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Dot the two Linear Alfvén wave equations with ${\mathbf{u}}_{\perp }$ and ${\mathbf{b}}_{\perp }/{\mu }_0$, respectively, and add. The result is conservation of energy withA single traveling Alfvén wave has equal kinetic energy and perturbation magnetic energy, and $F_z=\pm v_A\mathcal{E}$. This equality need not hold instantaneously at each point of a superposed standing wave.

As an original ideal-model example, stipulate $B_0=1.00\mathrm{T}$, $\rho =10^4\mathrm{kg}{\mathrm{m}}^{-3}$ and a uniform path $L=0.20\mathrm{m}$. Using ${\mu }_0\simeq 4\pi \times 10^{-7}\mathrm{H}{\mathrm{m}}^{-1}$, the Alfvén speed is $8.92\mathrm{m}{\mathrm{s}}^{-1}$ and the transit time $L/v_A$ is $22.4\mathrm{ms}$. Doubling the magnetic field halves that time. A sinusoidal velocity with peak amplitude $0.010\mathrm{m}{\mathrm{s}}^{-1}$ has magnetic peak amplitude $1.12\mathrm{mT}$ and mean energy flux $⟨F_z⟩=\rho v_A{U}^2/2=4.46\mathrm{W}{\mathrm{m}}^{-2}$ in the positive direction. These are stipulated ideal parameters, not historical measurements or a prediction including vessel walls.

For an isolated bounded conductor of uniform magnetic diffusivity $\eta >0$, contained in a sphere of radius $R$ and matched to a decaying potential field in an insulating exterior, dynamo action requires maximum stretching rate $S\ge \eta {\pi }^2/R^2$. Here $S$ bounds the largest eigenvalue of the rate-of-strain tensor throughout the flow and time. Use fluid boundary conditions eliminating the stretching boundary term, for example a no-slip boundary condition, and no imposed energy input. The proof combines the magnetic free-decay spectral bound with the magnetic energy equation. This is a necessary condition, not a sufficiency criterion; changing magnetic boundary conditions changes the spectral constant.

For a barotropic fluid, an energy per mass $e$ can be defined by $de/d\rho =p/{\rho }^2$. For the isothermal equation of state, $e=c_s^2\ln (\rho /{\rho }_0)$. Adding kinetic, gravitational and magnetic energy gives the ideal magnetic energy flux $\mathbf{u}(E+p+B^2/(2{\mu }_0))-\mathbf{B}(\mathbf{u}\cdot \mathbf{B})/{\mu }_0$. This is a mechanical barotropic energy identity; it does not say that an isothermal gas is thermally isolated.

For fixed cloud shape, gravitational energy scales as $-C_{g}G{M}^2/L$ and magnetic energy as $C_B{\Phi }^2/({\mu }_{0}L)$. The magnetic-criticality threshold for negligible pressure is $M/\Phi >\sqrt{C_B/C_g}/\sqrt{{\mu }_{0}G}$. Ideal flux freezing keeps this ratio fixed during homologous collapse; geometry determines the numerical coefficient.

A small weakly susceptible sphere has induced magnetic energy $-\chi V{B}^2/(2{\mu }_0)$ and vertical force $\chi VB{B}^{\prime }/{\mu }_0$. In field magnitude $B=B_0{r}_0/z$, balancing gravity gives the title height for $\chi <0$. The vertical force varies as $z^{-3}$, giving a restoring derivative at that equilibrium. Three-dimensional trapping cannot be established from an axial field magnitude alone; lateral field geometry and Maxwell constraints also matter.

Electrical reactance $X$ is the signed imaginary part of electrical impedance $Z=R+\mathrm{i}X$, measured in ohms. It describes the part of the terminal electric potential difference in phase quadrature with the electric current in a sinusoidal steady state. In the $e^{\mathrm{i}\omega t}$ convention, an ideal inductor has $X_L=\omega L>0$ and an ideal capacitor has $X_C=-1/(\omega C)<0$ for $\omega >0$. A large $|X|$ reduces electric current at fixed applied voltage, but an ideal purely reactive component returns the energy it receives over each cycle. This is why electrical reactance and electrical resistance have the same units yet describe different physical behavior. ^[ref].

