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Choose the upward orientation of the paraboloid cap. Its boundary is the circle $x^2+y^2=4$ at $z=1$, oriented counterclockwise as viewed from above. By Stokes theorem,On the boundary, $dz=0$ and $A\cdot dr=ydx-3xydy$. Parametrize by $x=2\cos \theta$, $y=2\sin \theta$, $0\le \theta \le 2\pi$. ThenThe second term integrates to zero over a full period, while the first contributes $-4\pi$. Equivalently, replace the cap by the upward disk at $z=1$: there $(\nabla \times A{)}_z=-3y-1$, whose integral is $-4\pi$ by symmetry.

An orientation is not stated in the original question. The displayed result uses the upward normal, also the outward normal on the curved top of the enclosed cap; the opposite orientation gives $4\pi$.

For a smooth oriented surface $S$ with positively oriented boundary $C=\partial S$, Stokes theorem statesHere $d\mathbf{S}=\mathbf{n}dA$ uses the chosen unit normal vector, while $d\mathbf{r}$ is tangent to the directed boundary. The boundary orientation follows the right-hand rule: curling the fingers along $C$ makes the thumb point in the normal direction. The field must be continuously differentiable on a neighbourhood of the surface, with the usual piecewise-smooth allowance for its boundary.

For the upper hemisphere with outward normal, the induced equatorial direction is counterclockwise as viewed from positive $z$.

First calculate the curl:By Stokes theorem, use the positively oriented equator $\mathbf{r}(\varphi )=(a\cos \varphi ,a\sin \varphi ,0)$, $0\le \varphi \le 2\pi$. Its tangent is $a(-\sin \varphi ,\cos \varphi ,0)$, while the restricted field is $(0,3a^2\cos \varphi \sin \varphi -a\cos \varphi ,0)$. ThusThe first term integrates to zero over a full period and the second uses ${\int }_0^{2\pi }{\cos }^2\varphi d\varphi =\pi$.

One can also calculate directly on the curved surface. With spherical coordinates $0\le \theta \le \pi /2$, $0\le \varphi \le 2\pi$,Substitution gives the scalar integrand $a^3{\sin }^2\theta \cos \theta \sin \varphi -a^2\sin \theta \cos \theta$. Its azimuthal sine term vanishes, and the rest gives $-2\pi a^2{\int }_0^{\pi /2}\sin \theta \cos \theta d\theta =-\pi a^2$, confirming the orientation and the same result.

For an oriented surface $S^{\prime }$ with compatible boundary direction, the Stokes theorem statesLet $H$ denote the removed patch and $S^{\prime }=S\setminus H$ the remaining cylinder. Keep the outward orientation inherited from the closed cylinder. Everywhere on the removed curved patch, the normal curl component is $-\alpha R$, so its flux is $-\alpha RA$, where $A$ is its actual curved area. Part (d) then givesConsequently the circulation around a hole in an oriented surface isThe positive direction here is induced by the remaining outward-oriented surface; locally it runs clockwise around the hole when viewed from outside. If the edge is instead traversed positively around the removed patch, the direction reverses and the answer is $-\alpha RA$. The question does not specify an edge direction, so the sign must be accompanied by this convention. The result depends only on the removed curved area, not on its shape. If the patch has several boundary components, use the sum of their induced line integrals. Since $[\alpha ]=1/(\text{length}\cdot \text{time})$, the result has the units of a velocity line integral, length squared per time.

The integrand field is exactly the curl computed in part (a), so use its polynomial vector potential $\mathbf{G}=(z^3,x^3,y^3)/3$. The upper cap meets the plane $z=0$ in an ellipse. Since the parameters are only assumed nonzero, set $a_0=|a|$ and $b_0=|b|$ to use positive semiaxes. Its correctly oriented boundary isOn that boundary, $G_x=0$ and $dz=0$, so only $G_{y}dy=x^{3}dy/3$ contributes. The Stokes theorem therefore givesThe last integral follows by expanding ${\cos }^{4}t=(3+4\cos 2t+\cos 4t)/8$. ThusThe result is independent of $c$. A sign check uses the divergence theorem: closing the cap with its base disk adds outward normal $-\hat{\mathbf{z}}$ there. Since $\nabla \cdot \mathbf{F}=0$, the cap flux equals ${\int }_{\mathrm{disk}}{x}^{2}dxdy>0$, consistent with the boxed expression and its absolute values.

