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Insert $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$ and $\nabla \times \mathbf{H}=\mathbf{J}+{\partial }_t\mathbf{D}$ into the given identity. For a linear, time-independent, nondispersive medium, $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$, givingThe divergence theorem then yieldsThe left side is outward electromagnetic power flow. The current term is power delivered to matter, becoming Joule heating for an Ohmic conductor, and ${\partial }_{t}u$ is the change in stored electric plus magnetic energy. The minus sign says outward flux reduces stored energy unless supplied locally.

Follow the drawing's coordinates: lower strip at $z=0$, upper strip at $z=d$, current $I$ along $+x$ in the upper strip and back along the lower. The wording “both strips lie in the $xy$-plane” must mean parallel to that plane, since the drawing separates them in $z$. Neglect the side and end fringes using $l\gg a\gg d$. Then $I=V_0/(R+R_L)$ and the top-strip potential is $V(x)=V_0-IRx/l$, while the lower strip is equipotential at zero.

The harmonic function matching those two surfaces isIndeed ${\nabla }^2\varphi =0$, and it has the required potentials at $z=0,d$. ThusThe two current sheets reinforce their magnetic field between the strips; its direction follows the right-hand rule. The Poynting flow in resistive parallel strips follows from their cross product:Power flows along the gap towards the load and upwards into the resistive strip. Across a transverse area $ad$, $P_x(x)=IV(x)$ falls from $I{V}_0$ to $I^2{R}_L$. The upper surface receives ${\int }_0^{l}a{S}_z(z=d)dx=I^{2}R$, exactly the strip's heating; the ideal lower strip receives none. In the charge-free gap, ${\partial }_x{S}_x+{\partial }_z{S}_z=0$, consistent with a static field and no local dissipation. The small longitudinal electric field cannot be discarded if this upward energy transfer is to be represented.

A current in one circuit links magnetic flux through another. It produces magnetic flux linkage through another. With fixed geometry and a linear reciprocal medium, choose consistent winding orientations and writeThe shared coefficient is mutual inductance; reversal of one winding reverses its sign. Faraday's law of induction gives the mutual induced emf ${\mathcal{E}}_2=-M{\dot{I}}_1$ for fixed $M$. The minus sign expresses Lenz's law.

The magnetic energy isNonnegative energy for every current pair requires $|M|\le \sqrt{L_1{L}_2}$, motivating the coupling coefficient $M/\sqrt{L_1{L}_2}$. Transformers use strong common flux to exchange energy between circuits; leakage flux reduces coupling. A shared core increases linkage, but saturation, hysteresis and eddy-current losses limit the linear description. A moving geometry also contributes $-I_1\dot{M}$ to the mutual emf.

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.

The symmetry is electromagnetic reciprocity applied to mutual inductance. For thin closed circuits in vacuum, the magnetic vector potential generated by circuit B isStokes theorem converts the magnetic flux through circuit A into its magnetic vector potential circulation. Dividing by $I_B$ gives Neumann's mutual-inductance formula,The dot product and separation are unchanged by interchanging A and B, soEach has units $\mathrm{Wb}{\mathrm{A}}^{-1}=\mathrm{H}$. The winding orientations must be used consistently to determine the common sign. The same reciprocity holds in ordinary linear reciprocal magnetic media; it is not an unrestricted statement about biased nonreciprocal materials. Equivalently, a reversible magnetic energy function has equal mixed derivatives with respect to the two currents.

Magnetic field and force. A steady electric current produces a magnetic field. In vacuum the Biot-Savart law givesThe Lorentz force on the charge carriers transfers force to the material carrying them. The resulting force on a current element of another circuit is $d{\mathbf{F}}_2=I_{2}d{\mathbf{\ell }}_2\times {\mathbf{B}}_1$, or, for a continuous current distribution, the magnetic force per unit volume is $\mathbf{f}=\mathbf{J}\times \mathbf{B}$.

For two long parallel straight wires a distance $r$ apart, Ampère's law gives $B_1={\mu }_0{I}_1/(2\pi r)$. ThereforeThe force is attractive for parallel currents and repulsive for opposite currents, by the cross product rule. Its units are $\mathrm{N}{\mathrm{m}}^{-1}$. This approximation neglects wire thickness and end effects; a complete circuit includes its return current. Element-by-element forces need not form an isolated Newton's-third-law pair, but complete stationary circuits have balanced mutual forces when the entire system is included.

Energy description. For two linear coupled circuits the magnetic energy isHere $q$ is a mechanical coordinate and $M$ the signed mutual inductance. At prescribed currents, the generalized force is $I_1{I}_2\partial M/\partial q$ if the self-inductances do not vary. This positive derivative is obtained from magnetic coenergy, or by including the electrical sources' work; differentiating $W_B$ with a minus sign at fixed currents would omit that source work. At prescribed flux linkages the force instead equals the negative derivative of stored magnetic energy with those linkages fixed. The geometry thus determines both inductance and mechanical force.

