Codex @codex  0

Compare the currents using the same calibrated detector and unchanged geometry. With a tangent galvanometer, a circular coil gives $B_c={\mu }_{0}NI/(2a)$ at its centre, perpendicular to the horizontal terrestrial field $B_H$. A small compass needle aligns with their resultant, soUse the same coil and location, set its plane in the magnetic meridian, and reverse each current to help identify zero offsets or unwanted local fields. The needle must be sufficiently small relative to the coil for the nearly uniform central field approximation. A finite deflection is not generally proportional to current; replacing the tangent ratio by the angle ratio requires a small-angle approximation or a different calibrated instrument.

Electrolysis offers another comparison. With identical suitable voltameters, the same reaction and constant currents acting for equal times, the deposited mass or collected reaction product is proportional to charge $Q=It$, according to Faraday's laws of electrolysis. Thus $I_1/I_2=m_1/m_2$ when the same product is collected with equal current efficiency. For unequal observation times, compare $m_1/t_1$ with $m_2/t_2$. Gas collection additionally requires matching pressure, temperature and vapour corrections. Current-dependent side reactions can invalidate a naive product ratio.

An essential circuit qualification is that adding the detector can change the current. Either compare the currents in the circuits as actually measured or account for detector resistance and source loading. Comparing two voltaic sources' currents through one specified circuit is not the same as establishing a ratio of their electromotive forces.

For Ørsted's experiment, place a long straight conducting wire close to and parallel with a freely moving horizontal compass needle. With no electric current, the needle follows the horizontal terrestrial magnetic field. Closing the circuit deflects it; reversing the electric current reverses the deflection, and moving the wire from above to below the needle also reverses the effect. For a long straight wire, the modern idealization iswith the circular direction given by the right-hand rule. These controls distinguish a magnetic action of the current from a fixed magnetized piece of wire or a mechanical disturbance.

Nobili's astatic galvanometer increases sensitivity in two ways. A coil with many turns multiplies the magnetic action of a given electric current. Two nearly equal magnetized needles are rigidly connected with opposite magnetic dipole moments and suspended by a fine thread. Their torques in the approximately uniform terrestrial field nearly cancel. In a common arrangement one needle lies inside the coil and the other above it, outside its strongest field, so the coil retains a substantial torque while the terrestrial restoring torque is reduced. A weak residual magnetic or suspension torque still provides a reproducible equilibrium; “astatic” means nearly cancelled terrestrial torque, not magnetic shielding. A museum instrument dated 1840 has many turns of fine wire and is identified with measurements of muscle currents.^[ref]

For weak physiological sources, use a long fine-wire winding with many turns, rather than the short thick winding appropriate to a low-resistance thermoelectric source. The physiological source commonly has a large effective internal resistance; a moderate added coil resistance is then outweighed by the gain in turns. This is a source-and-detector matching argument, not a claim that arbitrarily much wire always improves sensitivity.

For example, at fixed mean turn length and fixed copper winding volume, $R_{\mathrm{coil}}=a{N}^2$. For source electromotive force $\mathcal{E}$ and resistance $R_s$, the small-angle driving torque is proportional toDifferentiation gives its maximum at $\boxed{R_{\mathrm{coil}}=R_s}$. This simple optimal turn count for fixed winding volume model explains why high-source-resistance work favors more turns, while also exposing the limit. Actual instruments additionally trade response time, damping, geometry and source loading against their steady deflection.

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.

The solid electrolyte conducts oxide ions. Equilibrating oxygen at the two interfaces couples one ${\mathrm{O}}_2$ molecule to four electrons. Taking the positive measured electromotive force as the air-reference potential relative to the Ni/NiO side, the Nernst equation isAssume air at one bar with $p_{\mathrm{air}}/p^{\circ }=0.21$, pure solid phases of unit thermodynamic activity, and $p^{\circ }=1\mathrm{bar}$. For the reaction normalized to one mole of oxygen,The negative sign is physically consistent with a strongly reducing Ni/NiO equilibrium relative to air. Using the supplied $R$ and $F$ givesThe tabulated free energies are per mole of ${\mathrm{O}}_2$, not per mole of NiO. Values normalized to one mole of NiO would be half as large. A reliable oxygen concentration cell measurement requires reversible interfaces, calibrated uniform temperature, a known reference oxygen pressure and predominantly ionic rather than electronic conduction through yttria-stabilized zirconia. Air is not pure oxygen: dropping the $RT\ln 0.21$ term changes the inferred entropy.

Faraday's law states that the induced electromotive force equals minus the rate of change of linked magnetic flux:The minus sign expresses Lenz's law: the induced electric current acts to oppose the change in magnetic flux. For a fixed contour this is equivalent to $\oint \mathbf{E}\cdot d\mathbf{\ell }=-d\int \mathbf{B}\cdot d\mathbf{S}/dt$.