For a smooth oriented surface $S$ with piecewise smooth boundary $C=\partial S$ and a continuously differentiable vector field $\mathbf{G}$, the Stokes theorem statesThe boundary orientation is fixed by the right-hand rule: curling the fingers along the positive boundary direction makes the thumb point along the surface normal. For an outward upper cap, its boundary ellipse is traversed counterclockwise when viewed from the positive $z$ direction. The figure shows two possible caps with the same oriented boundary, anticipating the last part.

The modified upper surface has exactly the same ellipse at $z=0$ and the same boundary orientation. The field retains the global polynomial vector potential from part (a). By the Stokes theorem, both surface fluxes equal the same boundary line integral; the cap's height and interior shape do not enter. HenceThis is an instance of the fact that solenoidal flux depends on the oriented boundary, with the domain condition automatically satisfied by the smooth polynomial field on all of ${\mathbb{R}}^3$. When $a,b$ are chosen positive, the formula simplifies to $\pi a^{3}b/4$; the absolute-value form respects the original allowance of arbitrary nonzero signs.

The upward unit vector extracts the vertical curl component:It is constant over the unit disc, soThis agrees with part (d) by Stokes theorem: the boundary circle has counterclockwise orientation when viewed from the upward normal, including for the parametrizations $2\pi t^n$.

For $r>0$, the chain rule gives $\partial r/\partial x_i=x_i/r$. Differentiating the radial factor gives $d\varphi /dr=e^{ikr}(ikr-1)/(4\pi r^2)$, so in Cartesian coordinatesThe origin is excluded because the scalar field is singular there.

The Stokes theorem states ${\int }_S(\nabla \times \mathbf{F})\cdot \mathbf{n}dS={\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{\ell }$ for a sufficiently smooth vector field and oriented surface with piecewise smooth boundary. The boundary orientation follows the right-hand rule relative to $\mathbf{n}$.

For the convective acceleration identity, use the Levi-Civita symbol contraction ${ϵ}_{ijk}{ϵ}_{k\ell m}={\delta }_{i\ell }{\delta }_{jm}-{\delta }_{im}{\delta }_{j\ell }$. It givesRearranging proves the required vector identity.

For orthogonal coordinates, define the scale factors of orthogonal coordinates $h_i=|\partial \mathbf{r}/\partial q_i|$ and tangent unit vectors ${\mathbf{e}}_i=h_i^{-1}\partial \mathbf{r}/\partial q_i$. They form an orthonormal basis, and $d\mathbf{r}={\sum }_i{h}_i{\mathbf{e}}_{i}d{q}_i$. Equating $d\psi ={\sum }_i{\partial }_{q_i}\psi d{q}_i$ with $\nabla \psi \cdot d\mathbf{r}$ identifies each component. Thus the gradient in orthogonal coordinates isThis holds on a regular coordinate chart where all scale factors are positive.

The outward area element is $d\mathbf{S}=a^2\sin \theta d\theta d\varphi {\mathbf{e}}_r$. ThereforeSplit the sphere into the northern cap $0\le \theta \le \alpha$ and the southern cap $\alpha \le \theta \le \pi$. Their induced boundary orientations are $+C$ and $-C$, respectively. The Stokes theorem applies to ${\mathbf{A}}^{(+)}$ on the north and ${\mathbf{A}}^{(-)}$ on the south. Adding their fluxes gives $2\pi (1-\cos \alpha )+2\pi (1+\cos \alpha )=4\pi$, agreeing with direct integration.

