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For the prescribed displacement along $z$, the source acceleration is $\mathbf{a}(t_r)=-a{\omega }^2\cos (\omega t_r)\hat{\mathbf{z}}$. If $\theta$ is the angle from the $z$ axis, $|\mathbf{a}{|}^2-(\mathbf{n}\cdot \mathbf{a}{)}^2=a^2{\omega }^4{\cos }^2(\omega t_r){\sin }^2\theta$. Averaging ${\cos }^2$ over one wave period givesThe angular distribution of electric dipole radiation has zero intensity on the oscillation axis and maximum intensity in the equatorial plane, with no azimuthal dependence. Its total power isThe condition $a\omega \ll c$ is the required nonrelativistic condition. The plotted radial coordinate below is angular power, rather than field wave amplitude.

Since $\mathbf{E}\cdot \mathbf{n}=0$, the Poynting vector isThe area on a distant sphere corresponding to solid angle $d\Omega$ is $r^{2}d\Omega$. ConsequentlyThe average is over the source motion at retarded time. At leading nonrelativistic order, differences between source-time and reception-time averaging are beyond the approximation. The units are power per solid angle, $\mathrm{W}\mathrm{s}{\mathrm{r}}^{-1}$.

The acceleration covariance in nonrelativistic radiation is another way to organize this result: with $C_{ij}=⟨a_i{a}_j⟩$, the angular factor is $\mathrm{tr}C-{\mathbf{n}}^{T}C\mathbf{n}$. Integrating over all observation directions uses $\int [|\mathbf{a}{|}^2-(\mathbf{n}\cdot \mathbf{a}{)}^2]d\Omega =(8\pi /3)|\mathbf{a}{|}^2$ and gives the averaged Larmor formula, $⟨P⟩=q^2⟨a^2⟩/(6\pi {ϵ}_0{c}^3)$.

The consistent Liénard–Wiechert potential obeys $\mathbf{A}=\mathbf{u}\phi /c^2$. To first order in $\mathbf{\beta }=\mathbf{u}/c$, its large-distance form isTerms varying as $1/r^2$ give near fields and are negligible for the radiated power at large $r$. Keep the first-order velocity term in $\phi$: although small, its retarded-time derivative contributes a leading acceleration field. To the required order, $\partial t_r/\partial t\simeq 1$ and $\nabla t_r\simeq -\mathbf{n}/c$. Holding the observation direction fixed at leading order givesThus the outgoing electric field is transverse:Here $\mathbf{a}=d\mathbf{u}/d{t}_r$. The magnetic field follows from $\nabla \times \mathbf{A}$, or from the outgoing plane electromagnetic wave relation:The minus sign is essential. Both fields fall as $1/r$, are mutually orthogonal and obey $|\mathbf{E}|=c|\mathbf{B}|$. The corresponding Liénard-Wiechert field also contains nonradiative terms, omitted in this leading far-field calculation.

In Rayleigh scattering, a small bound system develops an induced electric dipole moment $\mathbf{p}(t)=\mathrm{Re}[\alpha (\omega ){\mathbf{E}}_0{e}^{-i\omega t}]$. Well below its binding resonances, the electric polarizability is approximately its static value $\alpha (0)$, independent of frequency. The mean radiated power and incident light intensity areDividing gives the total scattering cross sectionThis Rayleigh scattering wavelength dependence comes from the two time derivatives of an oscillating electric dipole moment, whose radiated field is proportional to ${\omega }^2{p}_0$. It requires the same scatterer with approximately constant low-frequency electric polarizability; resonances and strong optical dispersion invalidate the simple fourth-power law.

Thomson scattering describes the low-recoil scattering of electromagnetic radiation by a free charge. For charge $q$ and mass $m$, the weak incident field gives peak acceleration $|q|E_0/m$. The mean Larmor formula power is thereforeThere is no binding force to introduce a characteristic frequency. Dividing by the same incident light intensity gives the Thomson cross-section:For an electron, the length inside parentheses is the classical electron radius. The cross section has units ${\mathrm{m}}^2$ and is independent of wavelength in this classical limit. The driven velocity must remain small compared with $c$, and $\hslash \omega \ll m{c}^2$ must make Compton recoil kinematics negligible.

