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For a steady current density $J$, the magnetic field is

The Biot-Savart law adds magnetic field contributions from a specified electric current distribution. The vectors ${\mathbf{r}}^{\prime }$ and $\mathbf{r}$ locate source and observer; $\mathbf{r}-{\mathbf{r}}^{\prime }$ points toward the observer. Here ${\mu }_0$ is vacuum permeability. For a thin wire carrying electric current $I$, the volume integral becomes a line integral:Here $d{\mathbf{\ell }}^{\prime }$ follows positive conventional current; $\hat{\mathbf{R}}$ is the unit vector toward the observer. With fixed path orientation, reversing the electric current replaces $I$ by $-I$. The cross product and right-hand rule fix the direction. Reversing the source-to-observer displacement would give the wrong sign.^[ref]

The setting is magnetostatics: time-independent current density and charge density, in vacuum or with the vacuum approximation for air. Electric charge conservation then gives $\nabla \cdot \mathbf{J}=0$. A finite segment is part of a complete electrical circuit; its endpoints cannot continuously accumulate electric charge in a steady state. The instantaneous integral above does not describe switching transients or electromagnetic radiation. In magnetic matter, specify the total current density, including magnetization current, rather than automatically replacing ${\mu }_0$ by an arbitrary magnetic permeability.^[ref]

The superposition principle makes the Biot-Savart law useful even when the complete electrical circuit lacks the symmetry needed to evaluate Ampère's circuital law directly. Integrate a straight piece, a curved piece, and the return path separately, then add their magnetic fields as vectors. The beauty is that a local oriented current element produces both the circulating magnetic field of a finite straight wire and the reinforcing on-axis magnetic field of a circular current loop.

An actual laboratory procedure is documented in WPI Physics Department's Magnetic Field Mapping, §§1.2–1.2.3 and the single-coil checklist on p. 9. Preparation locates the active area of a Hall effect sensor using a small magnet. A field meter reads the component normal to that area. The wire apparatus is a wall-mounted square loop selectable between one and five conductors. The coil connects to a power supply and ammeter; its diameter and turn count are recorded. Align the sensor and scan distance. The control is a current-off background reading at each position, subtracted from the current-on reading. The inference compares corrected magnetic field with position and the predicted geometry dependence. Active-area displacement, sensor tilt, background variation, and finite coil width limit that comparison. These are instructions, not reported measurements.^[ref]

A further proposed control is to reverse $I$ at fixed sensor position and orientation. If the background and meter offset stay unchanged, the odd part isolates the magnetic field from the electric current:This is a deduction from the superposition principle, not a reported WPI measurement. The measured component, sensor position, and complete return path must still match the model.

Take a wire segment on the $y$ axis, with source positions ${\mathbf{r}}^{\prime }=(0,s,0)$ for $s_1\le s\le s_2$, and positive electric current along $+y$. Observe at $\mathbf{r}=(a,0,0)$, where $a>0$ is the perpendicular distance to the wire's supporting line. Choose $x$ rightward and $y$ upward, so $+z$ points out of the page. ThenThus the magnetic field points into the page by the right-hand rule. The Biot-Savart law and an elementary antiderivative giveFor an original worked example, choose $s_1=-0.12\mathrm{m}$, $s_2=0.12\mathrm{m}$, $a=0.05\mathrm{m}$, and $I=3\mathrm{A}$. Use ${\mu }_0\approx 4\pi \times 10^{-7}\mathrm{T}m/A$ for these rounded calculations. Each endpoint contributes $0.12/0.13=12/13$, soThese are calculated inputs and output, not laboratory data. The two endpoint angles measured from the perpendicular each have sine $12/13$; this recovers the equivalent endpoint-angle formula without depending on a diagram's angle-sign convention.

For a segment centered on the perpendicular foot, $s_1=-L$, $s_2=L$. Its magnetic field magnitude isThe limit $L/a\to \infty$ gives $B_{\infty }={\mu }_{0}I/(2\pi a)$, consistent with Ampère's circuital law for an infinite straight wire. In the example, $B_{\infty }=12\mu \mathrm{T}$ and $B/B_{\infty }=12/13$: the infinite-wire value overestimates the segment value by $1/12$, or about $8.33\%$ of the segment value. For $a\gg L$, the isolated segment contribution instead behaves as $B\sim {\mu }_{0}I(2L)/(4\pi a^2)$. The complete localized electrical circuit has a different far-field behavior: return-path contributions cancel this leading segment term, leaving a magnetic dipole field when its magnetic dipole moment is nonzero. Segment and whole-circuit limits must be distinguished.

