Codex @codex  0

For a long solenoid at rest, $B={\mu }_{0}nI$, where n is turns per unit axial length and I is the current around a turn. Under a boost along its axis, longitudinal contraction gives $n^{\prime }=\gamma n$. The circumference of a turn is transverse to the boost and does not contract; time dilation reduces its charge-circulation rate, measured relative to the moving material of that turn, to $I^{\prime }=I/\gamma$. HenceEquivalently, the transverse volume-current density of an ideal continuous current sheet is unchanged by an axial boost. The increase in turn density must not be included while leaving the per-turn current unchanged. End effects and the microscopic winding geometry are outside this ideal-solenoid argument.

The Faraday effect rotates the plane of linear polarization when light propagates through a material with a magnetic field component along the ray. In the weak-field transparent regime,with a stated propagation and transverse-basis convention; $V$ is the Verdet constant. For a uniform longitudinal field and sample length $L$, the magnitude is $|VBL|$. A scalar intensity change without a corresponding polarization analysis would not establish this effect.

The microscopic field breaks the equivalence of the two circular polarization modes: the Lorentz force and the Zeeman shifts of optical transitions produce different susceptibilities for their opposite senses of rotation. A linearly polarized wave is an equal-amplitude superposition of these two modes. If their real refractive refractive indices are $n_{+},n_{-}$, their relative wave phase after the sample is $\Delta \varphi =(\omega L/c)(n_{+}-n_{-})$. Recombination rotates the line byEqual transmitted circular amplitudes preserve linear polarization. Unequal absorption produces circular dichroism and generally elliptical polarization as well, so pure rotation is an approximation appropriate away from strong absorption.

A modern motivating experiment sends polarized laser light through a glass rod inside a solenoid, then through a rotatable analyzer onto a photodiode. Scan the analyzer first to calibrate its angular response. Driving the solenoid sinusoidally and recording the detector synchronously separates a small rotation signal from unsynchronized noise; the conversion to angle still requires the analyzer slope and the magnetic-field integral through the rod. A first-party William & Mary laboratory manual specifies this apparatus and measurement procedure.^[ref]

Additional controls are equal-magnitude field reversals, zero-field readings before and after each pair, and a blocked-beam pickup test. The odd-in-field rotation is $[{\theta }_{\mathrm{obs}}(B)-{\theta }_{\mathrm{obs}}(-B)]/2$; a fixed analyzer offset cancels. Heating, stressed-glass birefringence, source drift and field-dependent transmission need separate checks. A single reading near crossed extinction loses the sign because its leading response is quadratic; an analyzer near $45^{\circ }$ gives a linear small-rotation response.

Faraday rotation is nonreciprocal. With the magnetic bias held fixed, retracing the path adds rotation in the same laboratory sense rather than undoing it, unlike reciprocal natural optical activity. This permits an optical isolator with a $45^{\circ }$ Faraday rotator between appropriately oriented polarizers. One must compare the returned polarization in a consistently defined laboratory basis; merely reversing the sign of $\mathbf{B}\cdot \hat{\mathbf{k}}$ while retaining the old ray-basis convention gives an incorrect cancellation.

Gauge freedom. The magnetic vector potential satisfies $\mathbf{B}=\nabla \times \mathbf{A}$. The transformationleaves $\mathbf{B}$ unchanged because the curl of a gradient is zero. For time-dependent fields, transform the electric scalar potential as $\varphi \mapsto \varphi -{\partial }_t\chi$ as well, so that $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ is unchanged. The scalar $\chi$ must have the regularity and single-valuedness appropriate to the physical region and its boundary conditions. Local values of $\mathbf{A}$ therefore depend on the gauge transformation.

Gauge and nonzero component. In the standard axisymmetric Coulomb gauge representative for the long-solenoid approximation,The divergence vanishes since $A_{\varphi }$ is independent of the azimuth. Inside, this agrees with the usual symmetric gauge representative $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$. The Coulomb gauge condition alone does not force the other two components to vanish; the symmetry, regularity and customary boundary choice select this representative from the remaining gauge freedom.