The transmitting antenna is matched to its input transmission line, so its real terminal electrical resistance is $Z_i$. The identical receiving antenna has the same electrical impedance at the operating frequency. Its available received power for a matched load is the incident energy flux times its antenna effective area:The real receiving load is $Z_t$, so its voltage-wave reflection coefficient relative to $Z_i$ is ${\Gamma }_L=(Z_t-Z_i)/(Z_t+Z_i)$. The fraction of available power delivered to the output line isConsequently, neglecting path losses, mismatch of polarization of an electromagnetic wave and line losses,This is the Friis transmission equation with a receiving impedance-mismatch factor. For fixed $P_i,G_a,d,Z_i,Z_t$, it scales as ${\lambda }^2$. That scaling is conditional on holding antenna gain fixed: a fixed physical aperture can instead have $G_a\propto {\lambda }^{-2}$, and any wavelength dependence of the impedances also enters the expression. The matched-output check $Z_t=Z_i$ gives $P_t=P_i\mathcal{F}$.

Distinguish reflection at the input terminal, reflection at the receiving load, and a returned signal that traverses the free-space link twice. The input antenna is initially matched, so its direct terminal reflection is zero.

In the usual weak-coupling, matched-port two-port network model for the antenna pair, let the forward and reverse wave amplitude transfer coefficients be $t_{21}$ and $t_{12}$. Electromagnetic reciprocity gives equal power transfer factors $|t_{21}{|}^2=|t_{12}{|}^2=\mathcal{F}$, where $\mathcal{F}=(G_a\lambda /(4\pi d){)}^2$ from part (a). The receiving-load reflection returns wave amplitude $t_{12}{\Gamma }_L{t}_{21}$ to the original input port. The load-mismatch contribution to the returned power is thereforeFor fixed gains and impedances this returned power scales as ${\lambda }^4$: it contains two link traversals. By contrast, the reflected power at the receiving load before that return traversal is $P_i\mathcal{F}|{\Gamma }_L{|}^2\propto {\lambda }^2$. Calling that local reflected power the power already back in the input line would omit one propagation loss.

The original wording does not specify an antenna scattering model. The boxed result is the load-dependent echo in an ideal matched-port link, with structural scattering and residual input coupling neglected. The actual total input return can include coherent scattering from the receiving antenna even when ${\Gamma }_L=0$; gains and effective areas alone do not specify that background. If “reflected back” means only the direct reflection at the stated matched transmitting antenna, its value is $0$. These meanings should not be conflated; the ${\lambda }^4$ result requires the two-pass echo interpretation and its stated model.

After field electron emission, let $h$ denote the plate separation of the parallel-plate capacitor, which the question does not specify. The uniform capacitor electric field has magnitude $E_c=|V|/h$, and a nonrelativistic electron of mass $m_e$ has constant acceleration magnitude $a=e{E}_c/m_e$. The sign of its charge affects the direction of the radiation field but not its radiated power.

In the supplied nonrelativistic Liénard-Wiechert field, $\dot{\mathbf{\beta }}=\mathbf{a}/c$. At polar angle $\theta$ to the acceleration, $|\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{a})|=a\sin \theta$, henceThe outgoing Poynting vector has magnitude $S={ϵ}_{0}c{E}_{\mathrm{rad}}^2$. Its radiated power per solid angle is thereforeUsing $\int {\sin }^2\theta d\Omega =8\pi /3$ gives the Larmor formula and the requested constant power during the uniform-field part of the flight:The brackets in the given field require the acceleration at retarded time; the radiation reaches the far observer after the propagation delay. This derivation assumes that radiation reaction is a small correction to the acceleration and that the flight is nonrelativistic, for example $e|V|\ll m_e{c}^2$ for release from rest. Entry, emission and impact transients are not described by the uniform-field model. A result solely in terms of $V$ cannot be obtained without a plate spacing or a specified internal field.

The Johnson-Nyquist noise formula gives the open-circuit mean-square voltage in a small frequency band $d\nu$. When a source electrical resistance $R$ drives a load $R$ with impedance matching, the load sees half the source voltage, so the available noise power isThe factor of four in the open-circuit formula is therefore not an extra factor in the delivered power.

