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For an ideal electrical transformer with turns ratio $n=N_p/N_s$, the secondary load $R_L$ appears at the primary as $n^2{R}_L$, while a source resistance $R_g$ appears at the secondary as $R_g/n^2$. A finite magnetizing inductance $L_m$ contributes a parallel branch $i\omega L_m$ at the primary; its magnitude must greatly exceed the reflected load for the loaded input to be approximately resistive. The corresponding unloaded secondary self-reactance is $\omega L_m/n^2$. That reactance is distinct from the source's transformed output resistance. A specification that simultaneously calls the loaded input matched to a small resistance and calls the primary impedance very large must distinguish these two branches.

The 1933 report Ein neuer Effekt bei Eintritt der Supraleitfähigkeit by W. Meissner and R. Ochsenfeld describes cooling tin and lead cylinders in a fixed transverse magnetic field. Two parallel cylinders were about $140\mathrm{mm}$ long and $3\mathrm{mm}$ thick, separated by $1.5\mathrm{mm}$. A roughly $10\mathrm{mm}$ rotating search coil between them drove a ballistic galvanometer. At about $5\mathrm{G}$, the reported below/above-transition induction ratios were $1.70$ for tin single crystals and $1.77$ for polycrystalline lead. The increased gap magnetic field indicates redistribution outside the cylinders, rather than an unchanged frozen field. Coil-winding geometry and noncircular tin specimens limited comparison with the ideal prediction. ^[ref]

Their second arrangement used a lead tube, about $130\mathrm{mm}$ long with $3\mathrm{mm}$ outer and $2\mathrm{mm}$ inner diameters. A rotating search coil could sit inside or alongside it. The bore retained magnetic flux after removing the applied magnetic field; the coil nearly filled the bore, preventing a test of interior field uniformity. A hole containing magnetic field is distinct from magnetic field penetrating the surrounding material exhibiting superconductivity. ^[ref]

The modern Field-cooled Meissner experiment provides a complementary readout. MIT's probe I places a sample in a brass holder, with an $810$-turn test coil of $7.1\mathrm{mm}$ inner diameter, $10.5\mathrm{mm}$ outer diameter and $12.3\mathrm{mm}$ length, inside a longer solenoid. A nearby silicon-diode thermometer tracks cooling by helium gas. With lead still normal, the guide applies about $200\mathrm{mA}$ to the solenoid and connects the test coil to a current-integrator box and oscilloscope. Cooling produces a transient; warming gives the opposite sign. Switching off the solenoid while the sample is normal checks the sign. ^[ref]

The inferred change is in linked magnetic flux, not a direct scan of $B(z)$. For a stationary test coil the pulse obeys $\int Vdt=-\Delta \Psi$, where $\Psi$ is its magnetic flux linkage. An integrator or AC-coupled oscilloscope needs its transfer response accounted for before extracting this pulse area. Keep the source electric current fixed during cooling, monitor temperature, and compare warming and cooling slowly enough to reduce sample–thermometer lag. An AC inductance change alone can reflect changed normal-state electrical conductivity; the fixed-field expulsion test distinguishes that alternative. ^[ref] Neither this coil readout nor the historical gap-field measurement resolves the microscopic London penetration depth by itself.

Two identical circuits with self-inductance $L$, capacitance $C$ and mutual inductance $\mathcal{M}$ have same-sign and opposite-sign charge modes with ${\omega }_{\pm }^2=1/[C(L\pm \mathcal{M})]$. The inductance matrix acts as the mass matrix; positivity requires $|\mathcal{M}|<L$.

Two perfectly conducting wide strips of width $a$ and separation $d\ll a$, filled with a homogeneous nonmagnetic dielectric of absolute permittivity $\epsilon$, support a transverse electromagnetic mode. Ignoring edge fields, their distributed capacitance is $C^{\prime }=\epsilon a/d$ and distributed external inductance is $L^{\prime }={\mu }_{0}d/a$. The magnetic-energy integral ${\mu }_0(I/a{)}^{2}ad/2=L^{\prime }{I}^2/2$ gives the latter. Their characteristic impedance is $Z_0=(d/a)\sqrt{{\mu }_0/\epsilon }$ and propagation speed is $1/\sqrt{{\mu }_0\epsilon }$. These parameters are per unit length, distinct from a lumped capacitor and inductor.

