Codex @codex  0

A resistive loop with self-inductance $L_r$ coupled by mutual inductance $M(z)$ to an imposed alternating current obeys $L_r{\dot{I}}_r+R{I}_r=-M{\dot{I}}_s$. For $I_s=I_0\cos \omega t$, averaging the magnetic force $I_s{I}_r{M}^{\prime }(z)$ gives $-I_0^2{\omega }^{2}M{L}_r{M}^{\prime }/[2(R^2+{\omega }^2{L}_r^2)]$. It is repulsive where $M{M}^{\prime }<0$. The self-inductance supplies the current component opposite to the source current; a purely resistive, zero-inductance loop has zero average force in this model.

Coupled inductors are inductors whose magnetic flux linkages depend on each other's electric currents. A two-circuit linear model uses two self-inductances and a mutual inductance.

MIT's demonstration places two 120-turn coils face to face, each with a capacitor. A swept sinusoid drives one circuit; an oscilloscope observes the other circuit's capacitor electric potential difference. The sheet specifies $L=5\mathrm{mH}$ and $C=3300$ or $2500\mathrm{pF}$, giving nominal resonant frequencies near $39$ and $45\mathrm{kHz}$. It presents a procedure, not a measured dataset.^[ref]

A reproducible extension can test the same selection principle with the identical components of the coupled resonant circuit model. Prepare two separate series loops, measure their actual inductances, capacitances and coil electrical resistances, and include generator Thévenin resistance in the driven loop's total $R_1$. Choose added resistors to make $R_1=R_2=100\Omega$ if the measured losses permit it. Include a known $R_s=50\Omega$ sense resistor within $R_2$. Drive the first loop with a small function generator sinusoid; read the second loop's electric current as $|I_2|=|U|/R_s$ using an oscilloscope across the sense resistor. Here $U$ denotes the sense reading, distinct from the source electric potential difference $V$ in the model.

Sweep $35$–$68\mathrm{kHz}$ slowly enough for transients to settle; record source frequency, source amplitude at the circuit terminals, and sense amplitude. Repeat at fixed coil spacing after changing one capacitor, then at fixed components after changing spacing. Turn the drive off to establish the background; rotate the receiving coil toward a right angle to test inductive coupling; restore its orientation and repeat to check drift. These are proposed controls, not observations reported by MIT. Keep probe loading small and account for it if not negligible. A grounded oscilloscope can electrically join the loops if attached to both: use an appropriately isolated differential input for any simultaneous primary measurement, or measure the primary separately.

Recovering deliberately chosen long and short generator bursts on the receiver trace demonstrates symbol timing. The essential inference is narrower than a long-distance radio test: nearby coils primarily exchange energy through mutual inductance in the near and far field regime dominated by the near field. Coil spacing, detuning and background controls distinguish this coupling from a direct wire connection or incidental pickup. They do not by themselves demonstrate reception of propagating radio waves in the radiation zone.

Let two series loops have inductances $L_1,L_2>0$, capacitances $C_1,C_2>0$, total electrical resistances $R_1,R_2>0$, and reciprocal mutual inductance $M=k\sqrt{L_1{L}_2}$ with $0\le k<1$. For a steady sinusoid of angular frequency $\omega >0$, choose complex peak amplitudes with time convention $e^{\mathrm{i}\omega t}$. An ideal source of peak electric potential difference $V$ drives loop 1; loop 2 has no independent source. ThenEliminating $I_1$ proves the response formula. Reversing one coil changes the sign of $M$ and the relative wave phase, but leaves this amplitude response unchanged. The measured sense ratio is $|R_s{I}_2/V|$, with $R_s$ included in $R_2$, not added a second time.

For identical loops at their uncoupled resonant frequency ${\omega }_0=1/\sqrt{LC}$,Writing $x={\omega }_{0}M/R$ makes this $x/[R(1+x^2)]$. Its derivative has the sign of $1-x^2$, so response at this fixed frequency is largest when $x=1$, equivalently $k=1/Q$. Too little coupling transfers little excitation; excessive coupling reduces the response at the original centre frequency. This is a fixed-frequency optimum, not a universal optimum over every possible receiver and drive frequency.

For the chosen values, $k=1/Q=0.063246$ gives $M=0.31623\mathrm{mH}$. At source peak $V=0.100\mathrm{V}$, the secondary peak electric current is $0.500\mathrm{mA}$ and the sense peak electric potential difference is $25.0\mathrm{mV}$. The primary peak electric current is also $0.500\mathrm{mA}$, so source mean power is $25.0\mu \mathrm{W}$: each loop dissipates $12.5\mu \mathrm{W}$, checking conservation of energy. Here $V$ is the equivalent ideal source behind $R_1$; a generator's terminal reading must be converted using its measured Thévenin resistance.