In a series RLC circuit, electrical impedances add because the same electric current passes through every component. For ideal positive inductance $L$, capacitance $C$, and total electrical resistance $R_{\Sigma }>0$,Here $\varphi$ is voltage wave phase minus current wave phase. Below ${\omega }_0=1/\sqrt{LC}$, $X<0$ and electric current leads terminal voltage; above it, $X>0$ and electric current lags. At ${\omega }_0$, the two signed electrical reactances cancel, leaving $Z=R_{\Sigma }$ and the maximum electric current at fixed terminal voltage. This is the resonant series RLC response. Electrical resistance does not cancel: the toy circuit gives $f_0={\omega }_0/(2\pi )=1591.55\mathrm{Hz}$ and $I_{\mathrm{rms}}=9.091\mathrm{mA}$ there. These numbers follow directly from the stipulated component values. ^[ref].

Cancellation concerns terminal voltage, not the disappearance of stored energy. In the toy circuit at resonance, the individual inductor and capacitor voltages are each $0.9091\mathrm{V}$ in root mean square magnitude, with opposite wave phases, while their sum is zero. Their peak stored magnetic energy and capacitor energy are each $0.8264\mu \mathrm{J}$, attained at different times. With smaller positive $R_{\Sigma }$, either component voltage can exceed the applied voltage. This ability to select a frequency while sustaining internal energy exchange connects electrical reactance to tuning and frequency response.

For an ideal inductor or capacitor, the absorbed instantaneous power equals the rate of change of stored energy:Positive power stores energy; negative power returns it. Over a periodic steady state, each stored energy returns to its initial value, so its cycle-averaged absorbed power is zero. An ideal resistor instead has $p_R=R{i}^2\ge 0$, transferring energy into Joule heating. More generally, the average terminal power of this series RLC circuit isZero average power in an ideal reactive component does not imply zero instantaneous power, zero electric current, or zero stored energy. Real coils and capacitors have losses, so zero ideal reactive loss is a model assumption rather than a claim about every component. ^[ref].

For smooth ideal magnetohydrodynamics of constant mass density, define $I_{\pm }={\int }_V|{\mathbf{z}}^{\pm }{|}^{2}dV$. Taking the dot product of the Elsässer variable equation with $2{\mathbf{z}}^{\pm }$ givesTherefore each integral is constant when its outward boundary flux vanishes, for example when both velocity and magnetic field are tangent to the fixed boundary, or for periodic boundary conditions. The kinetic energy plus magnetic energy is $\rho (I_{+}+I_{-})/4$, while the cross-helicity is $\sqrt{{\mu }_0\rho }(I_{+}-I_{-})/4$.

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

For a fixed Newtonian gravitational potential, total energy density is $\mathcal{E}=\rho (u^2/2+\Phi )+p/(\gamma -1)+B^2/(2{\mu }_0)$. Its flux combines advected kinetic/potential energy, specific enthalpy and the Poynting vector. Dotting the ideal magnetohydrodynamic momentum equation with velocity, adding the adiabatic internal energy equation and the magnetic energy equation cancels the magnetic work. The pressure terms combine into $-\nabla \cdot (p\mathbf{u})$, proving the conservation law. A time-dependent imposed potential instead contributes $\rho {\partial }_t\Phi$.

Inductance relates electric currents to the magnetic flux linkage they produce. For a single linear coil with fixed geometry and magnetic response, its self-inductance $L$ is the constant inHere $\Lambda$ is magnetic flux linkage from the coil's own electric current $I$. The SI unit of inductance is the henry. Knowing inductance lets us predict the induced electromotive force when electric current changes, and the magnetic energy stored by an inductor. ^[ref].

For example, a coil of self-inductance $0.5\mathrm{H}$ carrying electric current $3\mathrm{A}$ has magnetic flux linkage $1.5\mathrm{Wb}$. If its $100$ turns link the same magnetic flux, each turn has magnetic flux $0.015\mathrm{Wb}$. Keeping the geometry and magnetic response fixed while doubling electric current doubles magnetic flux linkage and quadruples the inductor's magnetic energy.

The comparison with capacitance is useful: a linear capacitor has $Q=CV$ and stores capacitor energy $C{V}^2/2$ in its electric field, whereas a linear inductor has $\Lambda =LI$ and stores magnetic energy $L{I}^2/2$ in its magnetic field. The RLC circuit brings both forms of energy storage together.