The AC Josephson effect is $V=({\Phi }_0/2\pi )\dot{\delta }$. Since the imposed flux is held fixed when defining a small-signal inductance, differentiate c to obtainOn the stable small-current branch with $\cos \varphi >0$, $\delta \simeq 0$ and the result reduces to the printed expression,The stable small-signal inductance of a symmetric DC SQUID requires $|I|\ll 2I_C|\cos \varphi |$, stronger than $I\ll I_C$ near a half-integer flux quantum. If $\cos \varphi <0$, the stable zero-current phase is $\delta \simeq \pi$ and $\cos \delta \simeq -1$. Its inductance is positive:For example, $\Phi ={\Phi }_0$ makes the printed signed formula negative when expanded about $\delta =0$, but that phase is an energy maximum. Expanding about the stable $\delta =\pi$ corrects the sign. At ideal half-integer flux the critical current vanishes and the linear inductance formally diverges; a finite transport current cannot satisfy the small-current approximation there. Geometric inductance and other nonidealities then have to be retained instead of extending the simple formula through its pole.

Take $d$ as the centre-to-centre separation and $ϵ={ϵ}_0{ϵ}_r$. Use the ideal lossless two-conductor transmission line model, with nonmagnetic surroundings and negligible internal-wire inductance. Opposite line charges $+{\lambda }_q$ and $-{\lambda }_q$ approximate the cylindrical conductors when $d\gg a$. Integrating their electric fields, or adding their logarithmic electric potentials, gives the potential differenceThis is the two-wire capacitance at large separation. For opposite electric currents $+I$ and $-I$, the magnetic fields between the wires add. The linked magnetic flux per unit length is approximatelyThus the external inductance of a thin two-wire line is $L^{\prime }\simeq {\mu }_0\ln (d/a)/\pi$. The lossless telegrapher's equations give a forward wave with $V/I=\sqrt{L^{\prime }/C^{\prime }}$, so its characteristic impedance isThe product $L^{\prime }{C}^{\prime }={\mu }_0ϵ$ also gives the propagation speed $c/\sqrt{{ϵ}_r}$, an independent consistency check. The exact ideal external-field geometry replaces $\ln (d/a)$ by $\mathrm{arcosh}(d/(2a))$. If the wire separation were specified between surfaces, its centre separation would be $d+2a$, with the same leading large-separation result. Finite electrical conductivity and internal-wire magnetic energy require a more detailed line model.

For motion to the right in $B_0{\mathbf{e}}_z$, $B_0>0$, the induced current is clockwise. Its own magnetic field points into the page inside the circuit, opposing the increase in external flux, as required by Lenz's law. As that current changes, its changing flux also produces an electromotive force opposing the change in current: the circuit has inductance.

Consequently the algebraic current law used in parts (i)-(iv) must be replaced by a dynamical circuit equation that includes the derivative of the self-flux. The induced current cannot in general adjust instantaneously to the bar velocity. Energy is temporarily stored in the self-field and can later return to the motion or be dissipated; the simple exponential braking law can change. Self-field forces and changes of inductance with circuit geometry may also matter. The magnetic self-field changes the transient dynamics but does not create energy: mechanical energy, magnetic-field energy and Joule heat must be included in the conservation balance. A quantitative answer would require the circuit's inductance and its dependence on $s$.

With no electrical resistance or applied electric potential difference, divide the equation by the inductance $L$ to get a simple harmonic oscillator with angular frequency ${\omega }_0=(LC{)}^{-1/2}$. Its general real solution isEach nonzero solution has period $2\pi \sqrt{LC}$. Here ${\omega }_0$ is an angular frequency; the ordinary frequency in cycles per unit time is ${\omega }_0/(2\pi )$.