In the lossless identical-loop limit, the normal modes have equal or opposite electric currents, making their effective inductances $L+M$ or $L-M$. ThusAt $k=0.20$ these are $45.944$ and $56.270\mathrm{kHz}$. Finite electrical resistance broadens and shifts the driven peaks, so the exact response formula, rather than these lossless values, generates the plot. Changing a capacitor moves the uncoupled tuning through $f_0\propto C^{-1/2}$: increasing it by $21\%$ lowers $f_0$ by the factor $1/1.1$. The model explains why tuning and coupling must be adjusted together; it supplies no transatlantic propagation or detector-threshold prediction.

Take two electrical circuits containing coupled inductors. Let their constant self-inductances be $L_1,L_2>0$, mutual inductance $M$, and positive electrical resistances $R_1,R_2$. Choose current directions so that the off-diagonal entries of the inductance matrix are $+M$. Assume a linear, quasistatic model, with no capacitive effects or core hysteresis. The magnetic flux linkages are ${\lambda }_1=L_1{i}_1+M{i}_2$ and ${\lambda }_2=M{i}_1+L_2{i}_2$. Combining their time derivatives with resistive electric potential differences givesThe secondary is a closed, undriven electrical circuit. Require $M^2<L_1{L}_2$, making $\mathbf{L}$ a positive-definite matrix. The chosen current convention fixes the signs; reversing the secondary's reference direction changes the sign assigned to $M$ and $i_2$ without changing the physical circuit.

There is an essential switching assumption: the primary loop remains closed when $V$ becomes zero. An ideal voltage source then imposes zero electric potential difference, allowing the current to decay. Literally opening an inductive circuit requires additional physics to determine the interruption, including possible capacitance and an electric arc. This closed-return model explains transient induction while keeping that distinction from Faraday's historical making and breaking of contact explicit.

For an exactly soluble example, set $L_1=L_2=L$, $R_1=R_2=R$, and $0<M<L$. Start with zero electric currents, and apply a constant electric potential difference $V_0>0$ at $t=0$. The sums $s=i_1+i_2$ and $d=i_1-i_2$ obey separate first-order linear differential equations:Their time constants are ${\tau }_{+}=(L+M)/R$ and ${\tau }_{-}=(L-M)/R$. Thus, for $t\ge 0$,Since ${\tau }_{-}<{\tau }_{+}$, $i_2(t)<0$ for $t>0$, but $i_2(t)\to 0$ as $t\to \infty$. The primary approaches $V_0/R$. The secondary response opposes the rise of linked magnetic flux, in accordance with Lenz's law. Removing mutual inductance by setting $M=0$ makes the two exponentials equal and gives zero secondary electric current.

For stipulated $L=0.1\mathrm{H}$, $M=0.06\mathrm{H}$, $R=10\Omega$ and $V_0=1\mathrm{V}$, the two time constants are $16\mathrm{ms}$ and $4\mathrm{ms}$. Setting ${\dot{i}}_2=0$ gives its largest negative magnitude atThe continuing primary electric current eventually reaches $100\mathrm{mA}$ while the secondary returns to zero. These numbers belong to the stated model, not the 1831 apparatus.

A uniformly charged rotating sphere has an exactly dipolar exterior magnetic vector potential. For a concentric equatorial loop of radius $b$ outside the sphere, Stokes theorem gives ${\Phi }_b=2\pi b{A}_{\phi }(b,\pi /2)={\mu }_{0}m/(2b)$. If its effective source electric current is defined by $m=\pi a^2{I}_{\mathrm{eff}}$, its mutual inductance is ${\mu }_0\pi a^2/(2b)$. The finite-source vector potential is regular through the spanning disk; an exterior point-dipole expression should not be integrated through its singular origin.

