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For a loop of area $A$ and self-inductance $L$, negligible resistance conserves $LI+{\Phi }_{\mathrm{ext}}$. With zero trapped total flux and ${\Phi }_{\mathrm{ext}}=A{B}_0\sin (\omega t)\cos \theta$, the electric current is $I=-A{B}_0\sin (\omega t)\cos \theta /L$. The magnetic torque generalized to the angle between loop normal and field is ${\tau }_{\theta }=-IAB\sin \theta$. For an orientation varying slowly over a field cycle, averaging gives the displayed torque and effective potential $U_{\mathrm{av}}=A^2{B}_0^2{\cos }^2\theta /(4L)$. Its stable orientations have loop normal perpendicular to the applied field. Without damping, a freely released loop librates around such an orientation instead of settling there.

Cool a thick superconducting ring through its transition in a controlled external flux, remove or change the applied field while keeping the ring superconducting, and measure the persistent trapped flux using a calibrated pickup coil and sensitive magnetometer. Repeating while varying the cooling flux selects successive winding sectors. The measured flux is close to steps separated by ${\Phi }_0$ when the kinetic-inductance correction is negligible. A SQUID-based flux readout can resolve that spacing; scanning local magnetic imaging offers another way to observe the trapped flux without identifying the ring current with an ordinary dissipative transport current.

The experiment tests the phase-winding/fluxoid relation in a. Self-inductance, London penetration depth and the size of the ring determine how nearly the measured geometric flux follows integer steps. A periodic response alone should not be used to infer an absolute trapped flux without calibration and the relevant inductive corrections.

TODO — source figure gap: the surviving ordinary-ring and two-junction-ring outlines and labels are readable, but page 9's figure form calls 52 absent raster objects, /R514 through /R565. The original archived parent has identical damage. An intact figure is needed to recover any missing image-specific details; those raster components have not been inferred or recreated. The observation method above answers the physical request without requiring the unavailable artwork; the TODO concerns verification of the input illustration rather than a missing part of the solution. The original circuit schematic used in c is an independent diagram of the stated model, not a claimed reconstruction of the unavailable artwork.

Let ${\delta }_1$ and ${\delta }_2$ be gauge-invariant phase drops across the two identical junctions, oriented between the same terminal phases. Their currents obey the DC Josephson effect: $I_1=I_C\sin {\delta }_1$ and $I_2=I_C\sin {\delta }_2$. For negligible loop self-inductance and no additional phase drops in the arms, fluxoid consistency fixes their difference modulo $2\pi$:Define the mean terminal phase $\delta =({\delta }_1+{\delta }_2)/2$ and $\varphi =\pi \Phi /{\Phi }_0$. Then ${\delta }_{1,2}=\delta \pm \varphi$ and the two parallel branch currents add:Consequently the maximum magnitude of the current is $2I_C|\cos \varphi |$, not a negative critical current when the cosine changes sign. The phase shifts by $\pi$ between such stable branches. Junction asymmetry or appreciable self-generated flux modifies the simplest interference pattern, so the stated idealization is required before treating $\Phi$ as an externally prescribed flux.

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.

In Cartesian coordinates the magnetic field is $(bx,by,-2bz)$. Its divergence is $b+b-2b=0$, and all components of its curl vanish. Thus it satisfies the static vacuum magnetic equations $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{B}=0$, with no electric current in the region.

Fix clockwise orientation as viewed from above, looking down the positive $z$-axis. Its positive area normal is $-{\mathbf{e}}_z$. The loop's external magnetic flux is thereforeWith the opposite normal its flux has the opposite sign. On the circle $\mathbf{B}=ba{\mathbf{e}}_r-2bz{\mathbf{e}}_z$ and clockwise line element is $d\mathbf{\ell }=-a{\mathbf{e}}_{\phi }d\phi$. The Lorentz force element isThe radial terms cancel around the circle, soThis fixes the current/flux/force signs together.

Neglect self-inductance initially, so Faraday's law and Ohm's law give $RI=-K\dot{z}$. Including gravity then givesFor $R>0$ and $K\ne 0$, the terminal solution is $z=z_0-vt$ withThe general velocity relaxes exponentially to $-v$ on time scale $mR/K^2$. In the terminal state $I=Kv/R$, and the Joule-heating rate isThus all the lost gravitational potential energy becomes resistive heat when kinetic energy is constant.

Within the zero-inductance approximation $v\to 0$ as $R\to 0$; finite-speed motion would require an unbounded current. This limit is singular, so it is important to describe the physical ideal-conductor limit separately. With finite self-inductance $\mathcal{L}$, the circuit equation is $\mathcal{L}\dot{I}+RI=-K\dot{z}$. At $R=0$, total linked flux $\mathcal{L}I+Kz$ is conserved and there is no Joule heating. If $I=0$ initially at $z_0$, thenThis lossless limit of magnetic loop braking has undamped oscillations about $z_0-mg\mathcal{L}/K^2$, with magnetic energy storing and returning mechanical energy. It has no generic dissipative terminal descent. Thus the vanishing terminal speed is the formal resistive-model limit, not a justification for dropping inductance in a perfectly conducting moving loop.