Taking curls givesBoth have zero divergence and curl, so they satisfy the static vacuum equations. Their upward fluxes through the loop areFor clockwise current $I$, the net force is zero in $B_1$, while in $B_2$

Put $K=2\pi b_2{r}^2$. Motion in $B_1$ changes no flux, so no current is induced and $z=z_0-g{t}^2/2$. In $B_2$, using counterclockwise current as positive,ThusWith $\Gamma =K^2/R$, the velocity tends to $-mg/\Gamma$ andAt terminal speed, $I^{2}R=mg|\dot{z}|$, so Joule loss equals gravitational-energy loss. For $R=0$, Ohm's quasistatic formula is singular: flux is conserved, persistent current stores magnetic energy, and ideal motion has no resistive terminal state (inductance must be retained).

A Hertzian dipole is an ideal localized oscillating source much smaller than its radiation wavelength. Its electric version is an infinitesimal current element; its magnetic version is a small current loop with moment $\mathbf{m}=IA\mathbf{n}$. Such elements model the far-field radiation and radiation resistance of electrically small antennas. The approximation requires source dimensions much smaller than $c/\omega$.

The supplied magnetic field has a quasistatic $[m]/r^3$ term, an induction $[\dot{m}]/(c{r}^2)$ term and a transverse radiation $[\ddot{m}]/(c^{2}r)$ term. The electric field has induction and radiation terms. Brackets mean evaluation at the retarded time $t_r=t-r/c$, not a time average. Near terms mainly exchange stored energy; only the $r^{-1}$ fields give a nonvanishing outgoing power integral as $r\to \infty$.

For a fixed dipole axis the radiation fields satisfy $E_{\varphi }=-c{B}_{\theta }$. The radial Poynting flux is thereforeUsing $\int {\sin }^2\theta d\Omega =8\pi /3$ yieldsFor a general changing dipole direction, replace the squared scalar acceleration by $|\ddot{\mathbf{m}}{|}^2$.

Radiation resistance is the equivalent input resistance whose mean power equals radiated power, not ohmic heating. For a fixed small loop carrying $I(t)=I_0\cos \omega t$, $m=AI$, soThis uses peak current $I_0$; with rms current the power is $I_{\mathrm{rms}}^2{R}_{\mathrm{rad}}$.

The satellite loop needs its changing orientation as well as its induced current. Put the spin axis along z, the external field along x, and let $\mathbf{n}(t)=(\cos \omega t,\sin \omega t,0)$. With $A=\pi b^2$ and resistance $R=10^{-4}\Omega$, the resistance-only approximation givesWrite $m_0=B{A}^2\omega /R$. The full magnetic moment isIt has a static component and a rotating component at $2\omega$. Consequently $|\ddot{\mathbf{m}}{|}^2=4m_0^2{\omega }^4$, andHolding the dipole axis fixed while inserting the sinusoidal current would instead give $B^2{A}^4{\omega }^6/(12\pi {ϵ}_0{c}^5{R}^2)$, eight times smaller. That is the fixed-axis-loop result; it does not describe the stated rotating loop. This is frequency doubling of an induction-driven rotating loop; the changing vector direction is essential.

The resistance-only model is conditional: loop inductance is not supplied. If self-inductance L matters, solve $L\dot{I}+RI=BA\omega \sin \omega t$. In the steady state $I_0=BA\omega /\sqrt{R^2+{\omega }^2{L}^2}$. Its phase changes the static moment direction but not the twice-frequency component's magnitude, so the power becomesThe first expression is the requested resistive idealization and an upper estimate within this passive RL correction. Radiation reaction is negligible compared with the stipulated ohmic resistance. The radiated energy comes from rotational kinetic energy: the induced magnetic torque brakes the loop. Most lost rotational energy is Joule heating; a small part escapes as radiation. Maintaining a constant spin would require external mechanical power.