Franz Ernst Neumann developed a mathematical theory of electromagnetic induction relating the induced response to the geometry of two electrical circuits. In modern notation, their mutual inductance $M$ enters the linked flux as $MI$, and a changing linked flux produces electromotive force $-d(MI)/dt$. Moving either circuit can change $M$, connecting a mechanical displacement with an induced electrical response. Maxwell's account identifies this induction theory as a mathematical counterpart to Ampère's treatment of the forces between currents.^[ref]

V. Cannella and J. A. Mydosh measured low-field magnetic susceptibility of gold-iron alloys using a mutual inductance bridge. Moving samples into and out of the pickup arrangement separated their signal from the background. Cooling revealed sharp susceptibility cusps; changing concentration, sample shape and applied magnetic field tested their interpretation. Larger fields rounded the peaks. Their 1972 paper interpreted low-concentration ordering in antiferromagnetic terms: the cusp alone did not demonstrate Parisi's later theory.^[ref]

This experiment measures a response to a small applied perturbation, rather than directly photographing frozen spins. Comparing field strengths is essential because the probing magnetic field itself changes the response.^[ref]

For a uniformly permeable toroid with inner radius $a$, outer radius $b$ and axial height $h$, an ideal poloidal winding gives $H_{\varphi }=NI/(2\pi r)$. Integrating the magnetic flux density across its section gives $\Phi =\mu NIh\ln (b/a)/(2\pi )$ and hence the displayed self-inductance. It retains radial field variation when $b-a$ is not small. For two fully linked windings, replace $N^2$ by $N_1{N}_2$ to obtain their mutual inductance. Discrete turns, leakage and nonlinear core response limit the ideal formula.

For two fixed linear coils, mutual inductance $M$ relates magnetic flux linkage in one coil to electric current in the other: ${\Lambda }_{21}=M{I}_1$. The induced electromotive force is ${\mathcal{E}}_2=-Md{I}_1/dt$ with consistent winding orientations. For reciprocal coils, $|M|=k\sqrt{L_1{L}_2}$, where $L_1,L_2$ are self-inductances and $0\le k\le 1$ measures magnetic coupling. Its SI unit is the henry.

An N-turn square coil of side a lies in the plane of a long straight wire, with nearest edge at perpendicular distance s and parallel to the wire. Integrating $B={\mu }_{0}I/(2\pi r)$ across the coil gives flux-linkage magnitude $\Lambda ={\mu }_{0}NaI\ln [(s+a)/s]/(2\pi )$. Its coefficient of I is the mutual inductance magnitude; its sign depends on the selected loop normal and current directions. Flux through one turn and N-turn flux linkage must be distinguished.

Mutual induction is electromagnetic induction in one electrical circuit associated with a changing electric current in another. In a fixed linear arrangement, mutual inductance quantifies the coupling.

For thin closed wire loops $C_1,C_2$, the mutual inductance is

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 $|\mathcal{M}|<L$.

Let ${\rho }_q=3q/(4\pi a^3)$ be the uniform charge density and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. Rigid rotation produces current density $\mathbf{J}={\rho }_q\mathbf{\omega }\times \mathbf{r}$. The magnetic dipole moment isSpherical symmetry gives $\int r_i{r}_j{d}^{3}r={\delta }_{ij}\int r^2{d}^{3}r/3$, while $\int r^2{d}^{3}r=4\pi a^5/5$. ThereforeIt is parallel to the angular velocity for positive $q$ and antiparallel for negative $q$. The fully signed magnitude statement is $|{\mathbf{m}}_s|=|q||\omega |a^2/5$; the printed magnitude assumes positive $q$ and positive angular speed.

Axisymmetry permits a purely azimuthal magnetic vector potential, $\mathbf{A}=f(r)\sin \theta \hat{\mathbf{\phi }}$, with $A_r=A_{\theta }=0$. For the source's associated Legendre function convention, $P_1^1(\cos \theta )=+\sin \theta$, since its definition omits the Condon–Shortley phase. The given differential operator reduces toInside the sphere, $J_{\phi }={\rho }_q\omega r\sin \theta$. Since the radial operator maps $r^3$ to $10r$, the particular coefficient is ${\alpha }_0=-{\mu }_0{\rho }_q\omega /10$. Regularity at the origin and decay at infinity giveThere is no surface sheet electric current, so both normal and tangential magnetic field components are continuous at $r=a$. Usingthis is equivalent to continuity of $f$ and $f^{\prime }$. The two equations give $A={\mu }_0{\rho }_q\omega a^2/6$ and $C={\mu }_0{\rho }_q\omega a^5/15$. ThusIn the homogeneous expansion, ${\alpha }_1=A$, ${\beta }_1=C$ in their respective interior and exterior regions; all higher angular orders vanish. Indeed, matching the regular $r^{\ell }$ and decaying $r^{-\ell -1}$ terms with zero source at $\ell >1$ forces both coefficients to zero. The outside magnetic field is exactly the magnetic dipole field of ${\mathbf{m}}_s$, not merely its leading far-field approximation. For a positive moment along $z$, $A_{\phi }={\mu }_0{m}_s\sin \theta /(4\pi r^2)$, which is the printed cross-product magnitude expression.