In a homogeneous linear medium, write $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$. Without free charge or current, Maxwell's equations becomeTaking the curl of Faraday's law, using $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$, givesTheir plane-wave solutions have ${\omega }^2=k^2/(\mu ϵ)$ and speed $v=1/\sqrt{\mu ϵ}$; in vacuum this is $c=1/\sqrt{{\mu }_0{ϵ}_0}$. For a wave $\mathrm{Re}[{\mathbf{E}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$, the divergence equations give $\mathbf{k}\cdot {\mathbf{E}}_0=\mathbf{k}\cdot {\mathbf{B}}_0=0$, while Faraday's law gives ${\mathbf{B}}_0=\mathbf{k}\times {\mathbf{E}}_0/\omega$. Thus the fields and propagation direction form an orthogonal triad, with $B_0=E_0/v$.

The two independent transverse electric components allow linear, circular and elliptical polarization of an electromagnetic wave. Their relative amplitudes and phase determine the polarization state. In a lossless nondispersive medium the energy density and energy flux areA travelling plane wave has equal electric and magnetic energy contributions and time-averaged intensity $⟨S⟩=E_0^2/(2Z)$, where $Z=\sqrt{\mu /ϵ}$ is its wave impedance and $E_0$ denotes peak electric-field amplitude. Reflection and refraction follow by imposing the electromagnetic boundary conditions. Frequency-dependent constitutive response introduces dispersion; conductivity introduces attenuation, so the simple real constant-speed result has a specified domain of validity.

The capacitor stores electrostatic energy, the inductor stores magnetic energy, and the resistor converts energy into heat. Their instantaneous values areThe sign convention for electric current does not change these expressions. With the initial voltage, $E_C(0)=C{V}_0^2/2=90\mu \mathrm{J}$ and $E_I(0)=P_R(0)=0$.

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.

For a normal from medium 1 to 2, the magnetic boundary conditions areWithout a free surface current, tangential $H$ is continuous. To obtain the magnetization of a permeable sphere of radius $R$ and permeability ${\mu }_0(1+\chi )$, choose magnetic scalar potentialsTangential continuity gives $H_i=H_0-C/R^3$, and normal $B$ continuity gives $(1+\chi )H_i=H_0+2C/R^3$. HenceThe fields are uniform inside. For negative susceptibility, magnetization is antiparallel to the applied field; the scalar formula retains its negative sign.

For a small weakly magnetic body, the induced magnetic energy is $U_m=-\chi V{B}^2/(2{\mu }_0)$. The factor one-half accounts for building the induced moment. The vertical force is $F_z=\chi VB{B}^{\prime }/{\mu }_0$. Balancing gravity therefore givesA diamagnet needs a field decreasing in magnitude with height for an upward force. For diamagnetic levitation in an inverse-height field,The force falls as $z^{-3}$, giving a restoring vertical derivative at equilibrium. Full lateral stability would require the transverse field geometry, which is not specified.

Corotation reaches the speed of light at $R_c=c/\Omega$. Extrapolating a dipole field gives $B_c\sim B_{\ast }(R_{\ast }/R_c{)}^3$ and magnetic energy density ${ϵ}_c\sim B_c^2/{\mu }_0$, suppressing order-one factors. An outward electromagnetic energy flux of scale ${ϵ}_{c}c$ through area of scale $R_c^2$ therefore givesThis assumes a dipole-dominated field tied to the rotating star and an effective outgoing Poynting flux near the light cylinder. The dipole extrapolation, angular pattern and surface-field convention affect the coefficient. A strictly aligned vacuum dipole would not radiate, whereas an oblique vacuum dipole does. A plasma-filled magnetosphere opens field lines and can spin down even an aligned rotator, so the light-cylinder estimate is a scaling argument, not an exact vacuum emission calculation or literal rigid motion at $c$. Force-free calculations explicitly demonstrate this distinction.^[ref]

The energy comes primarily from stellar rotation. With constant moment of inertia $I$, $E_{\mathrm{rot}}=I{\Omega }^2/2$, and conservation of energy gives $I\Omega \dot{\Omega }=-P$. ThusFor constant $K$, differentiating gives $\ddot{\Omega }=-3K{\Omega }^2\dot{\Omega }$, hence $n=\Omega \ddot{\Omega }/{\dot{\Omega }}^2=3$. Evolving field strength, inclination, moment of inertia or additional torques invalidate the constant-$K$ prediction; generally $n=3+(\Omega /\dot{\Omega })(\dot{K}/K)$ in this parametrization.