For a moving circuit, Faraday's law takes the formHere $C(t)$ is the instantaneous wire contour, $\mathbf{v}$ its local material velocity, and $S(t)$ any spanning surface. The contour orientation and surface normal vector obey the right-hand convention. The integral on the right is the magnetic flux, differentiated with both the field and circuit position allowed to vary. The electric field term is the electric force per unit electric charge around the wire, and $\mathbf{v}\times \mathbf{B}$ supplies the motional electromotive force. Their sum is the electromotive force driving charge around the moving circuit. The condition $\nabla \cdot \mathbf{B}=0$ makes the magnetic flux independent of the spanning surface.

Choose the initial loop normal vector to be $+{\mathbf{e}}_y$, so rotation about $+{\mathbf{e}}_z$ gives $\mathbf{n}(t)=(-\sin \Omega t,\cos \Omega t,0)$. The spatially uniform magnetic field givesUsing Ohm's law for the resistive loop,Positive electric current is measured in the contour orientation associated with $\mathbf{n}(t)$. Reversing that convention reverses the signed answer. When $\Omega =\omega$, the magnetic flux is constant and the induced electric current vanishes. This uses the usual thin resistive-circuit approximation with self-inductance neglected.

Use the ideal long-solenoid approximation: negligible pitch and end effects, vacuum permeability ${\mu }_0$, and negligible displacement current. By cylindrical symmetry the magnetic field is axial. An Ampère's law rectangular loop with one axial side inside and the other outside gives $B_{\mathrm{in}}-B_{\mathrm{out}}={\mu }_{0}NI$. A loop wholly outside shows the outside axial field is constant with radius; requiring it to vanish far away givesThe axial sign is determined by the winding direction and the right-hand rule. The small axial component of current from the spiral pitch is neglected in this idealization.

There are $Nl$ turns, and the magnetic flux through each is $\pi a^2{\mu }_{0}NI$. The flux linkage is $LI$, so the self-inductance isThe induced electromotive force opposes increasing current by Faraday's law. Together with Ohm's law, the circuit equation is $L\dot{I}+RI={\mathcal{E}}_0$. Solving this first-order ordinary differential equation with $I(0)=0$ givesThe time constant is $L/R$, and the limiting current is ${\mathcal{E}}_0/R$.

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.

Write the wire diameter as $d_w$ to distinguish it from other geometric labels. Its cross-sectional area is $A_w=\pi d_w^2/4$, and the total wire length is $4a$. From the electrical resistivity, its electrical resistance isWhile the loop straddles the boundary, let $h$ be the height of each vertical side remaining in the magnetic field. The linked magnetic flux has magnitude $Bah$, with $\dot{h}=-v$. Faraday's law therefore gives $|\mathcal{E}|=Bav$. Neglecting the loop's self-inductance, Ohm's law givesThe into-page magnetic flux is decreasing, so Lenz's law requires a clockwise electric current as viewed in the diagram. If the loop is wholly inside a uniform magnetic field, or wholly outside it, its magnetic flux is constant and this induced electromotive force is zero. During a rapid transient one should instead solve $L\dot{I}+RI=Bav$; at the assumed steady speed the same boxed relation holds.

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 $L{I}^2/2$.

Let $A=\pi r^2$ and choose the rotating area normal to coincide initially with $+z$. At time $t$ its angle to the applied magnetic field is $\omega t$, so the external magnetic flux is ${\Phi }_B=BA\cos \omega t$. Faraday's law therefore givesIn the high-electrical resistance approximation, neglecting the loop's self-inductance, Ohm's law givesThe displayed signs specify positive circulation by the chosen normal; reversing the reference circulation reverses both signs. The high-electrical resistance assumption is quantitatively $\omega L\ll R$ if the loop has self-inductance $L$. Otherwise the current lags the induced electromotive force and must be obtained from $L\dot{I}+RI=\mathcal{E}$.

The magnetic dipole moment is $\mathbf{\mu }=IA\hat{\mathbf{n}}$ and its magnetic torque is ${\mathbf{\tau }}_{\mathrm{mag}}=\mathbf{\mu }\times \mathbf{B}$. Its component along the rotation axis opposes the imposed rotation. For positive $\omega$, the external driving torque needed to keep the angular velocity constant isAveraging over a complete rotation, with $⟨{\sin }^2\omega t⟩=1/2$, givesAs an independent sign and dimension check, $\omega ⟨{\tau }_{\mathrm{drive}}⟩=R⟨I^2⟩$. Thus the mechanical input power equals the mean Joule heating; a passive loop cannot supply a sustained driving torque. This average assumes the same negligible-self-inductance approximation as the current expression.