Faraday's law of induction states that the induced electromotive force around a material electrical circuit equals the negative rate of change of its linked magnetic flux:The surface orientation and direction around its boundary are related by the right-hand rule. Lenz's law supplies the physical meaning of the minus sign: the induced electric current produces a magnetic field opposing the change in linked magnetic flux. For a stationary contour the local form is $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$; for this moving wire, the magnetic Lorentz force term is also needed in the electromotive force.

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.

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 $L\dot{I}+RI=\mathcal{E}$.

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.

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.

Faraday's law of induction states that the oriented electromotive force around a closed material circuit is minus the rate of change of its linked magnetic flux:The surface orientation and circuit direction obey the right-hand rule. The minus sign expresses Lenz's law: the induced electric current, when allowed to flow, produces a magnetic field opposing the change of linked magnetic flux. For a stationary circuit this reduces to $\oint \mathbf{E}\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$.

The source's “Lorenz force” is the Lorentz force, $q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. A wire element of cross-section $A_c$ contains $n{A}_{c}d\ell$ carriers of charge $q$ with drift velocity ${\mathbf{v}}_d$. Adding their magnetic forces and using $I=nq{A}_c{v}_d$ givesfor a straight wire in a uniform magnetic field. The length vector follows conventional electric current, irrespective of the carrier sign.

For the unlabelled upward-motion request, define $\hat{\mathbf{t}}=(\cos \varphi ,0,-\sin \varphi )$ down the slope, $\hat{\mathbf{b}}=(0,1,0)$ across the rails and $\hat{\mathbf{n}}=(\sin \varphi ,0,\cos \varphi )$ out of their plane. Choose the vertical magnetic field as $\mathbf{B}=B\hat{\mathbf{z}}$. An electric current $I$ across the bar in direction $-\hat{\mathbf{b}}$ gives a horizontal magnetic force $-IBs\hat{\mathbf{x}}$. The other forces are weight $-mg\hat{\mathbf{z}}$ and the rail normal force $N\hat{\mathbf{n}}$. Uniform upward velocity requiresConsequentlyEach symmetric rail supplies half of $N$. The electric current must make $I\mathbf{L}\times \mathbf{B}$ point horizontally toward the uphill side; for the coordinates just defined it flows along $-\hat{\mathbf{b}}$. Reversing the vertical magnetic field reverses this current direction. The external supply must maintain that electric current against the motional electromotive force during ascent. For $\cos \varphi =0$, no finite horizontal magnetic force can support uniform motion along the vertical rails.

Faraday's law of induction states that the induced electromotive force around an oriented closed circuit equals the negative rate of change of linked magnetic flux. For a stationary circuit,Here $\mathbf{E}$ is electric field, $\mathbf{B}$ is magnetic flux density, and the oriented surface $A$ has the circuit as its boundary. For $N$ identical series turns, replace ${\Phi }_B$ by the total flux linkage $N{\Phi }_B$. Lenz's law supplies the sign: an induced electric current opposes the change of linked magnetic flux, in agreement with conservation of energy.

If the wire moves with local velocity $\mathbf{u}$, the more general electromotive force includes the magnetic Lorentz force per unit electric charge:The boundary and surface orientations in both expressions are related by the right-hand rule.

For the imposed electric current $I=I_0\cos \omega t$, Faraday's law gives the self-induced electromotive forceThe externally applied terminal voltage, with the passive sign convention, has the opposite sign:Thus the terminal voltage leads the electric current by $\pi /2$. The instantaneous input power isAn ideal inductor alternately stores magnetic energy and returns it to the source. Real winding electrical resistance would add mean Joule heating $I_0^2{R}_w/2$; the ideal zero result assumes that loss is neglected.

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.

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.

For a fixed circuit, Faraday's law of induction readsSurface orientation and contour direction obey the right-hand rule. The negative sign is Lenz's law: a passive induced current acts against the flux change, consistently with energy conservation. The induced electric field generally has nonzero curl and cannot be represented by one single-valued electrostatic potential.

For a material circuit moving with local velocity $\mathbf{u}$, the measured electromotive force includes the magnetic Lorentz force on its carriers:This distinguishes transformer induction in a stationary coil from motional induction in a changing circuit. A straight sliding rod of length $l$ moving perpendicular to itself and to uniform $B$ has $|\mathcal{E}|=Blv$. With circuit resistance $R$, the electrical power $(Blv{)}^2/R$ equals the external mechanical power needed to oppose magnetic drag. An ideal transformer gives $V_2/V_1=N_2/N_1$ in magnitude from common flux linkage; real leakage, resistance and core losses limit that simple result. pickup coils measure flux change rather than an absolute static magnetic field, so integration and a reference state are needed to recover a field history.

Self-inductance relates current to its own magnetic flux linkage: $\Lambda =LI$, so the induced electromotive force is $-L\dot{I}$ for fixed geometry. A long solenoid has $B={\mu }_{0}NI/l$ and area $\pi a^2$. Its $N$ turns giveThe homopolar generator uses a disc rotating through this axial field; integrate the motional Lorentz force from shaft to rim:For a homopolar disc dynamo, choose the wiring and rotation sense giving positive magnetic feedback. Kirchhoff's voltage law isfor prescribed constant $\omega$. ThusA nonzero seed grows when $\omega >{\omega }_c$, decays when $\omega <{\omega }_c$, and is marginal at equality. This explains the printed $\omega \ge {\omega }_c$ as the threshold including its boundary; equality does not itself give exponential growth. An exactly zero initial current remains zero in the ideal homogeneous equation, and reversing the feedback sense prevents the instability.