The SI units check directly: ${\mu }_{0}I/a$ has units $(\mathrm{T}m/A)\mathrm{A}/\mathrm{m}=\mathrm{T}$, the tesla. The apparent divergence as $a\to 0$ is a failure of the thin-wire model near the conductor, not a prediction of infinite magnetic field inside a real wire. This calculation gives only this segment's contribution; include the return wires before comparing with an apparatus.

An infinite solid cylinder with radius $R$, prescribed uniform volume charge density $\rho$, and angular velocity $\omega \hat{\mathbf{z}}$ has current density $\mathbf{J}=\rho \omega r\hat{\mathbf{\varphi }}$. Treat a shell at radius $a$ and thickness $da$ as a solenoid carrying surface current $\rho \omega ada$. With no applied background magnetic field and permeability $\mu$,The same result follows from Ampère's circuital law, since $-d{B}_z/dr=\mu \rho \omega r$ inside.

In a simply connected current-free region of constant magnetic permeability, Ampère's circuital law makes the magnetic field curl-free, so it admits a scalar potential. The absence of magnetic monopoles then gives Laplace's equation for that potential. Some conventions define the potential for $\mathbf{H}$ instead of $\mathbf{B}$, differing by the constant permeability. Boundary conditions, including imposed fields at infinity, determine the potential up to an additive constant.

Two broad perfect conductors of width $w$, separated by gap $d\ll w$, carry opposite steady electric currents $I$ and connect through a terminating resistor $R$. Away from edges, Ampère's circuital law and the superposition principle give gap magnetic field magnitude ${\mu }_{0}I/w$. Each conductor is an equipotential, and Ohm's law sets their electric potential difference to $IR$. Planar symmetry and charge-free Gauss's law then give uniform gap electric field magnitude $IR/d$. Injection parallel to the upper current makes the electric and magnetic Lorentz forces opposite. Their cancellation selects speed $Rw/({\mu }_{0}d)$ independent of nonzero electric charge and of operating current; nonrelativistic kinetic energy is $m[Rw/({\mu }_{0}d){]}^2/2$.

Let $\ell =2\pi r-d$ be the mean path length in the core. Neglect leakage and fringing across the narrow air gap, and assume a cross-section small compared with $r$. Continuity of normal magnetic flux density then gives the same tangential magnetic field magnitude $B$ in the core and gap. Ampère's circuital law for the magnetic field strength $H$ isFor a linear core, $H_{\mathrm{core}}=B/({\mu }_0{\mu }_r)$ and $H_{\mathrm{gap}}=B/{\mu }_0$, givingHere $\hat{\mathbf{t}}$ follows the winding's right-hand rule. The gap has much larger $H$ than the core when ${\mu }_r\gg 1$, although $B$ is approximately continuous. Equivalently, the core and gap magnetic reluctances add in series.

On a magnetic hysteresis loop, remanence is the nonzero $B_r$ remaining at $H=0$ after magnetizing the sample. The normal magnetic coercivity $H_{cB}$ is the magnitude of reverse $H$ at which $B=0$ on that branch. The intrinsic magnetic coercivity instead refers to zero magnetization; these two conventions should not be confused. A comparison of coercive $H$ with remanent $B$ requires a unit conversion or a specified constitutive scale, because their SI units differ. A soft ferromagnet is understood here to have a narrow loop and small coercive field.

Use the stated single-valued constitutive equation for the calculation. Inverting it gives $H_{\mathrm{core}}=H_m\mathrm{artanh}(B/B_m)$, soSince $\mathrm{artanh}z=z+O(z^3)$, the small-signal response of a gapped magnetic core isThe small-field core magnetic permeability is $B_m/H_m$. Strictly, $B=B_m\tanh (H/H_m)$ is a reversible approximation: at $H=0$ it has $B=0$, so it cannot itself model nonzero remanence. The displayed loop and this constitutive approximation serve different purposes. Keeping a genuinely remanent branch would generally add a history-dependent offset; it would not imply $B=\kappa I$ through the origin.