The transmitter's antenna gain $G$ makes the axial spectral energy flux at range $d$Multiplication by the receiving antenna effective area $A$ gives $d{P}_r=k_B{T}_{t}GAd\nu /(4\pi d^2)$. Define the equivalent receiving antenna temperature $T_A$ by the available matched-load noise $d{P}_r=k_B{T}_{A}d\nu$. ThenThis is the transmitter-induced equivalent noise temperature of the receiver port. It does not determine the actual physical temperature of its internal electrical resistance, or any additional receiver and background noise. The calculation assumes the usual far-field link, negligible path loss beyond geometric spreading, matched polarization of an electromagnetic wave at the two antennas and the stated effective collecting area. No additional one-half polarization factor is needed for the matched polarization of radiation from one transmitting antenna.

Only the $1/r$ fields transport a nonzero outgoing power to infinity. Retain those terms from the Hertzian dipole fields:The brackets mean evaluation at the retarded time,for a point source at the origin in vacuum. They are not a time average. The radial Poynting vector in the radiation zone isIntegrating over a sphere usesThereforeNear-field terms either give inverse powers of $r$ after the surface integral or describe reversible exchange with the near field. The emitted power labelled by source time $t_{\mathrm{src}}$ is ${\mu }_0\ddot{p}(t_{\mathrm{src}}{)}^2/(6\pi c)$; the enclosing surface receives it one light-travel time later. For a changing dipole orientation, the general form is $P={\mu }_0|\ddot{\mathbf{p}}{|}^2/(6\pi c)$.

The given static induced surface charge and the dipole integral from part (b) giveHere $ϵ$ is the dimensionless relative permittivity, as required by the stated $ϵ\pm 1,2$ expression; ${\alpha }_E$ is the SI electric polarizability. Take ${\mathbf{E}}_{\mathrm{inc}}(t)=E_0\cos \omega t\hat{\mathbf{e}}$ and $k=\omega /c=2\pi /\lambda$. In the Rayleigh scattering regime the field is approximately uniform across the sphere. Its induced dipole radiates mean powerThe incident time-averaged light intensity in vacuum is $I_{\mathrm{inc}}={ϵ}_{0}c{E}_0^2/2$. Dividing power by intensity gives the total cross section,The units are ${\mathrm{m}}^2$; the comparison is with the geometric projected area $\pi R^2$, not with a length:The inequality holds for an ordinary nonabsorbing dielectric, with finite contrast factor and $kR\ll 1$. It explains why a sufficiently small transparent sphere scatters weakly and why its scattered power has the characteristic ${\lambda }^{-4}$ dependence. Near a material resonance where $ϵ+2$ is small, this simple inequality and static-polarizability approximation need reconsideration, including loss and radiative corrections. For the stated low-frequency ordinary dielectric the supplied polarizability is the intended approximation.

A gravitational wave produces a time-dependent tidal distortion. For a normally incident plus-polarized wave in the long-wavelength limit, two perpendicular freely suspended arms change by $\delta L_x/L=h_{+}/2$ and $\delta L_y/L=-h_{+}/2$. Their differential length is thereforefor a simple round-trip Michelson interferometer. LIGO uses kilometre-scale orthogonal arms with optical cavities that enhance the phase response, a stabilized laser and suspended mirrors in vacuum. As an illustrative scale, $h=10^{-21}$ across a 4 km arm corresponds to differential length $4\times 10^{-18}\mathrm{m}$. The measured quantity is returning light phase, not mirror displacement against a rigid external ruler. Optical storage and control loops give a frequency-dependent calibrated response.

Seismic and terrestrial disturbances dominate low frequencies, thermal motion of suspensions and mirror coatings contributes at intermediate frequencies, and photon shot noise limits high-frequency readout. Isolation and suspension systems reduce ground coupling; low-loss materials reduce thermal noise; vacuum suppresses gas scattering and forces. Greater optical power reduces shot noise but increases radiation-pressure fluctuations and thermal demands. Stable optical alignment and feedback are essential, yet their own sensor and actuator noise must be controlled. These are engineering requirements because the desired strain is extraordinarily small.^[ref]

A candidate should appear coherently at separated observatories with consistent propagation delay and antenna response. Auxiliary environmental monitors, calibration injections and time-shifted coincidence backgrounds test instrumental explanations. Matched filtering is powerful for modelled compact-binary chirps; coherent burst searches allow less predictable signals. Binary-pulsar orbital energy loss already provided indirect evidence for gravitational radiation before 2013, but it is distinct from directly measuring passing waves.^[ref] At the examination date, direct detection remained a search objective. Reliable detection requires both sufficient strain sensitivity and a demonstrated astrophysical signal rather than correlated or accidental noise.

