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A current loop is a closed path carrying electric current. For a thin wire following a closed curve $C$, the volume integral defining its magnetic dipole moment becomes a line integral:The vector area $\mathbf{A}$ is $A\hat{\mathbf{n}}$ for a planar current loop. For a nonplanar current loop, use the vector area, rather than the scalar area of an arbitrarily chosen spanning surface. Reversing the electric current reverses $\mathbf{m}$ without changing the geometry. ^[ref].

For $N$ closely spaced identical turns carrying the same electric current, the magnetic dipole moments add: $\mathbf{m}=NIA\hat{\mathbf{n}}$. Each turn contributes its own vector area; unequal areas or differently oriented turns require a vector sum. As a calculated example, twelve circular turns of radius $R=3.0\mathrm{cm}$ carrying $I=0.40\mathrm{A}$ give $m=NI\pi R^2=1.36\times 10^{-2}\mathrm{A}{\mathrm{m}}^2$. This is an idealized example, not an experimental measurement.

A magnetic dipole moment is a vector that summarizes the strength and orientation of a localized magnetic source. For a planar current loop, it is the electric current times the vector area: $\mathbf{m}=IA\hat{\mathbf{n}}$. The right-hand rule fixes the unit normal vector $\hat{\mathbf{n}}$: curl the fingers of the right hand along the conventional current, and the thumb points along $\mathbf{m}$. This single vector predicts both the distant magnetic dipole field and the turning effect of an imposed magnetic field.^[ref]

The SI unit is ampere times square metre, $\mathrm{A}{\mathrm{m}}^2$, equivalently joule per tesla. A magnetic dipole moment is neither a magnetic field nor a magnetic monopole: a closed current loop can have a nonzero magnetic dipole moment while its net outward magnetic flux through any closed surface remains zero, as required by Gauss's law for magnetism.

For a localized, steady electric current density $\mathbf{J}(\mathbf{r})$, the more general definition isHere the cross product is integrated over the complete source, including return paths. Steadiness requires $\nabla \cdot \mathbf{J}=0$, and localization requires sufficiently rapid decay for the boundary terms to vanish. Under these assumptions the localized steady-current moment identities give $\int \mathbf{J}{d}^{3}r=0$: shifting the origin by $\mathbf{a}$ changes the proposed magnetic dipole moment by $-\frac{1}{2}\mathbf{a}\times \int \mathbf{J}{d}^{3}r=0$. An isolated open wire segment does not satisfy these assumptions. ^[ref].

For $h,R>0$ and $0<\alpha \le 2\pi$, the curved sector $z=h(1-r^2/R^2)$, $0\le r\le R$, $0\le \theta \le \alpha$ has volume $\alpha h{R}^2/4$ beneath it. Its surface area isTo derive this, parametrize the surface by $(r\cos \theta ,r\sin \theta ,h(1-r^2/R^2))$. The upward vector area element is $(2h{r}^2\cos \theta /R^2,2h{r}^2\sin \theta /R^2,r)drd\theta$; its magnitude integrates to the stated surface area. Its integral gives the vector areaIn the shallow limit $h/R\to 0$, the surface area is $\alpha R^2/2+\alpha h^2/2+O(h^4/R^2)$, approaching the area of its planar sector projection.

Orient the boundary cyclically as $B_0,B_1,\dots ,B_{m-1},B_m=B_0$, and orient every fan triangle consistently as $(P,B_i,B_{i+1})$. Its vector area isChoose the sign convention in which a triangle with positive normal component along the unit vector $V$ has positive projected area. Then the total signed, or algebraic, projected area isThis is a signed sum, so overlaps and oppositely oriented pieces cancel. Expand the cross product:The sum of $B_i-B_{i+1}$ is zero around the closed polygon. ConsequentlyThis vector area of a polygonal boundary depends only on the oriented boundary, not on $P$, and does not require that the polygon be planar. Translating the boundary adds telescoping terms too, so the formula is independent of the choice of spatial origin. It is the polygonal instance of the vector area boundary integral $\frac{1}{2}\oint B\times dB$.

For $\mathbf{A}\ne 0$, the Cauchy-Schwarz inequality gives $A(V)\le |\mathbf{A}|$ because $|V|=1$. HenceThe opposite direction gives the minimum $-|\mathbf{A}|$; either direction maximizes the absolute algebraic area. Zero algebraic area occurs exactly for unit directions satisfying $V\cdot \mathbf{A}=0$, a great circle of directions perpendicular to the area vector. If $\mathbf{A}=0$, every direction gives zero and every direction is a maximizer, even though the sum of unsigned triangle areas may be positive. Reversing the initial orientation reverses every signed result consistently.