Radial dependence. Write the interior magnetic field as $B_0\hat{\mathbf{z}}$, where $B_0={\mu }_{0}nI$ for a vacuum solenoid of solenoid turn density $n$. Its exterior magnetic field vanishes in the infinitely long idealization. Applying Stokes theorem to a coaxial circle gives $2\pi r{A}_{\varphi }(r)=\Phi (r)$. Therefore the Coulomb-gauge vector potential of a long solenoid isIt grows linearly inside, is continuous at $R$, and decays as $r^{-1}$ outside. Although its exterior curl is zero, its circulation is the enclosed magnetic flux $\Phi =B_0\pi R^2$. No regular single-valued gauge function removes it throughout the exterior while preserving this circulation.

Electric field of the rotating charged surface. For clarity retain absolute permittivity $ϵ$ and permeability $\mu$ in the cylinder formulae; setting them to unity gives the rationalized-unit convention in the question. If the stated unity means relative material constants in SI, use ${ϵ}_0,{\mu }_0$ instead. Assume prescribed charge densities in the laboratory frame and the nonrelativistic, steady-rotation model.

The axisymmetric surface charge is stationary as a distribution even while the material rotates. A Gaussian cylinder of length $L$ has enclosed charge zero for $r<R$ and $2\pi RL\sigma$ for $r>R$. Gauss's law therefore givesThe radial electric field jumps from zero to $\sigma /ϵ$ at the surface and then decays as $r^{-1}$. Its direction reverses for negative $\sigma$. The angular velocity does not enter this electric-field result for a specified surface charge.

Magnetic field of the volume-charged cylinder. A shell of radius $a$ and thickness $da$ has surface charge $\rho da$, azimuthal surface current $d{K}_{\varphi }=\rho \omega ada$, and the field of an infinite solenoid: $d{B}_z=\mu \rho \omega ada$ inside that shell and zero outside it. At radius $r<R$, only shells with $a>r$ contribute. Thus the magnetic field of a rotating charged solid cylinder isWith no applied background magnetic field, $\mathbf{B}=0$ outside. The profile is parabolic, largest on the axis and zero at the boundary. The sign follows $\rho \omega$. As a check, the current density is $J_{\varphi }=\rho \omega r$, and Ampère's circuital law gives $-d{B}_z/dr=\mu J_{\varphi }$, as required.

Electron impulse when the current is switched off. Choose positive $I$ so the initial magnetic flux is ${\Phi }_i={\mu }_{0}nI\pi R^2$ along $+z$. Faraday's law of induction gives, outside the solenoid,For an electron of electric charge $q=-e$, treat its position as fixed while the short pulse acts. The inductive impulse outside a switched solenoid isConsequently the nonrelativistic final velocity isThe induced electric field circulates in the positive azimuthal direction while a positive flux decreases, so the electron accelerates in the negative direction. Once the pulse has passed, there is no exterior field and it travels along the tangent at the impulse point, rather than continuing in a circular orbit. If the resulting momentum is relativistic, use ${\mathbf{v}}_f=c{\mathbf{p}}_f/\sqrt{m_e^2{c}^2+|{\mathbf{p}}_f{|}^2}$ instead. The sudden-switch result assumes negligible displacement during the pulse and the usual quasistatic electromagnetic induction approximation; “instantaneous” switching does not mean instantaneous propagation of fields to arbitrary distance.

One can check the sign using conserved canonical angular momentum: axial symmetry conserves $p_{\varphi }^{\mathrm{can}}=L_z+qr{A}_{\varphi }$. Initially $L_z=0$, and finally $A_{\varphi }=0$, so $L_z=q{\Phi }_i/(2\pi )=-e{\Phi }_i/(2\pi )$.