The cosmic microwave background is relic thermal radiation from the hot early universe. Its temperature anisotropy is expanded in spherical harmonics, with angular power spectrum $C_{\ell }^{TT}$. Thomson scattering of a local radiation quadrupole generates linear polarization around last scattering and again during reionization. Polarization maps are decomposed into E-mode polarization and B-mode polarization, giving spectra such as $EE$, $TE$ and $BB$ in addition to temperature information.

Cosmic microwave background acoustic peak positions compare the sound horizon with the angular-diameter distance to last scattering, constraining geometry and expansion history. Relative peak heights constrain baryon loading and matter-radiation equality; the damping tail constrains diffusion and other early-universe physics. The scalar spectral index and primordial scalar amplitude constrain inflationary perturbations. Coherent acoustic structure and temperature-polarization correlations support predominantly adiabatic initial perturbations. Large-angle E-mode polarization measures the reionization optical depth, helping break its degeneracy with the scalar amplitude inferred from small-scale temperature anisotropy.

Under a specified cosmological model, these data constrain ordinary and cold dark matter densities, the Hubble constant, spatial curvature and cosmic age. Complementary distance information helps break the geometrical degeneracy: temperature peaks alone do not uniquely establish a cosmological constant. The Planck 2013 parameter analysis used Planck temperature data and, in important combinations, WMAP low-multipole polarization; it should not be described as a full Planck polarization release.^[ref]

Linear scalar cosmological perturbations generate E-mode polarization but no primordial B-mode polarization in the simplest treatment. tensor cosmological perturbations can generate B-mode polarization, making them a possible probe of inflationary gravitational waves. gravitational lensing converts some E-mode power into B-mode polarization, however, and Galactic dust and synchrotron radiation also contribute polarized foregrounds. Multifrequency observations, foreground modelling, beam calibration and control of temperature-to-polarization leakage are therefore necessary. Polarization does not by itself establish a primordial tensor signal. temperature and polarization jointly test the expansion history and primordial perturbations, while complementary data and foreground control determine how uniquely parameters can be inferred.

For a homogeneous canonical scalar field, ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. Potential-energy domination gives $P_{\varphi }\simeq -{\rho }_{\varphi }$, accelerated expansion and a decreasing comoving Hubble radius. Hubble friction can permit slow roll down a sufficiently flat potential. Enough inflation stretches an initially small causally connected region to encompass our observable universe, suppresses the relative spatial-curvature term and dilutes unwanted massive relics. quantum fluctuations supply possible seeds for cosmological structure; a successful model must also end inflation and create the hot post-inflationary universe. These statements do not imply that arbitrary initial conditions inevitably inflate.

Use the stated reduced-Planck units throughout the field calculation. Differentiating the Friedmann constraint and substituting the field equation givesDivision assumes the nontrivial expanding branch $H>0$; the vacuum endpoint follows continuously.

During rapid oscillations, require $m\gg H$ and $|\dot{A}|\ll mA$. To leading order, $\dot{\varphi }\simeq -mA\sin mt$, while the Friedmann constraint gives $H_m^2\simeq m^2{A}^2/6$. ThusAveraging over many oscillations on a timescale still short compared with $H^{-1}$ givesThe quadratic-inflaton Hubble oscillations obey, to leading order in $H_m/m$,Differentiating this expression reproduces the cosine term, with smaller corrections from the slowly varying amplitude. Instantaneously $H$ remains nonincreasing because its exact derivative is $-{\dot{\varphi }}^2/2$; an oscillatory correction does not mean oscillatory growth of the total Hubble parameter.

The mean equation integrates toThe shift $t_{\ast }$ is set by matching to the earlier evolution, not necessarily the Big Bang origin of the full solution. Averaged kinetic and potential energies are equal, so $⟨P_{\varphi }⟩\simeq 0$, $⟨{\rho }_{\varphi }⟩\propto a^{-3}$ and $A\propto a^{-3/2}$. The coherent quadratic oscillations behave as pressureless matter, not radiation, even though they are a scalar field rather than a gas of cold particles. radiation would instead give $a\propto t^{1/2}$. The averaged approximation is controlled only after oscillations are fast compared with expansion.

