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Choose the electric current counterclockwise as viewed from $+z$; reversing it reverses the magnetic field. Parameterize the circular wire byThe cross product in the Biot-Savart law isConsequently the transverse terms integrate to zero over $0\le \varphi \le 2\pi$, while the axial term is constant:ThusThe direction is $+z$ on both halves of the axis for this electric current orientation, including $z<0$; it is not “away from the plane” on both sides. The right-hand rule gives the same direction.

Read the original diagram as a clockwise arc carrying electric current from its upper endpoint to its lower endpoint, with tangent semi-infinite leads extending to the right. Their crossing is insulated, so there is no extra junction. Choose $+z$ out of the page. At the centre the arc produces a negative axial magnetic field, and each straight lead a positive one, by the right-hand rule.

For the arc, every element is at distance $R$, with $d\mathbf{l}$ perpendicular to its displacement to the centre. The Biot-Savart law therefore givesFor each straight lead, write $s\ge 0$ for distance along the tangent measured from its endpoint. Its perpendicular distance from the centre is $R$, soThe integral equals $1/R$ because an antiderivative is $s/[R\sqrt{R^2+s^2}]$. Adding both leads and the arc,Using the full infinite-wire magnetic field for either lead would incorrectly double its contribution.

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.

For a regular hexagon of side $a$, the perpendicular centre-to-side distance is $d=\sqrt{3}a/2$. Each side subtends endpoint angles with $\cos \beta =1/2$ and $\cos \alpha =-1/2$. Its centre field is consequently ${\mu }_{0}I/(4\pi d)$.

All six contributions have the same normal direction, soThe direction follows the current's right-hand rule. Equal side lengths alone would not fix an arbitrary distorted hexagon's field; the usual regular-coil geometry is used here.

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].

Let $S$ be an oriented smooth surface with piecewise smooth boundary, and let $F$ be a continuously differentiable vector field on a neighborhood of $S$. With the compatible boundary direction determined by the right-hand rule,For $G=(-y,x,0)$ on the horizontal unit disc with upward normal, the left side is $2$ times its area, giving $2\pi$. The counterclockwise boundary has tangential field of magnitude one and circumference $2\pi$, giving the same line integral. Reversing the surface normal reverses the boundary direction and both signs. The theorem replaces a circulation calculation with a surface integral of the curl, whichever is easier.

Torque measures a force's turning effect about a specified origin. The geometry in the angular momentum diagram applies with force replacing momentum: transverse forces give positive or negative torque, while a radial force gives zero. A longer perpendicular lever arm makes the same force more effective, explaining why pushing near the end of a handle is useful.

For a force $\mathbf{F}$ applied at position $\mathbf{r}$ from a fixed origin $O$ in an inertial frame, its torque, or moment of force, isHere $F=|\mathbf{F}|$, $\varphi$ is the angle between position and force, and $\ell =r\sin \varphi$ is the perpendicular lever arm. Direction follows the right-hand rule; in the $xy$ plane ${\tau }_z=x{F}_y-y{F}_x$, positive counterclockwise as viewed from $+z$. The International System of Units unit is $\mathrm{N}\mathrm{m}=\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2$. Although torque and energy have the same physical dimensions, torque is an axial vector and energy is a scalar; their roles differ.

For $\mathbf{r}=(2,0,0)\mathrm{m}$ and $\mathbf{F}=(0,3,0)\mathrm{N}$, ${\tau }_z=+6\mathrm{N}\mathrm{m}$. Applying the same force at $(1,0,0)\mathrm{m}$ gives $+3\mathrm{N}\mathrm{m}$; reversing it gives a negative torque; directing it along $\mathbf{r}$ gives zero. More generally, sum ${\mathbf{r}}_i\times {\mathbf{F}}_i$ to find the total torque about one common origin.

For a constant-mass point mass, the product rule and Newton's second law show why net torque changes angular momentum:The first cross product vanishes because its factors are parallel. For a collection with canceling internal torques, the remaining sum is external torque, giving conservation of angular momentum when it vanishes.^[ref]

A nonzero torque can turn the direction of angular momentum as well as change its magnitude. Indeed, the dot product gives $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$: a perpendicular torque changes direction without instantaneously changing the magnitude. This is the geometry behind gyroscope precession. For a rapidly spinning symmetric top with a fixed tip, centre of mass distance $d$ from the tip, mass $M$, and spin-dominated angular momentum magnitude $L\simeq I\omega$, the slow steady precession approximation gives $\Omega \simeq Mgd/L$, where $g$ is gravitational acceleration. ^[ref] This is an approximation with $\Omega \ll \omega$; the uniform precession of a heavy symmetric top retains contributions omitted by the spin-dominated estimate.

Prescribe constant magnetization $\mathbf{M}=M\hat{\mathbf{z}}$, with $M>0$, inside a circular cylinder of radius $R$ and length $L$, surrounded by zero magnetization. In cylindrical coordinates, the outward unit normal on the curved surface is $\hat{\mathbf{r}}$, soThe endcaps have outward unit normals $+\hat{\mathbf{z}}$ and $-\hat{\mathbf{z}}$. Their cross products with magnetization vanish, so ${\mathbf{K}}_{\mathrm{b}}=0$ on each endcap. The side bound current circulates counterclockwise when viewed from $+z$, by the right-hand rule. Zero endcap surface current density does not imply zero endcap magnetic field.

For an original example, let $M=800A/m$, $R=0.020\mathrm{m}$, and $L=0.10\mathrm{m}$. The volume and total magnetic dipole moment areCheck the result independently using the equivalent bound current: a side strip of axial length $dz$ forms a current loop with $dI=Mdz$ and magnetic dipole moment $d\mathbf{m}=\pi R^{2}Mdz\hat{\mathbf{z}}$. Its integral over $L$ gives the same $M\pi R^{2}L\hat{\mathbf{z}}$. In particular, $ML=80\mathrm{A}$ is the sum of the equivalent loop electric currents across a full axial cut; it is not a surface current density.