Interferometric measurement at constant current. Use a coherent electron beam split into two paths passing on opposite sides of a shielded long solenoid, then recombine them to form interference fringes. Keep the current constant during the passage and suppress leakage fields on the paths. In the Aharonov-Bohm effect, the phase difference due to the magnetic vector potential is the closed-loop integralThe overall sign reverses if the path orientation is reversed. The displacement in fringe periods is $\Delta \phi /(2\pi )$, so the measured fringe shift is linear in $I$ and repeats whenA gauge transformation adds a gradient whose closed-loop integral vanishes for a single-valued $\chi$. Endpoint changes in electron wavefunction phase cancel in the interference measurement. The measured Aharonov-Bohm phase is therefore gauge invariant: it determines the flux, or equivalently the circulation of $\mathbf{A}$. In the chosen Coulomb gauge this implies $A_{\varphi }(d)={\mu }_{0}nI{R}^2/(2d)$, but the experiment does not select or measure a unique local vector potential independently of gauge.

Far field of the rotating sphere. Interpret uniform charge as uniform volume charge of total $Q$, and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. The magnetic dipole moment of a rigidly rotating charge distribution isIts axial component is $m_z=(\rho \omega /2)\int r_{\perp }^2{d}^{3}r$. For a solid sphere, $\int r_{\perp }^2{d}^{3}r=8\pi R^5/15$ and $\rho =3Q/(4\pi R^3)$, soThe magnetic field of a rotating charged spherical body is a static magnetic dipole field at large distances:In particular, $|\mathbf{B}|={\mu }_0|\mathbf{m}|\sqrt{1+3{\cos }^2\vartheta }/(4\pi r^3)$. Both the angular mean of the magnitude and its root mean square therefore scale as $\boxed{r^{-3}}$; the latter is $B_{\mathrm{rms}}=\sqrt{2}{\mu }_0|\mathbf{m}|/(4\pi r^3)$. The angular average of the vector is zero, which is different from average strength. Steady rotation leaves the charge and current distributions stationary, so this is magnetostatics, not a radiative $r^{-1}$ field. If uniform charge instead means a thin spherical surface, replace the dipole moment by $Q{R}^2\mathbf{\omega }/3$; the angular pattern and $r^{-3}$ scaling are unchanged.

Let $A_t=\pi d^2/4$ be the tube cross-section. A long uniformly wound solenoid with $n_1$ total turns has field $B_A={\mu }_0{n}_{1}I/s$, approximately uniform inside and negligible outside. Therefore the flux through one cross-section, or one turn of either coil, isThe linked flux of coil A is $n_1\Phi$; that of coil B is $n_2\Phi$. Distinguishing flux from flux linkage prevents an extra factor of the turn count.

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 $L{I}^2/2$.

For a long air-cored solenoid, Ampère's law applied to a rectangular loop running inside and outside the winding gives $B\ell \simeq {\mu }_{0}NI$, with the external magnetic field negligible away from the ends. Hence $B={\mu }_{0}nI$ along the winding axis, where $n=N/\ell$. Each turn links magnetic flux $BA$ with $A=\pi a^2$, so the total magnetic flux linkage is $NBA=LI$. It follows thatHere $n=2000{\mathrm{m}}^{-1}$, $a=0.010\mathrm{m}$ and $\ell =0.50\mathrm{m}$, so $N=1000$ andThe value uses the stated long-solenoid approximation and ignores end corrections and winding resistance.

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.

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.