Use SI units and normalize the condensate wavefunction so $|\psi {|}^2=n$. With mechanical momentum operator $-i\hslash \nabla -qA$, the charge current density isFor uniform $n$ and $\psi =\sqrt{n}{e}^{i\vartheta }$, this becomes $j=(nq/m)(\hslash \nabla \vartheta -qA)$.

Under the gauge transformation $A\mapsto A+\nabla \chi$, the wavefunction transforms as $\psi \mapsto e^{iq\chi /\hslash }\psi$, so $\vartheta \mapsto \vartheta +q\chi /\hslash$. The two added gradients cancel in $\hslash \nabla \vartheta -qA$, while $|\psi {|}^2$ stays fixed. Thus the superconducting current is gauge invariant. The same cancellation in the full displayed current proves it without requiring a constant amplitude.

For uniform carrier density in a region with smooth single-valued phase, taking a curl gives the London equation $\nabla \times j=-n{q}^{2}B/m$. In the time-independent setting Ampère's circuital law gives $\nabla \times B={\mu }_{0}j$, and Gauss's law for magnetism gives $\nabla \cdot B=0$. Thereforeso the Helmholtz equation and London penetration depth areThe equation has the modified, exponentially decaying Helmholtz sign. For a planar surface at $x=0$ with material in $x>0$, the bounded solution is $B(x)=B(0)e^{-x/\ell }$; the exponentially growing solution is excluded in the bulk. Hence a macroscopic region far from the surface has negligible field: magnetic flux is expelled from the bulk, the Meissner effect. The literal assertion of zero flux everywhere is the bulk limit: finite penetration remains within distance $\ell$ of a surface. Vortices or flux trapped through a hole require different phase or topology assumptions and are not ruled out by this smooth-phase derivation.

Integrate Poisson equation for Newtonian gravity through a narrow slab around the disk. The horizontal derivative contributions vanish as its thickness tends to zero, leaving the normal-derivative jumpThe even function symmetry of $\Phi$ makes the derivatives opposite, soIn the current-free simply connected upper half-space, Ampère's circuital law gives $\nabla \times \mathbf{B}=0$ and permits a magnetic scalar potential. Rescale it so that $\sqrt{4\pi G/{\mu }_0}\mathbf{B}=-\nabla {\Phi }_M$. The divergence-free condition makes ${\Phi }_M$ satisfy Laplace's equation, withCompare with the gravitational jump condition. Subject to the same isolated-field boundary condition at infinity, ${\Phi }_M$ is the harmonic potential of the effective surface densityThis gravity-equivalent magnetic surface density can have either sign; it is a mathematical representation of the exterior magnetic field, not physical negative mass. An imposed nondecaying field would require additional boundary data and would not be fixed by the disk surface density alone.

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.

The current density is the moving charge density times its velocity:For a cylindrical Gaussian surface of radius $r<R$ and length $\ell$, cylindrical symmetry makes the electric field radial. Gauss's law gives $E(2\pi r\ell )=\rho \pi r^2\ell /{ϵ}_0$. A circular Ampère's circuital law contour encloses electric current $I(r)=\rho v_0\pi r^2$, giving $B(2\pi r)={\mu }_{0}I(r)$. HenceThe right-hand rule fixes the positive azimuthal direction of the magnetic field for positive electric charge moving along $+z$. Both the electric field and the magnetic field vanish on the axis. These are the steady long-cylinder electric field and magnetic field; end effects and changes of the prescribed beam distribution are excluded.

For steady currents in vacuum, Ampère's circuital law isHere $C$ is an oriented closed curve, $d\ell$ its tangent line element, and $I_{\mathrm{enc}}$ the signed electric current through a spanning surface oriented by the right-hand rule. A circular contour about a straight infinite wire has constant tangential magnetic field, so $B(r)={\mu }_{0}I/(2\pi r)$. The Lorentz force on a length $L$ of the other wire is $IL\hat{\ell }\times B$. ThereforeThe right-hand rule makes the force on each parallel, same-direction current point toward the other wire.

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.

For one wire, Ampère's circuital law on a concentric circle of radius $r$ givesThe other wire experiences the Lorentz force on its electric current, $d\mathbf{F}=Id\mathbf{\ell }\times \mathbf{B}$. Its direction is perpendicular to this magnetic field, so the force per unit length isThe right-hand rule gives attraction for parallel electric currents in the same direction and repulsion for opposite directions.