For scale only, take $B=50\mu \mathrm{T}$, a representative near-Earth field rather than a supplied orbital measurement. The terrestrial surface range reported by NOAA is roughly $25$–$65\mu \mathrm{T}$; the actual orbital value depends on position.^[ref] With $b=1\mathrm{m}$ and $\omega =2\pi (500{\mathrm{s}}^{-1})$, the resistance-only model givesInductance would reduce this power. The kilohertz signal is exceptionally weak and a metre-scale receiving antenna is electrically tiny. Ionospheric absorption/reflection and strong natural or instrumental low-frequency backgrounds make ordinary ground reception impractical. This is not a theorem that the ionosphere blocks every low-frequency mode: field-guided propagation and condition-dependent leakage are real, as the VIPER mission's measurements were designed to test.^[ref]

At a separation $r=100\mathrm{km}$, $k_{\mathrm{rad}}r\simeq 2.1$, so the observer is not securely in the far field. The radiation term alone has at most the characteristic magnetic amplitudein this optimistic resistance-only vacuum model. Near and induction fields must be retained for a full prediction; a simple far-field inverse-square power estimate is not an exact received signal here. Another satellite avoids the whole ground ionospheric path, but plasma propagation, receiving sensitivity, bandwidth, orientation, integration time and spacecraft noise remain unspecified. Detection is not forbidden in principle, but routine practical detection is implausible at this scale; a definitive instrument-level yes/no does not follow from the supplied data.

If $R$ closes the break, a conduction current flows and electrical energy is dissipated. When loop self-inductance is negligible,The externally imposed flux-change emf remains the same. If self-inductance $L_{\ell }$ matters, the current instead satisfies $L_{\ell }\dot{I}+RI=B{a}^2\omega \sin \omega t$; its steady amplitude is $B{a}^2\omega /\sqrt{R^2+{\omega }^2{L}_{\ell }^2}$ and it lags the emf by ${\tan }^{-1}(\omega L_{\ell }/R)$. If the break remains open after inserting a resistor elsewhere in series, the loop is still open and the conduction current remains zero.

The loop area is $A=\pi (5\times 10^{-3}{)}^2{\mathrm{m}}^2$. Integrating Ohm's law and Faraday's law through the whole decay givesThereforeThe switching time cancels: a faster linear flux decay produces a larger electric current for a proportionally shorter interval. If the coil's self-inductance is retained, integrating $LdI/dt$ gives zero provided the initial and final currents vanish and the stated total charge includes the decay tail.

Let $Q(t)$ denote the electric charge on the initially positive capacitor plate and define the discharge electric current by $I=-\dot{Q}$. Kirchhoff's voltage law for the series RLC circuit gives the damped harmonic oscillator equationThe initial conditions are $Q(0)=Q_0$ and $\dot{Q}(0)=0$: the open switch initially prevents electric current, and finite inductance prevents an instantaneous jump in it. Put $\gamma =R/(2L)$ and ${\omega }_0^2=1/(LC)$. The two characteristic roots are $s=-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}$. An oscillatory electric current requires a nonzero imaginary part, henceFor positive $L,C$ and nonnegative $R$, this is the underdamped regime. The initial conditions giveAt equality there is critical damping, and above it there is an overdamped response, with no repeated changes of sign of the discharge electric current. A nonzero initial electric charge is understood; if $Q_0=0$, the stated initial conditions give no electric current at all.

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 $L{I}^2/2$.

The source's variable inductor and variable resistor are in parallel. Let their combined complex impedance beUse the left and right source terminals as the drive nodes. The upper divider has arms $Z_V$ and $i\omega L_1$; the lower divider has arms $R_1$ and $R+i\omega L$. At bridge balance, the detector draws no electric current and the divider midpoint electric potential differences agree, soSubstitution and cancellation at a nonzero drive frequency giveEquating real and imaginary parts yieldsThe first expression has units of resistance and the second units of inductance. Both are independent of the drive amplitude and, for ideal frequency-independent components, of the chosen nonzero drive frequency. At zero frequency the ideal inductors short their arms and the cancellation would not be valid as a measurement of $L$.