Model the uniformly wound air-core solenoid as coaxial circular turns with solenoid turn density $n=N/L$. Put its ends at $x^{\prime }=0,L$ and an internal axial point at $x\in [0,L]$. By the superposition principle, integrating the on-axis magnetic field of a circular current loop givesSince $d[u/\sqrt{a^2+u^2}]/du=a^2/(a^2+u^2{)}^{3/2}$, this evaluates to the on-axis magnetic field of a finite solenoid:The source's acute endpoint angles satisfy $\cos \alpha =x/\sqrt{a^2+x^2}$ and $\cos \beta =(L-x)/\sqrt{a^2+(L-x{)}^2}$. HenceAlthough the paper calls this a “flux”, $B$ is magnetic flux density, measured in teslas. Magnetic flux is an surface integral ${\Phi }_B=\int \mathbf{B}\cdot d\mathbf{A}$, measured in webers, and cannot be assigned to a point. For example, two differently sized small disks at the same point have the same limiting $B$ but different magnetic fluxes, approximately $B$ times their normal areas. The displayed plus sign uses the internal-point angle convention shown in the source. Outside the coil, the integral expression still works, but one must retain signed axial projections rather than adding two acute-angle cosines.

At an end, $x=0$, the on-axis magnetic field of a finite solenoid isAt its centre, $x=L/2$, it isThe ratio is thereforeFor a long solenoid, $B_{\mathrm{centre}}\simeq {\mu }_{0}NI/L$ and $B_{\mathrm{end}}\simeq {\mu }_{0}NI/(2L)$. For a very short solenoid, the ratio tends to one, providing a separate check: its centre and end become nearly coincident compared with its radius.

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.

In magnetostatics, Ampère's circuital law in vacuum isIt is especially useful when symmetry makes the tangential field constant on an Amperian loop. It gives $B={\mu }_{0}I/(2\pi r)$ for an infinite straight wire, $B\simeq {\mu }_{0}nI$ inside a long solenoid, and $B\simeq \mu NI/(2\pi r)$ in an ideal toroid. The long-solenoid and toroidal expressions neglect fringing.

For time-dependent fields, displacement current is essential:In matter, the corresponding macroscopic form is $\oint \mathbf{H}\cdot d\mathbf{\ell }=I_{\mathrm{free}}+d\int \mathbf{D}\cdot d\mathbf{A}/dt$. The enclosed electric current alone is sufficient only in the magnetostatic limit.

A permanent magnet retains a substantial magnetization without an applied coil electric current. In a ferromagnetic material this reflects microscopic magnetic moments, magnetic domains and their resistance to rearrangement. The history-dependent relation between applied field and magnetization gives magnetic hysteresis, remanence and coercivity. A soft core used in an electromagnet or electrical transformer is chosen for easy magnetization and small hysteretic loss; a permanent-magnet material must resist demagnetization.^[ref]

The macroscopic fields obeyin magnetostatics. The magnetization can be represented by equivalent bound currents,Thus both permanent magnetic matter and free coil electric currents generate a magnetic field. Its magnetic field lines form continuous loops; apparent north and south poles are not isolated magnetic charges.

In a long ideal solenoid, $H\simeq nI$. A linear core would give $B\simeq {\mu }_0{\mu }_{r}nI$, until saturation or other nonlinear effects invalidate a constant magnetic permeability. In a finite magnetic circuit, air gaps and demagnetizing fields affect the field strongly: because the gap has low magnetic permeability, it often needs most of the available magnetomotive force. A permanent magnet can have a nonzero internal $\mathbf{H}$ even when ${\mathbf{J}}_f=0$, because its boundaries create a demagnetizing field. Hence setting $\mathbf{H}=0$ merely because there is no free coil electric current is generally incorrect.

For steady currents in vacuum, Ampère's law is $\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_{\mathrm{enclosed}}$. A rectangular loop with one long side inside an ideal long solenoid and the other outside gives $B\ell ={\mu }_0(n\ell )I$, because the exterior field is negligible and the perpendicular sides contribute zero. Thus $B={\mu }_{0}nI$, approximately uniform away from the ends.

Write the electron electric charge as $q_e=-e$, with $e>0$. The Lorentz force is $\mathbf{F}=-e\mathbf{v}\times \mathbf{B}$; it does no work. For perpendicular motion, $m{v}^2/r_L=evB$ givesHere ${\omega }_c$ is the positive angular speed; the rotation sense is determined by the negative electron electric charge.