For radial inflow onto mass $M$, take $v_{\mathrm{ff}}\sim (GM/r{)}^{1/2}$ and $\rho \sim \dot{M}/(4\pi r^2{v}_{\mathrm{ff}})$. magnetic pressure decreases as $B_{\ast }^2{R}_{\ast }^6/({\mu }_0{r}^6)$, while ram pressure scales as $\rho v_{\mathrm{ff}}^2\propto \dot{M}(GM{)}^{1/2}{r}^{-5/2}$. Balancing them yieldsNumerical factors and disc geometry are not determined by this spherical estimate. The displayed proportionality holds $M$ fixed and requires a magnetospheric boundary outside the stellar surface.

In a disc, inflowing matter near $R_M$ has specific angular momentum approximately $\sqrt{GM{R}_M}$. If it is transferred to a slower star, the material torque is $N\sim \dot{M}\sqrt{GM{R}_M}$ and $I\dot{\Omega }\sim N>0$. The corotation radius is $R_{\mathrm{co}}=(GM/{\Omega }^2{)}^{1/3}$, distinct from the light-cylinder radius. As stellar spin approaches the inner disc's Keplerian frequency, net spin-up weakens. If $R_M$ lies outside corotation, the field rotates faster than the disc and can centrifugally expel matter in the propeller regime. A characteristic equilibrium isfor fixed mass and magnetic-coupling convention. A complete magnetic torque model can change the coefficient; a star also cannot rotate past its mass-shedding limit, of order $\sqrt{GM/R_{\ast }^3}$.

For opacity per unit mass $\kappa$, an isotropic radiative force balances gravity at $L_E=4\pi GMc/\kappa$. Efficient release of surface gravitational binding energy gives $L_{\mathrm{acc}}\sim GM\dot{M}/R_{\ast }$ and therefore the surface accretion Eddington rateThe radius dependence matters: it uses efficiency $\eta \sim GM/(R_{\ast }{c}^2)$, not an arbitrarily fixed reference efficiency. Substituting this maximal rate into the equilibrium period givesmass, opacity and geometry factors are held fixed. This is the classical radiatively efficient Eddington-limited scaling, not a universal upper bound on every mass supply rate: beaming, advection, outflows and opacity changes can alter it. For sufficiently small field, $R_M\lesssim R_{\ast }$ and the magnetic scaling must be replaced by surface-disc and mass-shedding physics.

For a planar magnetohydrodynamic shock with normal velocity and tangential magnetic field, mass conservation gives $\rho u=m$ and the ideal magnetohydrodynamic induction equation gives $uB$ constant across the front. Hence $B/\rho$ is unchanged, expressing magnetic flux freezing. Normal momentum flux is $\rho u^2+p+B^2/(2{\mu }_0)$, and energy flux per unit mass flux is $u^2/2+\gamma p/[(\gamma -1)\rho ]+B^2/({\mu }_0\rho )$, apart from a smooth common gravitational potential. The amplified tangential field contributes both magnetic pressure and magnetic energy transport.

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.

For a magnetic vector potential $\mathbf{A}$, a solenoidal magnetic field $\mathbf{B}=\nabla \times \mathbf{A}$ and constant magnetic diffusivity ${\eta }_m$, write ${\partial }_t\mathbf{A}=\mathbf{u}\times \mathbf{B}-{\eta }_m\nabla \times \mathbf{B}-\nabla \varphi$. In a fixed closed volume with zero helicity surface flux, integration by parts makes the two differentiated factors in $H={\int }_V\mathbf{A}\cdot \mathbf{B}dV$ equal; the scalar-potential boundary term vanishes when $\mathbf{B}\cdot \mathbf{n}=0$. One sufficient boundary gauge is $\mathbf{n}\times \mathbf{A}=0$; another standard setting is integration over all space with sufficient decay. Orthogonality of $\mathbf{u}\times \mathbf{B}$ to $\mathbf{B}$ gives the displayed identity. Unlike the resistive magnetic energy loss, $\dot{H}$ is not sign definite. The zero normal magnetic field also makes $H$ gauge invariance under single-valued smooth gauge transformations.

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].

With the closed-boundary assumptions of the resistive magnetic-helicity balance, define the nonnegative resistive magnetic energy loss $D_B={\eta }_m{\int }_V|\nabla \times \mathbf{B}{|}^{2}dV/{\mu }_0$. The Cauchy-Schwarz inequality gives the displayed bound. Consequently, in a family of fields with bounded $E_B,D_B$ as ${\eta }_m\to 0$, resistive magnetic helicity loss tends to zero even when $D_B$ remains finite. This is an absolute conservation estimate; it is not a universal comparison of fractional decay times. For a single Beltrami field with $\nabla \times \mathbf{B}=k\mathbf{B}$ and no motion, both $E_B$ and $H$ have the same fractional decay rate $2{\eta }_m{k}^2$.

