isolated superconducting islands can have the same superconducting phase fraction $f$ while leaving $b=0$. For the illustrative disconnected geometry we stipulate that the normal conducting network retains $R_n$.

To plot a chosen connectivity sequence, set $b(f)=0$ for $f<f_c$ and $b(f)=1$ for $f\ge f_c$, with $r=0$ on the connected branch. The two displayed values of $f_c$ describe different chosen architectures. A real percolation calculation would need an explicit network and spatial distribution; the scalar phase fraction $f$ does not determine $b$ by itself.

An instrument with unresolved electric potential difference bounded by $\delta V$ reports only $R_{\mathrm{eff}}\lesssim \delta V/|I|$. For chosen $I=1\mathrm{mA}$ and $\delta V=100\mathrm{nV}$ this floor is $10^{-4}\Omega$, assuming offsets and noise are included in the bound. The figure clips smaller model values to that floor. Increasing $|I|$ improves this formal bound only while Joule heating and the path's superconducting critical current remain controlled. None of these transport statements determines the superconducting phase fraction or establishes complete Meissner effect shielding.

The passive sign convention defines terminal electric potential difference $V=\varphi (A)-\varphi (B)$ and positive electric current entering at $A$. Then $VI$ is absorbed electrical power: positive means the component receives energy, negative means it supplies energy. Reversing both reference directions leaves the product unchanged. For a positive electrical resistance obeying Ohm's law, both $V$ and $I$ reverse together and Joule heating remains $I^{2}R\ge 0$.

Let $E_B={\int }_V{B}^2/(2{\mu }_0)dV$ be the total magnetic energy. Multiply the curl form of the resistive induction equation by $\mathbf{B}/{\mu }_0$ and integrate over the fixed volume. The identityfollows from the divergence theorem. The stated zero boundary fields remove its surface term. Applying it to the motional and resistive contributions yields the magnetic energy balance in resistive magnetohydrodynamics:The first term is the negative of the mechanical power delivered to the fluid by the Lorentz force density. It transfers energy between fluid motion and the magnetic field, with either sign. The second is the nonpositive magnetic contribution to Joule heating. A nonzero positive electrical resistivity therefore dissipates magnetic energy at rate $\eta {\int }_V{J}^{2}dV$; it does not force the total magnetic energy to decrease when fluid motion supplies a larger positive power.

Write $\mu ={\mu }_0{\mu }_r$ and $ϵ={ϵ}_0{ϵ}_r$ for homogeneous scalar material constants, and use $\mathbf{J}=\sigma \mathbf{E}$. Taking the curl of Faraday's law of induction and substituting the Ampere-Maxwell law givesThe vector calculus identity for a double curl therefore yieldsThe second term represents electric current response, rather than displacement current. A dispersive $\sigma (\omega )$ is used in the frequency domain; its general time-domain counterpart is a temporal response convolution.

For the convention $e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, the plane wave equation isDecompose the electric field into components parallel and perpendicular to $\mathbf{k}$. With $S=\mu ϵ{\omega }^2+i\mu \sigma \omega$, the longitudinal eigenvalue is $S$ and each transverse eigenvalue is $S-k^2$; thus the determinant is $S(S-k^2{)}^2$. For a transverse electromagnetic wave, $\mathbf{k}\cdot {\mathbf{E}}_0=0$ andwhen ${\mu }_r=1$ and the background ${ϵ}_r\simeq 1$. A real dissipative electrical conductivity generally makes $k$ complex, with the decaying branch chosen for a passive medium.

For a cold, collisionless, unmagnetized plasma, treat ions as stationary and write $e>0$ for the magnitude of the elementary charge. The electron equation of motion is $-i\omega m_e\mathbf{v}=-e\mathbf{E}$, soThis reactive response includes no collision-induced Joule heating. Substitution into the transverse dispersion relation givesThe quantity ${\omega }_p$ is the electron plasma frequency. The refractive index is real above this cutoff; below it, the wave is evanescent. The electron response is represented here by $\sigma (\omega )$, so it must not also be counted a second time in the background permittivity.