Faraday's law of induction gives the electromotive force $\mathcal{E}=-d{\Phi }_B/dt$. For a fixed circuit, $\mathcal{E}=\oint \mathbf{E}\cdot d\mathbf{l}$; for moving material wire, include the motional term $\mathbf{u}\times \mathbf{B}$ in that line integral. Lenz's law fixes the sign: the induced response opposes the change of linked flux, rather than necessarily opposing the field itself.

For the long solenoid, Ampère's law gives $B\simeq {\mu }_{0}nI$ inside and negligible exterior field away from the ends. With $N=nD$ turns and area $\pi a^2$, the linked flux is $N{\Phi }_B={\mu }_0{n}^{2}D\pi a^{2}I$. Hence the self-inductance isThis assumes $D\gg a$, closely spaced turns, negligible end fringing and an air core with relative magnetic permeability close to one. A finite core permeability or significant fringing changes the answer.

Orient the ring's area normal with positive external magnetic flux. Faraday's law gives the applied electromotive forceThe ring's self-inductance opposes changes in its electric current. With positive electrical resistance $R$, Kirchhoff's voltage law therefore gives $L\dot{I}+RI={\mathcal{E}}_{\mathrm{ext}}$. Define $\varphi ={\tan }^{-1}(\omega L/R)$. The periodic steady state isThe second result follows from Joule heating, $P=R{I}^2$, and $⟨{\sin }^2⟩=1/2$. For unspecified initial conditions, add a transient $I_h=C\exp (-Rt/L)$; the displayed time average is over the periodic steady state after that transient has decayed. The inductor temporarily stores and returns energy, with no net storage change over one complete cycle.

Each long wire lies in the other's magnetic field, $B={\mu }_{0}I/(2\pi d)$. The magnetic Lorentz force per unit length has magnitudeFor the two-wire circuit's self-inductance, a nonzero wire radius is essential. Introduce a radius $a_w\ll d$. Between the wires their magnetic fields add. The external magnetic flux linkage per unit length through the connecting ribbon is, to leading order in $a_w/d$,It follows that the leading thin-wire loop self-inductance per unit length isIt is the loop value for the pair, not the contribution of just one conductor. Corrections from the wire interior depend on the current distribution: for nonmagnetic round wires with uniform direct electric current, the pair contributes another ${\mu }_0/(4\pi )$; for a surface current the internal contribution is zero. The leading logarithm is common to both models.

TODO (source data): the source gives no wire radius or current-distribution convention. There is no finite radius-independent value of the requested self-inductance: taking ideal zero-width wires sends $L_{\mathrm{ext}}^{\prime }$ to infinity. The force and the parameterized leading self-inductance above are determined, but a unique finite answer requires that additional specification. For example, changing $a_w$ to $a_w/2$ changes the leading result by ${\mu }_0\ln 2/\pi$ even at fixed separation.

Project Newton's second law down the slope. The normal reaction has no component along the rails, while the magnetic term opposes motion:This first-order linear differential equation has the release-from-rest solutionThe electrical self-inductance has been neglected; retaining it introduces a coupled electric current equation rather than this instantaneous resistive response.

The component of magnetic force along the slope has magnitude $IBs\cos \varphi$. The rate at which the sliding bar does work against this magnetic braking isBy Ohm's law, $\mathcal{E}=IR$, soThis exactly equals the Joule heating in the resistor in the negligible-self-inductance approximation. The normal force does no work because it is perpendicular to the velocity.

The gravitational power is instead $mgv\sin \varphi$. By the work-energy theorem,It equals the Joule heating only at constant speed. For a concrete counterexample, during the initial acceleration the kinetic energy increases, so some gravitational power remains mechanical.

Use the coordinates in 8. At downhill speed $v$, the moving bar has $\mathbf{v}\times \mathbf{B}=-Bv\cos \varphi \hat{\mathbf{b}}$. The electromotive force and induced electric current are thereforeHere $I$ is positive along $-\hat{\mathbf{b}}$ across the sliding bar, then returns through the rails and resistor. This direction makes the magnetic force oppose downhill motion, as required by Lenz's law. The factor $\cos \varphi$ is essential: only the component of magnetic field normal to the rail plane threads the circuit. Rail and bar electrical resistances and circuit self-inductance are neglected relative to the stated large $R$ model.

In the long-solenoid approximation, the internal magnetic flux density is uniform, $B={\mu }_{0}nI$, and the external magnetic field is negligible. This follows from the magnetostatic Ampère law applied to a rectangular loop running inside and outside the solenoid. Its number of turns is $N=nD$ and its cross-sectional area is $A=\pi a^2$. Each turn links magnetic flux ${\Phi }_B=BA={\mu }_{0}nI\pi a^2$. Therefore the flux linkage isUsing the definition of self-inductance, $\Lambda =LI$, givesThis neglects end leakage and winding thickness and uses vacuum permeability for the air core.

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.