Choose an origin near the bounded loop and write ${\mathbf{r}}^{\prime }$ for the source position. The magnetic vector potential and its far-field multipole expansion arewith the loop held fixed as $r=|\mathbf{r}|\to \infty$. The first term integrates to zero since $C$ is closed. Define its oriented vector area byThe second equality is Stokes theorem, applied componentwise to any oriented spanning surface. In particular the vector area is independent of the choice of spanning surface. From ${\oint }_{C}d(r_i^{\prime }{r}_j^{\prime })=0$, the matrix ${\oint }_C{r}_i^{\prime }d{r}_j^{\prime }$ is antisymmetric, which gives ${\oint }_C(\mathbf{r}\cdot {\mathbf{r}}^{\prime })d{\mathbf{r}}^{\prime }=\mathbf{S}\times \mathbf{r}$. HenceThe orientation follows the current by the right-hand rule. For $r\ne 0$, taking the curl of this leading magnetic vector potential gives the magnetic dipole fieldFor a planar loop, $\mathbf{S}$ is the signed area times the unit normal vector. For a nonplanar loop it is the oriented vector area, rather than the scalar area of an arbitrarily chosen surface.

For the triangular face, take the cross product of two edges and divide by two:The origin lies on the opposite side from this normal, since $(1,1,2)\cdot \overrightarrow{PO}=-3$. Thus this is already the outward vector area, withThe vector areas of a closed polyhedron sum to zero, so the sum for the remaining three faces is

Their individual outward vector areas are $(0,0,-3/2)$, $(1/2,-1,1)$, and $(-1,1/2,-1/2)$, which directly check that sum.

Use the same axes as in (a) and choose the outward orientation. On the curved spherical lune, $\mathbf{n}={\mathbf{e}}_r$ and $dA=a^2\sin \theta d\theta d\phi$. Its vector area isThe $y$ and $z$ components cancel. The $x$ component is $a^2(\pi /2)2\sin (\beta /2)$, henceHere ${\mathbf{e}}_x$ points along the azimuthal bisector; another coordinate choice rotates this vector accordingly. Its magnitude is the signed projected area, not the scalar curved area $2\beta a^2$. Reversing the orientation reverses its sign.

The upward orientation of ${\mathbf{X}}_r\times {\mathbf{X}}_{\theta }$ corresponds to traversing the boundary from the apex to the starting rim, around the rim in increasing azimuth, and back to the apex. This is the indicated $ABCA$ orientation and is outward from the solid. Integrate that vector area element directly:The radial integrals are $2aR/3$ in the horizontal components and $b^2/2$ in the vertical component. Hence the vector area isThe absolute value is needed only because the original statement permits negative $b$. Reversing the boundary orientation reverses the entire vector. For a full rotation the horizontal components cancel, leaving $\pi b^2\mathbf{k}$, the signed projected area of the base.

For the outward orientation of a spherical cap, its vector area element isIn the surface integral defining the vector area, the $x$ and $y$ components vanish by integration over a complete azimuth. The remaining component isConsequentlyThis also equals the area of the cap's projection onto the horizontal plane, directed upwards.

The direction must first be represented by a unit vector: $\hat{\mathbf{q}}=(1,1,1)/\sqrt{3}$. The scalar projection of the vector area is thereforeIf a vector projection is intended, multiply this scalar by $\hat{\mathbf{q}}$:

The boundary order fixes the orientation. Two edges issuing from $A$ are $\overrightarrow{AB}=(0,0,1)$ and $\overrightarrow{AD}=(1,1,0)$. Their cross product gives the oriented vector area:Its norm is $\sqrt{2}$, the ordinary rectangle area. The sign agrees with traversing $A\to B\to C\to D\to A$ by the right-hand orientation rule.

The traversal fixes the orientation of the vector area as half the cross product of the edges $AB$ and $AC$:Its Euclidean norm $1/\sqrt{2}$ is the area of the triangle, and its direction follows the right-hand rule.

The prescribed orientation orders the two edges from $A$ as $\mathbf{c}-\mathbf{a}$ followed by $\mathbf{b}-\mathbf{a}$. The vector area is half their cross product:The sign agrees with the right-hand rule for the traversal $A\to C\to B\to A$.