For a nonzero longitudinal component, taking the dot product of the matrix equation with $\mathbf{k}$ instead gives $S(\mathbf{k}\cdot {\mathbf{E}}_0)=0$. The same cold plasma response makes $S=({\omega }^2-{\omega }_p^2)/c^2$, henceAt nonzero $k$, the remaining equation forces ${\mathbf{E}}_0$ to be parallel to $\mathbf{k}$, and Faraday's law of induction gives ${\mathbf{B}}_0=0$. These are electrostatic Langmuir waves, with no $k$ dependence of angular frequency in the zero-temperature model. Equilibrium neutrality does not forbid their perturbed charge density; Gauss's law requires such a density when $\mathbf{k}\cdot {\mathbf{E}}_0\ne 0$.

Allow time dependence while keeping the constitutive relations linear, isotropic, nondispersive, and independent of time. The Maxwell equations in matter giveFor the Poynting vector $\mathcal{P}=\mathbf{E}\times \mathbf{H}$, the divergence identity yieldsThe first two terms are exact time derivatives. Taking the zero-field energy density as zero gives the electromagnetic energy densityConsequently the Poynting theorem is the continuity equationIn integral form,The left side is the rate of increase of stored electromagnetic energy plus outward Poynting flux. The right side accounts for work delivered to the free electric current; in an Ohmic medium ${\mathbf{J}}_f\cdot \mathbf{E}$ is the local rate of Joule heating.

In Cartesian coordinates the magnetic field is $(bx,by,-2bz)$. Its divergence is $b+b-2b=0$, and all components of its curl vanish. Thus it satisfies the static vacuum magnetic equations $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{B}=0$, with no electric current in the region.

Fix clockwise orientation as viewed from above, looking down the positive $z$-axis. Its positive area normal is $-{\mathbf{e}}_z$. The loop's external magnetic flux is thereforeWith the opposite normal its flux has the opposite sign. On the circle $\mathbf{B}=ba{\mathbf{e}}_r-2bz{\mathbf{e}}_z$ and clockwise line element is $d\mathbf{\ell }=-a{\mathbf{e}}_{\phi }d\phi$. The Lorentz force element isThe radial terms cancel around the circle, soThis fixes the current/flux/force signs together.

Neglect self-inductance initially, so Faraday's law and Ohm's law give $RI=-K\dot{z}$. Including gravity then givesFor $R>0$ and $K\ne 0$, the terminal solution is $z=z_0-vt$ withThe general velocity relaxes exponentially to $-v$ on time scale $mR/K^2$. In the terminal state $I=Kv/R$, and the Joule-heating rate isThus all the lost gravitational potential energy becomes resistive heat when kinetic energy is constant.

Within the zero-inductance approximation $v\to 0$ as $R\to 0$; finite-speed motion would require an unbounded current. This limit is singular, so it is important to describe the physical ideal-conductor limit separately. With finite self-inductance $\mathcal{L}$, the circuit equation is $\mathcal{L}\dot{I}+RI=-K\dot{z}$. At $R=0$, total linked flux $\mathcal{L}I+Kz$ is conserved and there is no Joule heating. If $I=0$ initially at $z_0$, thenThis lossless limit of magnetic loop braking has undamped oscillations about $z_0-mg\mathcal{L}/K^2$, with magnetic energy storing and returning mechanical energy. It has no generic dissipative terminal descent. Thus the vanishing terminal speed is the formal resistive-model limit, not a justification for dropping inductance in a perfectly conducting moving loop.

The resistor's Joule heating power is $I^{2}R=(B^2{\ell }^2/R)v_0^2{e}^{-2t/\tau }$. Integrating up to time $t$ givesThe remaining kinetic energy is $K(t)=\frac{1}{2}m{v}_0^2{e}^{-2t/\tau }$, so $\boxed{Q(t)+K(t)=\frac{1}{2}m{v}_0^2}$ at every time. In particular $\boxed{Q(\infty )=\frac{1}{2}m{v}_0^2}$: all initial kinetic energy becomes heat. The magnetic force's mechanical power is $-I^{2}R$, consistent with this energy balance.

For ${\mathcal{E}}_0=0$ and $B_0\ne 0$, $v=v_0{e}^{-t/\tau }$ and $I=-(B_{0}L/R)v$. The instantaneous Joule heating isHence the energy dissipated up to time $t$ and its infinite-time value areThe remaining kinetic energy is $\frac{1}{2}m{v}_0^2{e}^{-2t/\tau }$, so kinetic energy plus dissipated heat equals the initial kinetic energy at every time. The magnetic field redirects charge motion, and the circuit transfers mechanical energy to heat through the electric forces; there is no battery input in this case. If $B_0=0$, no current is induced, no energy is dissipated, and all the initial kinetic energy remains.