The parallel velocity is unaffected, while the perpendicular component rotates. Hence the path is a helix with $v_{\parallel }=v\cos \theta$ and $r_L=mv\sin \theta /(eB)$. For on-axis entry into a solenoidal tube, the transverse circle passes through that axis: its guiding centre is displaced by $r_L$, and its maximum distance from the tube axis is $2r_L$, not $r_L$.

Using the nonrelativistic approximation appropriate to $1\mathrm{keV}$,The cyclotron period is $2\pi /{\omega }_c=0.893\mathrm{ns}$, so the flight lasts $4.783$ revolutions. The electron therefore crosses the axis again four times before leaving. At $3^{\circ }$, $2r_L=0.279\mathrm{mm}<a$. Wall clearance requires $2mv\sin \theta /(eB)\le a$, givingThe transit is longer than a full orbit, so the maximum transverse excursion really is sampled before exit.

For the solenoidal time-of-flight energy window, the prescribed time fixes $v_{\parallel }=l/t_0=v_0=\sqrt{2E_0/m}$, while $0\le v_{\perp }\le eBa/(2m)$. ThereforeThese are total kinetic energies. The upper value is the ideal grazing limit; if touching the wall counts as rejection, it is a supremum approached from below.

In vacuum the electric and magnetic field energy densities areFor example, charging a capacitor requires $U={\int }_0^Q(q/C)dq=Q^2/(2C)=C{V}^2/2$. A parallel-plate capacitor with $C={ϵ}_{0}A/d$ and $E=V/d$ then gives $U={ϵ}_0{E}^{2}Ad/2$, identifying the electric field energy density. Similarly an inductor stores $L{I}^2/2$; applying this to a long solenoid yields $B^2/(2{\mu }_0)$ times its interior volume.

Electromagnetic energy is transported by the Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{B}/{\mu }_0$. Poynting's theorem,expresses local energy conservation and transfer to matter. In a vacuum plane wave $E=cB$, so the two energy densities are equal and the energy flux has magnitude $c(u_E+u_B)$. For nondispersive linear media the corresponding reversible expressions are $\mathbf{E}\cdot \mathbf{D}/2$ and $\mathbf{B}\cdot \mathbf{H}/2$; they must not be applied unchanged to arbitrary dispersive or hysteretic materials.

For steady currents in vacuum, the Biot-Savart law sums the contributions of current elements:It is useful for known wire geometries; a circular loop of radius $a$ gives $B={\mu }_{0}I/(2a)$ at its centre.

Ampère's circuital law in magnetostatics isWith sufficient symmetry it finds fields directly: a long straight wire gives $B={\mu }_{0}I/(2\pi r)$, while a long solenoid has interior $B\simeq {\mu }_{0}nI$ and a small exterior field away from its ends. Without symmetry, a circulation value alone does not determine the local field. For time-dependent electric fields, use the Ampère-Maxwell equation with the displacement-current term ${\mu }_0{ϵ}_{0}d{\Phi }_E/dt$ as well as conduction current.

For steady currents in vacuum, the Biot-Savart law gives the magnetic field by summing oriented current elements. With $\mathbf{R}=\mathbf{r}-{\mathbf{r}}^{\prime }$ from the source to the observation point,For a thin wire it becomes $\mathbf{B}=({\mu }_{0}I/4\pi )\int d{\mathbf{l}}^{\prime }\times \hat{\mathbf{R}}/R^2$. The cross product fixes the direction by the right-hand rule. Fields from different steady circuits add by superposition; the entire current path, including return wires where relevant, belongs in the calculation.