For a constant vector field, the flux integral is its dot product with the vector area. Thus

For a rigidly rotating prescribed vacuum path, use spatial coordinates attached to the apparatus and laboratory time $t$. At a path point $\mathbf{r}$, the local apparatus velocity is $\mathbf{v}=\mathbf{\Omega }\times \mathbf{r}$. A light displacement has instantaneous laboratory displacement $d\mathbf{r}+\mathbf{v}dt$, up to the common rotation of the spatial axes. The null condition is thereforeWrite $d\ell =|d\mathbf{r}|$. To first order in $|\mathbf{v}|/c$, the positive root isReversing the path reverses the dot product term but leaves the leading path-length term unchanged. Subtract the two return times and use the scalar triple product identity:This is a line integral version of the Sagnac area law; the vector area identity also follows from Stokes theorem. For a mirror-defined polygon the actual ray paths move slightly as the mirrors rotate; the stationary optical path at zero rotation gives the same first-order result. The prescribed-path calculation is an idealization, not an exact finite-speed formula for every moving-mirror layout.

The dot product explains the geometry: only the projection of the oriented vector area along the angular velocity contributes. A planar loop whose normal vector is perpendicular to $\mathbf{\Omega }$ has zero leading delay. For repeated traversals with the same orientation, the signed vector areas add; oppositely oriented lobes can cancel even when the total unsigned area is large. Equal perimeter alone does not imply equal response. A constant translational velocity contributes nothing to this leading closed-path line integral, since ${\oint }_{C}d\mathbf{r}=0$.

The Sagnac effect is a difference in return times, or an associated interference fringe displacement, for light sent in opposite directions around a closed optical path in a rotating apparatus. The light leaves one beam splitter, travels around the same loop in opposite senses, and returns to the rotating beam splitter and readout. For a slowly rotating apparatus in flat spacetime, the leading laboratory arrival-time difference isHere $c$ is the vacuum speed of light, $\mathbf{\Omega }$ is the apparatus's angular velocity relative to an inertial frame, and $\mathbf{A}$ is the vector area oriented by the chosen positive direction around the closed curve $C$. With positive rotation about the upward normal vector, the counterclockwise beam is labelled $+$ and arrives later. Reversing rotation, or reversing the orientation convention alone, changes the sign of $\Delta t$.

The surprise is that a small loop can reveal its rotation by comparing two beams locally, even though each local vacuum speed of light measurement gives $c$. A rotating reference frame is not a single inertial frame: a beam must catch the moving receiver after a full circuit. The Sagnac area law makes this difference a practical way to measure angular velocity, while the Relativistic Sagnac delay on a circular ring distinguishes laboratory coordinate time from the receiver's proper time.

A Sagnac interferometer uses a beam splitter to send light with optical coherence around a common closed optical path in opposite directions, then recombines the beams to read their phase difference. Mirrors or a guiding path define the loop; the source, beam splitter, loop and readout rotate together in the basic experiment. A small relative beam tilt makes spatial interference fringes whose displacement reveals a change of phase difference. At one fixed output, a photodetector can instead record the corresponding change in light intensity.

A spherical wedge is the portion of a solid sphere between two meridional half-planes meeting at a dihedral angle $\beta$, with $0<\beta <2\pi$. Put the common axis along $z$ and its azimuthal bisector along $x$. For radius $a$, use spherical coordinates $0\le r\le a$, $0\le \theta \le \pi$, $-\beta /2\le \phi \le \beta /2$. Its volume is $2\beta a^3/3$.

For constant mass density, reflection symmetries put the centre of mass on the bisector. The first moment isDividing by volume gives $\bar{x}=3\pi a\sin (\beta /2)/(8\beta )$. It decreases from the limiting value $3\pi a/16$ for an infinitesimal wedge to zero for a full sphere.

The curved boundary is a spherical lune. Its outward vector area is $\pi a^2\sin (\beta /2){\mathbf{e}}_x$. Indeed, its $x$ component has the same angular integrals as the first moment, multiplied by $a^2$; the other components cancel by symmetry. Its scalar surface area is $2\beta a^2$, which is generally different from the magnitude of its vector area.

For an oriented closed polygon, triangulate with a common vertex $P$. Its total vector area is $\frac{1}{2}{\sum }_i(B_i-P)\times (B_{i+1}-P)$; the terms involving $P$ telescope, giving the displayed boundary-only formula. Its signed projected area along a unit vector $V$ is $V\cdot \mathbf{A}$. The maximum is $|\mathbf{A}|$, in direction $\mathbf{A}/|\mathbf{A}|$ if nonzero, and zero projections have directions perpendicular to $\mathbf{A}$. If $\mathbf{A}=0$, every direction has zero signed projection despite possible nonzero unsigned surface area.

For the faces of a closed polyhedron, take outward unit normal vectors. Applying the divergence theorem to an arbitrary constant vector field $c$ givesSince $c$ is arbitrary, the vector areas sum to zero. This is an oriented statement; the positive scalar surface areas do not cancel. The same proof works for every piecewise smooth closed boundary to which the divergence theorem applies.