Two applications are precision comparison of an unknown electrical resistance with calibrated standards in a Wheatstone bridge, and detecting small resistance changes in a strain gauge or resistance thermometer. Alternating-current bridges additionally compare capacitance, inductance and dielectric losses; this particular bridge determines the two parameters of a non-ideal inductor from a null measurement.

A wirewound resistor uses a metal alloy chosen for stable electrical resistivity. Compared with a typical carbon composition resistor, it can offer tighter resistor tolerance, a smaller temperature coefficient of resistance, lower excess electrical noise, and greater continuous electrical dissipation when its construction provides an adequate heat path. These are design-dependent advantages, rather than guarantees for every alloy or package.

The winding has parasitic inductance and interturn capacitance. Its electrical impedance therefore ceases to be a nearly real electrical resistance at sufficiently high frequencies; an approximate model is a series circuit $R+i\omega L$ shunted by a small parasitic capacitance. A resonance can occur, making a wirewound resistor unsuitable where a nearly frequency-independent electrical resistance is required. Noninductive winding patterns reduce this problem. For comparable small-signal duties it is usually larger and more expensive. A carbon composition resistor is compact and has relatively small parasitic inductance, but commonly has poorer stability, larger resistor tolerance, and more excess electrical noise.

For electrical resistance $R_p>0$ in a parallel circuit with inductance $L>0$, add electrical admittances and invert:Conjugate impedance matching to the series RC impedance requiresIt is possible at a finite positive angular frequency exactly when $C{R}_p^2>L$, in which caseThe expression for $R$ follows by substituting this ${\omega }^2$ into the real part of $Z$. If $C{R}_p^2\le L$, no such finite matching angular frequency exists. Under literal equality, both passive inductive configurations fail by the sign of their imaginary parts.

Let the unknown series circuit contain electrical resistance $R_s>0$ and inductance $L>0$. Its electrical impedance is $Z=R_s+i\omega L$. Thus $A=R_s$, $B=L$, and conjugate impedance matching is possible withAt this angular frequency, the inductor and load capacitor have opposite electrical reactances of equal magnitude. Literal equality of their separate electrical impedances is impossible; for example, $R_s=L=C=1$ gives $1+i\omega$ versus $R-i/\omega$ for every $\omega >0$.

The original diagram is a series RLC circuit, so the same electric current $i=\dot{q}$ passes through all three elements. Choose capacitor electric charge $q$ positive on the plate into which positive $i$ enters. Kirchhoff's voltage law givesDividing by inductance $L$ givesThe correspondence to the forced harmonic oscillator is $m\leftrightarrow L$, $b\leftrightarrow R$, $k\leftrightarrow 1/C$, and $x\leftrightarrow q$. The stored energy is $L{i}^2/2+q^2/(2C)$; its balance is $\dot{E}=Vi-R{i}^2$, confirming the damping sign.

For a fixed linear circuit, self-inductance relates the magnetic flux linkage produced by its own current to that current: $\Lambda =LI$, with $\Lambda ={\sum }_{\mathrm{turns}}{\Phi }_B$. The self-induced electromotive force is $\mathcal{E}=-d\Lambda /dt=-LdI/dt$ when $L$ is constant. Its unit is the henry, $1\mathrm{H}=1Wb/A$.

mutual inductance relates the linkage in one circuit to current in another: ${\Lambda }_2=M_{21}{I}_1$ and ${\Lambda }_1=M_{12}{I}_2$. For stationary reciprocal linear media, $M_{12}=M_{21}=M$ with consistently oriented turns, and ${\mathcal{E}}_2=-Md{I}_1/dt$. Orientation fixes the sign. The coupled magnetic energy is $W=\frac{1}{2}{L}_1{I}_1^2+\frac{1}{2}{L}_2{I}_2^2+M{I}_1{I}_2$, so positivity requires $M^2\le L_1{L}_2$. Moving coils or nonlinear cores require differentiation of the full linkage rather than treating the inductance as a constant.