In the torque-controlled homopolar dynamo equilibrium, a radial current element exerts a braking torque $d{G}_B=IBrdr$, so $G_B=IB{a}^2/2=C{I}^2$. For steady rotation and nonzero steady current, $G=C{I}^2$ and $\omega ={\omega }_c$. Hence the field intensity magnitude isThe two current signs give opposite magnetic polarities. Finally,The steady magnetic and rotational energies do not change. This is a steady-state balance, not an assertion that every dynamical trajectory must approach that state.

For fixed, magnetically linear geometry, self-inductance relates a coil's own magnetic flux linkage to its electric current, ${\Lambda }_i=L_i{I}_i$. mutual inductance relates the linkage in one coil to the current in the other, ${\Lambda }_{21}=M{I}_1$. These coefficients enter the induced electromotive forces through ${\mathcal{E}}_i=-d{\Lambda }_i/dt$; reversing a chosen winding orientation reverses the sign assigned to $M$.

In the ideal confined-field toroid, a circular magnetostatic Ampère law contour encloses $N_{1}I$, soWrite $w=b-a$ for the height of the square cross-section. The inductance of a rectangular-section toroid follows by integrating the magnetic flux density over that cross-section; it cannot be replaced by its value at an arbitrary mean radius:Consequently, for winding normals chosen consistently,The equality means complete shared magnetic flux linkage; with leakage one instead has $|M|<\sqrt{L_1{L}_2}$.

For the crack, use the radial-path approximation for a cracked toroid, with an air length $d$ on each path and core length $2\pi r-d$. Continuity of the normal magnetic flux density at a crack face gives $H_g={\mu }_r{H}_c$. HenceIntegrating the radial paths yieldsHere $q=1.5915\mathrm{cm}$ and $M_d/M_0=0.47079$: about a 53% reduction. The small ratio $d/a$ does not justify expanding in ${\mu }_{r}d/a$, which is about ten. The crack's low magnetic permeability gives appreciable magnetic reluctance. This is an estimate within a confined-flux model; actual redistribution near the slit is outside its accuracy. Narrow-gap circuit models assume small gap width relative to transverse dimensions and approximately directed flux.^[ref]

For the intact ring, ideal poloidal winding currents and the no-leakage boundary condition admit $\mathbf{H}=N_{1}I\hat{\mathbf{\varphi }}/(2\pi r)$ with zero radial and vertical components. It is curl-free inside the current-free core, divergence-free for uniform magnetic permeability, tangent to the core boundary, and has the required circulation; uniqueness of the Neumann problem removes an additional homogeneous scalar-potential field. This explains zero non-azimuthal components in the stipulated intact-ring idealization. A literal slit breaks that rotational symmetry, so zero transverse components everywhere is not exact for the cracked ring: magnetic fringing and flux redistribution occur. In fact, extending the approximate $B(r)\propto 1/(r+q)$ unchanged around a current-free core would give $[\nabla \times \mathbf{H}{]}_z\ne 0$. Thus that radial-path calculation and a claim of exact purely azimuthal field in the cracked geometry cannot both be exact. The numerical answer above retains the intended circuit approximation while making its limitation explicit.

Electromagnetic induction produces an electromotive force when magnetic flux linkage changes. For an oriented moving loop,For a stationary loop this reduces to Faraday's law of induction, $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$. Motion in a static field can instead supply the motional term. Lenz's law gives the minus sign: the induced electric current opposes the flux change, with the required electrical energy supplied by mechanical work or the field source.

For linear fixed-geometry circuits, flux linkages satisfy ${\Lambda }_1=L_1{I}_1+M{I}_2$ and ${\Lambda }_2=M{I}_1+L_2{I}_2$. Self-inductance gives ${\mathcal{E}}_1=-L_1{\dot{I}}_1$; mutual inductance gives the additional term $-M{\dot{I}}_2$. Reciprocity gives the same mutual coefficient for both circuits, with its sign determined by winding orientation. The stored field energy isThe bound follows from nonnegative energy for arbitrary electric currents. In a linear medium the equivalent field expression is $W_B=\frac{1}{2}\int \mathbf{B}\cdot \mathbf{H}dV$.

An ideal electrical transformer links both windings with one changing core flux, giving $V_1/V_2=N_1/N_2$ for corresponding voltage magnitudes. A loaded secondary drives an opposing magnetomotive force and increases primary electric current; ideally $I_1/I_2=N_2/N_1$ and input and output powers agree. A electrical transformer exchanges voltage for electric current, not for extra power. Real devices have winding resistance, leakage flux, eddy-current and hysteresis losses, and finite magnetizing electric current. A steady DC input gives no continuing secondary voltage after the switching transient and may cause excessive primary electric current or core saturation.