An inductance matrix relates magnetic flux linkages to electric currents in a linear model of coupled inductors. For reciprocal two-coil coupling, $\mathbf{L}=\left(\begin{array}{cc}{L}_1& M\\ M& L_2\end{array}\right)$. Its stored magnetic energy is ${\mathbf{i}}^T\mathbf{Li}/2$. With $L_1,L_2>0$, it is a positive-definite matrix exactly when $M^2<L_1{L}_2$. Equality is a singular limiting model requiring separate treatment.

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.

An ideal source $V$ feeds a series resistance $2R$, followed by parallel branches $R+L$ and $2R$. The Thévenin equivalent seen by the inductor has voltage $V/2$ and resistance $R+(2R\parallel 2R)=2R$. From zero initial current, the inductor current is $V(1-e^{-t/\tau })/(4R)$, the other branch current is $V(1+e^{-t/\tau })/(8R)$, and the source current is $V(3-e^{-t/\tau })/(8R)$. Kirchhoff's current law holds at every time, and the final stored magnetic energy is $L{V}^2/(32R^2)$.

Extrapolate an anchored magnetic dipole field to the light cylinder $R_c=c/\Omega$ and estimate outward power as magnetic energy density times $c{R}_c^2$. The result fixes the displayed scaling but not the geometric coefficient. An aligned vacuum dipole does not radiate, while a current-carrying plasma magnetosphere can have nonzero aligned spin-down. Corotation at the light cylinder is a notional limiting construction, not a material surface moving rigidly at light speed.

The linear incompressible flow $\mathbf{u}=(-\omega x,-\omega y,2\omega z)$ compresses the transverse plane and stretches the axial direction. An axial magnetic field obeys $B_t-\omega r{B}_r=2\omega B+\eta {\Delta }_{\perp }B$. In ideal magnetohydrodynamics, its amplitude grows as $e^{2\omega t}$ while its transverse width decreases as $e^{-\omega t}$. Thus its magnetic energy per unit axial length grows as $e^{2\omega t}$, even when opposite-polarity angular sectors give zero signed magnetic flux. Finite magnetic diffusivity limits the amplification of localized finite-flux Gaussian fields.

In vacuum-permeability magnetohydrodynamics, the magnetic energy in a volume is $E_B=\int B^2/(2{\mu }_0)dV$. Its volume density equals the magnetic pressure.

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.

The general ideal magnetic-energy balance has outward-surface contribution ${\mu }_0^{-1}\oint [(u\cdot B)B-B^{2}u]\cdot dS$ and volume contribution $-\int u\cdot (J\times B)dV$. The latter vanishes for a force-free magnetic field. For an axisymmetric boundary rotating tangentially, $u\cdot B=\Omega f(\psi )$ and the directed flux of a thin tube is $2\pi d\psi$. Pair its exit and entry endpoints to obtain the displayed energy injection, counting each boundary endpoint once. Equal endpoint angular velocities inject no magnetic energy.

In a bounded regular domain, fix the global extension or circulation convention for the magnetic vector potential. A bounded inverse of curl on tangent solenoidal vector fields gives $\Vert \mathbf{A}{\Vert }_2\le C_D\Vert \mathbf{B}{\Vert }_2$. The Cauchy-Schwarz inequality then gives $|H_M|\le C_D\Vert \mathbf{B}{\Vert }_2^2=2{\mu }_0{C}_D{E}_B$. One construction extends the tangent field by zero outside the domain and uses the whole-space Biot-Savart law; its kernel of size $|\mathbf{x}-\mathbf{y}{|}^{-2}$ is integrable over a bounded region, giving the required $L^2$ bound. Nonzero conserved magnetic helicity therefore prevents complete decay of magnetic energy.

The shell-integrated magnetic energy spectrum distributes magnetic energy over wavenumbers. With a solenoidal isotropic spectral tensor normalized to ${\hat{C}}_{ii}=2H(k)$ and the inverse Fourier transform measure $d^{3}k/(2\pi {)}^3$, one has $\mathbb{E}|\mathbf{B}{|}^2={\pi }^{-2}\int k^{2}H(k)dk$. In Gaussian units, $E_B(k)=k^{2}H(k)/(8{\pi }^3)$ integrates to the energy density $\mathbb{E}|\mathbf{B}{|}^2/(8\pi )$. A spectrum per unit mass or normalized to $B^2/2$ has different constant factors. Scale-local fluctuation estimates give $\delta B_l^2$ proportional to $k{E}_B(k)$ at $k\sim l^{-1}$ when the spectrum is normalized consistently.