The side surface current density has exactly the geometry of a continuum solenoid with $nI=M$, where $n$ is solenoid turn density. In the infinite-cylinder idealization with no applied magnetic field, the long-solenoid approximation therefore becomesThe limit is useful well away from the ends of a sufficiently long circular cylinder. A finite circular cylinder instead has fringing magnetic field and a demagnetizing field. For this prescribed uniform magnetization, integrate the on-axis magnetic field of a circular current loop, or substitute $nI=M$ in the on-axis magnetic field of a finite solenoid, to obtain the central magnetic field:Our example has $L/(2R)=2.5$ and $B_z(0)/({\mu }_{0}M)=5/\sqrt{29}\approx 0.92848$, rather than unity. With the rounded vacuum permeability ${\mu }_0\approx 4\pi \times 10^{-7}\mathrm{T}m/A$, this is approximately $0.9334\mathrm{mT}$, compared with the infinite-cylinder value $1.0053\mathrm{mT}$. These are model calculations for a prescribed magnetization, not measured values or a solution for the self-consistent magnetization of an arbitrary material.

Use vorticity to distinguish the local spin of a fluid element from the motion of its center along a pathline. Circular streamlines can occur both with and without vorticity: the two velocity fields below even have the same circulation around the unit circle.

For a continuously differentiable velocity field $\mathbf{u}=(u_x,u_y,u_z)$, vorticity is the curl vectorEach partial derivative divides velocity by distance, so vorticity has units ${\mathrm{s}}^{-1}$. In a planar velocity field independent of $z$, only ${\omega }_z={\partial }_x{u}_y-{\partial }_y{u}_x$ survives. Positive ${\omega }_z$ points along $+{\mathbf{e}}_z$: the right-hand rule associates it with counterclockwise spin viewed from positive $z$ toward the plane. ^[ref]

The local rotational angular velocity is half the vorticity, ${\mathbf{\Omega }}_{\mathrm{spin}}=\mathbf{\omega }/2$. To see the factor, split the velocity gradient $L_{ij}={\partial }_j{u}_i$ into its symmetric rate-of-strain tensor $E=(L+L^T)/2$ and antisymmetric spin tensor $W=(L-L^T)/2$. A nearby displacement $\delta \mathbf{x}$ has relative velocity $L\delta \mathbf{x}$; its rotational part satisfies $W\delta \mathbf{x}=(\mathbf{\omega }/2)\times \delta \mathbf{x}$. This is a proof outline of the local spin interpretation. The rate-of-strain tensor can also change the angle of a particular material line, so that line's angular velocity need not equal $\mathbf{\omega }/2$.

For solid-body rotation, $\mathbf{u}=(-\Omega y,\Omega x,0)$ gives $\mathbf{\omega }=(0,0,2\Omega )$ and ${\mathbf{\Omega }}_{\mathrm{spin}}=(0,0,\Omega )$. Taking $\Omega =1{\mathrm{s}}^{-1}$, the vorticity is $2{\mathbf{e}}_z{\mathrm{s}}^{-1}$, although the angular velocity is only $1{\mathbf{e}}_z{\mathrm{s}}^{-1}$. For the potential line vortex, $\mathbf{u}=K{\mathbf{e}}_{\theta }/r$, the planar curl is $r^{-1}{\partial }_r(r{u}_{\theta })=r^{-1}{\partial }_{r}K=0$ wherever $r>0$. Its circular pathlines therefore coexist with zero local spin.

This distinction makes vorticity useful for recognizing shear flow as well as swirling motion; taking the curl of the Euler equations also removes a pressure gradient and leads to the vorticity equation. A historical landmark is Hermann von Helmholtz's 1858 paper on fluid vortex motion, an early study devoted to vortex dynamics. ^[ref]

Use a linear shear flow to see why straight pathlines can still carry local spin. Faster layers pull past slower ones, and a small symmetric paddle can turn clockwise while its center travels straight ahead.

For the planar simple shear flow $\mathbf{u}=(\gamma y,0,0)$ with constant signed shear rate $\gamma$, direct differentiation givesWith $\gamma =1{\mathrm{s}}^{-1}$, these are $-1{\mathrm{s}}^{-1}$ and $-0.5{\mathrm{s}}^{-1}$: both signs indicate clockwise orientation under the right-hand rule. Yet a particle initially at $(x_0,y_0)$ follows the straight pathline $x(t)=x_0+\gamma y_{0}t$, $y(t)=y_0$. The curl measures differences of velocity between neighboring particles, rather than bending of one particle's pathline.

The velocity gradient has $L_{xy}=\gamma$ as its only nonzero entry. Its spin tensor supplies clockwise rotation, while its rate-of-strain tensor supplies simultaneous deformation. For a material line making angle $\alpha$ to the horizontal, the equation $\dot{\delta \mathbf{x}}=L\delta \mathbf{x}$ gives $\dot{\alpha }=-\gamma {\sin }^2\alpha$. Initially horizontal and vertical material lines therefore have angular velocities $0$ and $-\gamma$; their mean is $-\gamma /2$. At $t=0.5\mathrm{s}$ with $\gamma =1{\mathrm{s}}^{-1}$, an initially vertical displacement becomes $(0.5,1)$ times its original length and has angle $\arctan 2\approx 63.435^{\circ }$, a clockwise change of $26.565^{\circ }$. The plotted cross deforms; an actual paddle's response also depends on its shape and interaction with the fluid. The exact local spin statement concerns the spin tensor, rather than every tracer's orientation.