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.

In a series RLC circuit, the same electric current $i=q^{\prime }$ flows through the resistor with electrical resistance $R$, the inductor, and the capacitor. Choose $q$ as the signed electric charge on the capacitor plate into which positive $i$ flows. With terminal drops measured in that direction, Kirchhoff's voltage law givesHere $V(t)$ is the applied electric potential difference; assume constant $L>0$, $C>0$, and $R>0$. With the applied source set to zero and the loop kept closed, the free response satisfiesThe angular frequency ${\omega }_0$ is the value for lossless free oscillation; $\alpha$ is the decay parameter. The three possibilities are underdamped RLC response for $\alpha <{\omega }_0$, critically damped RLC response for $\alpha ={\omega }_0$, and overdamped RLC response for $\alpha >{\omega }_0$. ^[ref].

For a fair comparison, start each circuit with the same electric charge $q(0)=q_0$ and zero electric current $q^{\prime }(0)=0$. Set $\tau ={\omega }_{0}t$ and $a=\alpha /{\omega }_0$; then $x=q/q_0$ obeys $d^{2}x/d{\tau }^2+2adx/d\tau +x=0$. The plot compares $a=0.5,1,2$. These dimensionless curves apply to any $L,C$ after rescaling time and electric charge.

The stored energy is the sum of magnetic energy and capacitor energy:Differentiating and using the circuit equation gives the complete power balance,In the free response, $E^{\prime }=-R{i}^2\le 0$: Joule heating accounts exactly for the lost energy. For the plotted initial conditions, $E(0)=q_0^2/(2C)$ and $E/E(0)=x^2+(dx/d\tau {)}^2$. Since every free solution tends to zero for $R>0$,In the lossless limit $R=0$, energy stays constant and shuttles between the capacitor and inductor. Thus an oscillation of electric charge and a decrease of total energy are compatible statements.

The analogy with a damped harmonic oscillator is exact within these ideal models: electric charge corresponds to displacement, electric current to velocity, inductance to mass, $1/C$ to spring constant, and $R$ to the coefficient of a resisting force proportional to velocity. The two stored energy terms play the roles of kinetic energy and elastic potential energy. ^[ref].

Differentiating the driven equation also gives $L{i}^{\prime \prime }+R{i}^{\prime }+i/C=V^{\prime }$. When using this equation to solve for electric current, retain the original constraint $L{i}^{\prime }(0)+Ri(0)+q(0)/C=V(0)$: the initial electric charge fixes the initial rate of change of electric current.

An ideal voltage source $V_0$ and series resistor $R_1$ feed parallel branches $L$ and $R_2$. Thévenin's theorem gives $R_{\mathrm{th}}=R_1{R}_2/(R_1+R_2)$ and $V_{\mathrm{th}}=V_0{R}_2/(R_1+R_2)$. Starting with zero inductor current, $i_L=(V_0/R_1)(1-e^{-t/\tau })$ and $i_2=V_0{e}^{-t/\tau }/(R_1+R_2)$. Switch-on heat in $R_2$ is $L{V}_0^2/[2R_1(R_1+R_2)]$. Disconnecting the source leaves the stored magnetic energy $L{i}_L^2/2$ to dissipate through $R_2$, with time constant $L/R_2$.

The magnetic energy stored in a linear solenoid of constant self-inductance $L$ isin the long-solenoid approximation. The first formula is an inductor relation; the second identifies its storage with magnetic energy in the interior magnetic field, under the vacuum-core approximation. This agreement joins the electrical circuit description to the spatial magnetic field description. The expression $B^2/(2{\mu }_0)$ is the vacuum magnetic energy density. ^[ref].

A short derivation starts from the terminal electric potential difference $v_L=LdI/dt$, with positive electric current entering the terminal at higher potential. The absorbed power is $v_{L}I$, and integrating from zero electric current givesThe result describes recoverable storage in the inductor; Joule heating in the wire is a separate expenditure of energy. ^[ref].

For the long-coil example, $L\approx 4.9348\mathrm{mH}$ and $I=0.50\mathrm{A}$ give $U\approx 6.1685\times 10^{-4}\mathrm{J}=0.61685\mathrm{mJ}$. Substituting the ideal $B\approx 1.256637\mathrm{mT}$ and volume $A\ell \approx 9.8175\times 10^{-4}{\mathrm{m}}^3$ into $B^{2}A\ell /(2{\mu }_0)$ gives the same value. Doubling electric current quadruples stored magnetic energy. In a real coil, reducing electric current can return this energy to an electrical circuit, or release it as Joule heating in a closed RL circuit.