reheating requires interactions beyond the isolated field equations above. The oscillating inflaton can decay into other particles, whose scattering and further decays establish a thermal bath. The perturbative inflaton reheating equations areExpansion redshifts the components differently; transfer becomes efficient around $H\sim \Gamma$, and radiation domination follows when the transferred relativistic energy dominates. For suitable couplings, nonperturbative preheating by parametric resonance can precede thermalization, with backreaction and rescattering affecting its completion. Neither the decay rate nor thermalization is specified by $m$ and $V$ alone. Instantaneous reheating idealizes all this as immediate conversion of the end-of-inflation energy to radiation.^[ref]

For instantaneous-reheating horizon matching, specify the comparison: take the initial patch to be of order $H_i^{-1}$, compare the same-shaped inflationary history with approximately fixed $H_i/H_{\mathrm{end}}$, and assume instantaneous reheating followed by adiabatic expansion. Ignoring changes in relativistic degrees of freedom, $a_0/a_{\mathrm{reh}}\propto T_{\mathrm{reh}}$, while energy conservation and the Friedmann equation give $H_{\mathrm{end}}\propto T_{\mathrm{reh}}^2$. The present size of this inflated patch therefore scales asAt fixed present Hubble radius, a higher instantaneous reheating scale requires more e-folds in this standard comparison, approximately $\Delta N\simeq \ln (T_2/T_1)$, up to slow scale and entropy-degree corrections. Hotter reheating gives more subsequent expansion, but the associated initial Hubble patch is smaller by a stronger power of temperature. If instead one artificially holds the initial physical patch or $H_i$ fixed while varying the end energy independently, the post-inflationary geometric effect alone would give the opposite trend. temperature without an inflation-scale comparison convention does not fix that alternative problem; this is why the matching assumptions must be stated.^[ref]

Hyperscaling adds the physical assumption that the singular free energy associated with one correlation volume is of order $k_B{T}_c$:At fixed scaled field $x=h/s^{\Delta }$, this predicts $f_s\sim s^{d\nu }$, whereas the given thermodynamic form has power $2-\alpha$. HenceThe product $g_f(x)[g_{\xi }(x){]}^d$ can be a finite dimensionless function of fixed $x$; it need not be constant. The homogeneity of $\xi$ alone is insufficient: the correlation-volume free-energy assumption supplies the connection. Above an upper critical dimension, a dangerously irrelevant coupling can invalidate this argument; at a marginal dimension logarithms may qualify pure powers.

A coarse-grained scalar order parameter $\varphi (\mathbf{x})$ describes a short-range Ising-like transition with sign symmetry. Write the dimensionless Landau-Ginzburg functional asHere $r=a_{t}t$, with the relevant critical-mass shift understood when fluctuations are included. Higher local powers and gradients are negligible only within a justified long-distance expansion. The partition function is $Z=\int \mathcal{D}\varphi e^{-\mathcal{H}}$ with a microscopic cutoff. The local functional is not already the fluctuation-integrated thermodynamic free energy.

Mean-field theory replaces this integral by its stable uniform saddle, $r{\varphi }_0+u{\varphi }_0^3=h$. At $h=0$,The branch curvature is $r_L=r+3u{\varphi }_0^2$: it is $r$ above and $2|r|$ below. HenceTogether with the finite heat-capacity jump from $f_{\min }$, this yieldsThe equal-$|t|$ susceptibility amplitude ratio is two. Uniform saddles also miss domain walls and spatial texture. For an $n$-component order parameter, symmetry allows a vector generalization; ordered transverse directions additionally carry Goldstone fluctuations.

To include small fluctuations set $\varphi ={\varphi }_0+\delta \varphi$. The Gaussian action is $\frac{1}{2}\int [c(\nabla \delta \varphi {)}^2+r_L(\delta \varphi {)}^2]$, givingThe displayed $G(q)$ is a spectral covariance per Fourier delta, avoiding an extra volume factor. As $r_L\to 0$, a growing correlation length lets increasingly large correlated regions fluctuate. The singular Gaussian heat correction contains $\int d^{d}q/(c{q}^2+r_L{)}^2$, which is infrared singular for $d<4$ and logarithmic at four. This signals failure of a simple saddle expansion; the Gaussian heat exponent obtained by extrapolation is not the exact interacting critical exponent.