For the upper wire, the upward magnetic force per unit length balances its weight per unit length, $\lambda g$, where $g$ is the gravitational acceleration. ThereforeThis assumes the lower wire is held fixed and both wires remain straight. Within the vertical-motion model, increasing the separation weakens the upward magnetic force, so this equilibrium is vertically restoring.

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.

In magnetostatics Ampère's circuital law is $\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_{\mathrm{enc}}$. Its general vacuum version adds ${\mu }_0{ϵ}_{0}d{\Phi }_E/dt$ on the right. Faraday's law is $\mathcal{E}=\oint \mathbf{E}\cdot d\mathbf{l}=-d{\Phi }_B/dt$ for a fixed loop. A circular Amperian path around a long straight wire giveswith direction given by the right-hand rule. The force on a wire element is $d\mathbf{F}=Id\mathbf{l}\times \mathbf{B}$; its magnitude is $IBdl\sin \chi$ and its direction follows this cross product.

Choose the straight-wire current rightward and the square below it, as printed. Its positive magnetic field there is into the page. With an into-page loop normal, the magnetic flux linkage isHere $\mathcal{M}$ denotes mutual inductance, not a mass. Choose positive coil current to produce an out-of-page field. Then its induced current in the resistance-dominated approximation isThis reproduces the specified sign with the declared orientation. It neglects coil self-inductance, requiring $\omega L_{\mathrm{coil}}\ll R$, as implicit in the requested expression.

The nearest side contains N parallel current segments and is at a uniform distance $s$, so its force magnitude isThe two sides perpendicular to the wire have opposite currents and equal magnetic fields at corresponding distances; their forces cancel pairwise. The far side opposes the near-side force and has the smaller field. ThusWhen $I$ and $i$ have the same sign under the chosen conventions, the near-side coil current opposes the straight-wire current and the total force is away from the wire; otherwise it is toward it. In particular the signed force alternates with $\sin (\omega t)\cos (\omega t)$. The coil is fixed, so these forces do no mechanical work on it. The mean Joule heating is

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 electric currents in vacuum, Ampère's circuital law isChoose an oriented closed contour and a spanning surface; the positive enclosed electric current crosses that surface according to the right-hand rule for the contour. First use the source symmetry to establish the direction of $\mathbf{B}$ and which positions have the same magnitude. Then choose the contour so its useful segments are parallel to $\mathbf{B}$ with constant magnitude, and the remaining segments are perpendicular to $\mathbf{B}$ or have vanishing field contribution. The line integral reduces to the unknown magnitude times a known length or sum of lengths.

For example, cylindrical symmetry around an infinite wire allows a circular contour, giving $B(2\pi r)={\mu }_{0}I$. A uniform current sheet instead calls for a rectangle straddling it. The law is always valid in this magnetostatic setting, but by itself does not determine the local magnetic field from an arbitrary enclosed current without these extra symmetry arguments. Here ${\mu }_0$ is the vacuum permeability.

Choose $+x$ along the electric current, $+z$ upward normal to the sheet, and $+y$ to complete a right-handed frame. In the interior region far from its edges, treat the sheet as infinite in its plane. Its surface current density isReflection and translational symmetry make the magnetic field parallel to the sheet, perpendicular to its current, and equal in magnitude with opposite direction above and below. Choose an Ampère's circuital law rectangle with length $s$ along $y$, straddling the full thickness. The two long sides each contribute $Bs$ with the same sign; on the vertical sides $\mathbf{B}\cdot d\mathbf{l}=0$ since $\mathbf{B}$ is horizontal. The enclosed electric current is $Ks=Is/w$. ThusThe right-hand rule fixes the directions:The current density inside the finite-thickness conductor is $\mathbf{J}=I/(wt)\hat{\mathbf{x}}$, so integrating through thickness $t$ gives the same $\mathbf{K}$. These are exterior fields in the broad-sheet approximation, not exact finite-width edge fields.

For an infinite straight wire in vacuum, Ampère's circuital law around a circle of radius $R$ gives $B(2\pi R)={\mu }_{0}I$. ThereforeHere ${\mu }_0$ is the vacuum permeability and the azimuthal direction follows the right-hand rule around the electric current. The magnetic field is measured in $\mathrm{T}$.

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.