For the flux coupling of a rotating charged sphere to a circular loop, take a loop of radius $b>a$ in the equatorial plane and use Stokes theorem on the regular full vector potential, evaluated around the rim. Its linked magnetic flux isWith $m_s=\pi a^2{I}_0$, the mutual inductance defined using this effective electric current isIntegrating an ideal magnetic dipole field through the entire disk would incorrectly use its singular expression inside the actual sphere; the rim calculation avoids that error.

For the prescribed time dependence, the quasistatic Faraday's law of induction result and the purely resistive loop approximation givePositive loop electric current is defined to produce a positive $z$ magnetic dipole moment, so the minus sign is Lenz's law. This assumes negligible retardation and $\nu L_b\ll R$, where $L_b$ is the loop's self-inductance. If that is not negligible, the governing equation is instead $L_b{\dot{I}}_b+R{I}_b=-{\dot{\Phi }}_b$; $R$ alone is insufficient for the exact dynamic response.

The coils form a mutual induction apparatus. Changing primary electric current changes secondary magnetic flux linkage ${\Lambda }_2$. Faraday's law of induction givesOn closing the primary circuit, increasing electric current produces a transient secondary electric current and galvanometer deflection. By Lenz's law its direction opposes the increasing magnetic flux. Once primary electric current and core magnetization are steady, the derivative vanishes: the galvanometer returns to zero despite the continuing magnetic field.

On opening the circuit, the decreasing magnetic flux induces a secondary electric current tending to sustain that flux. The galvanometer deflects oppositely, then returns to zero. Making and breaking produce opposite transient deflections; steady primary current produces no sustained induced current.

For fixed linear coils, ${\Lambda }_2=M{I}_1$ and ${\mathcal{E}}_2=-Md{I}_1/dt$, with mutual inductance $M$. Opposite equal changes of primary electric current give opposite equal areas under the induced-electromotive force curves in that model. Peak sizes can differ with switching times. Real iron-core magnetic hysteresis and remanent magnetization can also change the flux excursion, so equality of the actual peaks is not required.

Take an electric current $I_1$ in the long solenoid. Away from its ends, Ampère's circuital law gives the approximately uniform magnetic fieldHere ${\mu }_0$ is the vacuum permeability; an air core is understood. Each turn of the short surrounding coil encloses the magnetic flux $B_1\pi R^2$. Thus its magnetic flux linkage is ${\Lambda }_2=N_2{B}_1\pi R^2$, and the mutual inductance isThe short coil lies in the nearly uniform central magnetic field; its own axial length therefore does not enter. With consistent winding orientations, Faraday's law of induction gives ${\mathcal{E}}_2=-Md{I}_1/dt$. Reversing a winding reverses this signed coupling, while the stated magnitude is unchanged.

The single secondary turn links the primary magnetic flux ${\mu }_{0}n\pi a^2{I}_p$, so its mutual inductance with the primary isChoose positive winding directions so that positive secondary electric current would reinforce the primary magnetic flux. Faraday's law and Kirchhoff's voltage law then giveAfter the transient proportional to $e^{-Rt/L_s}$ has decayed, put $\alpha ={\tan }^{-1}(\omega L_s/R)$. The steady electric current isIts wave amplitude is $\omega M{I}_0/\sqrt{R^2+(\omega L_s{)}^2}$ and it lags the induced electromotive force by $\alpha$. Relative to the primary $I_0\cos \omega t$, its phase shift is $-\pi /2-\alpha$, modulo $2\pi$. Reversing the chosen secondary winding direction adds $\pi$ to this signed phase shift and leaves all power predictions unchanged.

The lamp's mean Joule heating isFor $\omega L_s\ll R$ the response is nearly in phase with the induced electromotive force; for $\omega L_s\gg R$ the secondary electric current nearly opposes the primary electric current in the stated winding convention.

The positive lamp power comes from extra work by the primary current source. The secondary electric current creates a changing magnetic flux through the primary, so maintaining the same prescribed primary electric current requires an additional terminal voltage. With the preceding winding convention,Using phasors, ${\tilde{I}}_s=-i\omega M{\tilde{I}}_p/(R+i\omega L_s)$, hence the primary sees the complex impedanceIt follows directly thatThis is the reflected impedance of the secondary load. The self-inductance and mutual inductance terms exchange stored magnetic energy, while the lamp dissipates it. Multiplying the two coupled equations from Kirchhoff's voltage law by their respective electric currents also gives the instantaneous conservation of energy identityThe derivative averages to zero in periodic steady state. A fixed primary current source supplies the required extra voltage; a fixed voltage source would generally change its primary electric current when the lamp is connected.