The Ginzburg criterion compares the fluctuation of a correlation-volume average with ${\varphi }_0^2$, rather than using all ultraviolet fluctuations at a point. Keeping $q\lesssim {\xi }^{-1}$ below the transition gives, up to finite convention-dependent constants,Dividing by ${\varphi }_0^2=|r|/u$ yieldsFor $d<4$, writing $r=a_{t}t$ gives a nonuniversal Ginzburg reduced-temperature windowInside $|t|\lesssim \mathrm{Gi}$, interacting fluctuations must be treated. For example, in three dimensions $\mathrm{Gi}\sim u^2/(c^3{a}_t)$ in this normalization. Above four dimensions the long-wavelength ratio vanishes at criticality; at four it is marginal and interaction-generated logarithms require further analysis.

The spatial gradient term supplies correlations absent from uniform Landau theory. The Ginzburg test bounds where its mean-field saddle remains self-consistent; it does not compute nonclassical exponents or a universal material-independent window.^[ref]

Let $\dot{M}=500\mathrm{kg}{\mathrm{s}}^{-1}$ and $V=10\mathrm{m}{\mathrm{s}}^{-1}$. Incoming material has zero horizontal momentum, while incorporation requires horizontal momentum $VdM$. Thus the engine supplies force $F=V\dot{M}$, and its power isNo duration or loading distance is given, so a single total work cannot be assigned. Over an interval $t$ in seconds,The remaining $25,000t\mathrm{J}$ becomes internal energy in the inelastic collision that brings the falling material to the wagon's horizontal velocity. Vertical falling kinetic energy, if any, is a separate contribution to dissipation; it does not alter this horizontal work balance. The source alternates between stone and coal; the calculation depends only on the given loading rate.

Let the motor's constant power be $P>0$ and put $\omega =\dot{\theta }$. The braking coefficient in the proposed formula must describe a retarding torque $k\omega$. This constant-power rotational acceleration obeys the energy balance:Consequently the steady angular velocity is ${\omega }_0=\sqrt{P/k}$, and the ordinary rotational frequency is ${\omega }_0/(2\pi )$. Solving the linear equation for ${\omega }^2$ from rest givesThe printed time formula is missing a minus sign. Its logarithm is negative for $0<\omega <{\omega }_0$, whereas elapsed time must be positive. The terminal speed is approached only as $t\to \infty$.

A constant-power motor starting exactly from rest is an idealization: its driving torque $P/\omega$ would diverge. The energy equation has a regular solution for ${\omega }^2$, although the initial angular acceleration is singular. If $k\omega$ were literally a tangential force, a lever arm would also be needed; the dimensions of the requested expression identify $k$ as a rotational damping coefficient.

At terminal velocity, the upward magnetic force balances the weight. Substitute the electric current from the preceding solution:The geometric factors cancel, givingHere $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $B=1\mathrm{T}$, ${\rho }_m=8960\mathrm{kg}{\mathrm{m}}^{-3}$ and ${\rho }_c=1.7\times 10^{-8}\mathrm{\Omega }\mathrm{m}$. As numerical checks, the specified $a=0.08\mathrm{m}$ and $d_w=0.001\mathrm{m}$ give $M=2.252\times 10^{-3}\mathrm{kg}$, $R=6.926\times 10^{-3}\mathrm{\Omega }$ and $I_s=0.276\mathrm{A}$.

The gravitational power lost, $Mg{v}_s$, equals the Joule heating $I_s^{2}R$, consistent with energy conservation. This terminal velocity applies only while the top arm remains in the magnetic field; after it exits, the magnetic braking ceases, apart from any brief inductive transient.

For an ideal inductor of constant self-inductance $L$, the passive-sign electric potential difference is $V_L=LdI/dt$. The supplied power is $V_{L}I$, so increasing the electric current quasistatically from zero storesAny electrical resistance would additionally dissipate energy as heat; it is not included in this stored magnetic energy.

For a long vacuum solenoid with $N$ turns, axial length $\ell$ and cross-sectional area $A$, Ampère's law gives $B={\mu }_0(N/\ell )I$. Its flux linkage is $N\Phi =NBA$, soThe nearly uniform field occupies volume $A\ell$, with negligible external-field energy in the long-solenoid approximation. Dividing its stored energy by this volume yieldsThe vacuum and linear-inductance assumptions matter; a nonlinear magnetic material requires the appropriate field-energy integral rather than simply substituting a varying inductance into $L{I}^2/2$.