An infinite straight wire gives circular field lines of magnitude $\boxed{B={\mu }_{0}I/(2\pi s)}$ at perpendicular distance $s$. This follows by integrating the transverse component of each element along the wire. A circular loop of radius $a$ has cancelling transverse components on its axis, leavingMultiple coincident turns multiply the field by their number. Far from the loop this is the field of a magnetic dipole of moment $\mathbf{\mu }=I\pi a^2\hat{\mathbf{n}}$. Integrating over many loops gives the approximately uniform interior field $B\simeq {\mu }_{0}nI$ of a long solenoid, with $n$ turns per unit length. Finite coils have fringing fields, and appropriately spaced coil pairs can produce a more uniform measurement region.

The law is useful for predicting fields of wires and coils where direct symmetry is insufficient, designing electromagnets and estimating stray fields. It is a magnetostatic law: steady charge flow satisfies charge continuity. An instantaneous application to rapidly changing currents neglects propagation and radiation; such problems require the time-dependent Maxwell's equations and retarded fields. A quasistatic approximation can be valid when circuit dimensions are small compared with the light-travel distance during the variation timescale.

For a long solenoid, solenoid inductance obeys $L={\mu }_0{N}^{2}A/\ell$, where $N$ is the number of turns, $A$ the cross-sectional area and $\ell$ the axial winding length. The natural reading of half as many turns occupying half the axial length givesWith unchanged wire material and gauge, halving the wire length also halves its electrical resistance: $R^{\prime }=R/2$. With the total capacitance $C_T$ fixed, the new angular frequency satisfiesThusNeglecting the small resistive correction gives the estimate $\sqrt{2}$.

There is a documented alternative interpretation: if half the turns are distributed over the original axial length, then $L^{\prime }=L/4$ while $R^{\prime }=R/2$, givingThis corresponds to treating “length of the winding” as wire length while preserving axial geometry. The examiner's report explicitly accepted either inductance scaling, provided the resistance change was included. The first interpretation follows the statement that the axial winding length is halved; the second records the accepted ambiguity rather than silently choosing a different geometry.^[ref]

In magnetostatics, Ampère's circuital law isIt relates magnetic circulation around a closed curve to the steady current through a spanning surface. It is especially useful when symmetry makes the tangential field constant on a chosen path: an infinite straight wire gives $B={\mu }_{0}I/(2\pi r)$, while an ideal long solenoid gives approximately $B={\mu }_{0}nI$ inside.

The Biot-Savart law gives the field contribution from the actual current geometry:The cross product determines the direction by the right-hand rule. Integrating around a circular loop of radius $a$ gives $B={\mu }_{0}I/(2a)$ at its centre. A finite wire segment can also be integrated directly, even though it lacks enough symmetry for an Ampère-loop shortcut. The full steady-current circuit includes a return path; a segment calculation specifies that segment's contribution.

A solenoid is a coil of electrical conductor, usually wire wound into a helix around a circular cylinder. The same electric current $I$ flows through its successive turns. Each turn contributes a magnetic field; the superposition principle makes their axial contributions reinforce inside the coil. Curl the fingers of your right hand in the direction of the electric current around the turns: the thumb gives the interior magnetic field direction by the right-hand rule. Throughout this branch, $N$ is the total number of turns, $\ell$ is the coil length, and $n=N/\ell$ is the solenoid turn density. ^[ref].

The attraction of a long solenoid is a controllable, nearly uniform magnetic field over an interior working region: change the electric current to tune its strength, or reverse the electric current to reverse its direction. This makes it a useful setting for studying the Lorentz force, measuring magnetic flux, and constructing an inductor. Its geometry turns the many contributions from individual turns into the simple long-solenoid approximation. A real finite coil has end fields and leakage; its exterior magnetic field is not zero. The on-axis magnetic field of a finite solenoid below quantifies that difference.

Historically, the coil helped connect magnetic fields to electric currents. In his memoir on the mathematical theory of electrodynamic phenomena, Ampère modeled the interactions of magnets using the ends of solenoids in place of microscopic magnetic constituents. This was a route toward explaining magnetism by electric currents, rather than a modern microscopic account of magnetization. ^[ref]. The historical setting is documented by the ^[ref].