There is no contradiction with the flux of a globally smooth curl through a closed surface: neither potential is regular on the entire sphere. On their overlap they differ by $2{\mathbf{e}}_{\varphi }/(r\sin \theta )=2\nabla \varphi$, a local gauge transformation. The longitude is not a globally single-valued scalar, so that difference can have nonzero circulation around $C$.

Away from the axis, cylindrical coordinates put the field in the line vortex formThe only potentially nonzero component of its curl isThe other components vanish because there is no axial dependence and no radial or axial velocity. Thus the curl is zero for $\rho >0$; the field is singular on the axis, so a classical curl there is not defined.

For a spanning surface avoiding that axis, the smoothness hypotheses of Stokes theorem hold and give ${\oint }_C\mathbf{v}\cdot d\mathbf{r}=0$. If the enclosed planar region contains the axis, the singularity prevents applying that theorem directly. Excise a small disk about the origin. The resulting annulus has zero curl, so its outer circulation equals the counterclockwise circulation on the small circle:This sign assumes positive, counterclockwise orientation. Clockwise orientation gives $-\kappa$; more generally the value is $\kappa$ times the winding number about the origin. A curve through the origin is excluded because its integral is singular. In the distribution interpretation the missing vorticity is concentrated on the axis as $\kappa \delta (x)\delta (y){\mathbf{e}}_z$, explaining the nonzero circulation despite zero classical curl away from it.

Using Cartesian coordinates, the Levi-Civita symbol and the Einstein summation convention, the constant-vector assumption givesRecognizing the divergence and directional derivative provesHere the contractions use the Kronecker delta.

For an oriented piecewise smooth surface $S$ with boundary $C$, Stokes theorem stateswhere the orientations obey the right-hand rule and $\mathbf{Q}$ is continuously differentiable near $S$. Apply it to $\mathbf{Q}=\mathbf{F}\times \mathbf{a}$. The scalar triple product givesThe last integrand is $\mathbf{a}\cdot [(d\mathbf{S}\times \nabla )\times \mathbf{F}]$, as the component formulashows. Since $\mathbf{a}$ is arbitrary, the cross-product form of Stokes theorem follows:The derivatives in the surface expression act on $\mathbf{F}$, not on the surface element.

For the specified square, the path is clockwise when viewed from positive $z$, so the compatible unit normal is $-{\mathbf{e}}_z$. On the first and fourth edges, $d\mathbf{l}$ and $\mathbf{r}$ are parallel, giving zero cross product. On the second edge, $y=1$ and $x$ runs from $0$ to $1$, so $d\mathbf{l}\times \mathbf{r}={\mathbf{e}}_{z}dx$ contributes ${\mathbf{e}}_z$. On the third edge, $x=1$ and $y$ runs from $1$ to $0$, so the contribution is $-{\mathbf{e}}_z{\int }_1^{0}dy={\mathbf{e}}_z$. Thus the line integral is $2{\mathbf{e}}_z$.

For an independent surface integral, ${\partial }_i{F}_j={\delta }_{ij}$ and $\nabla \cdot \mathbf{F}=3$ for the position vector field. Its integrand is therefore $d\mathbf{S}-3d\mathbf{S}=-2d\mathbf{S}$. The square has unit area and downward orientation, giving

For an oriented piecewise smooth surface $S$ with boundary curve $C=\partial S$, and a continuously differentiable vector field $\mathbf{F}$ near $S$, Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal, $dS$ is positive surface area, and $d\mathbf{r}$ is the directed tangent line element. The positive traversal of $C$ is fixed by the right-hand rule relative to $\mathbf{n}$. Thus the flux of the curl through the surface equals the circulation along its oriented boundary.

The ellipse bounding the surface is $u=1$. With the upward normal, its positively oriented boundary is traversed by increasing $v$ from zero to $2\pi$:The given components are Cartesian components, so their ordinary dot product with this tangent isTherefore Stokes theorem givesA direct check uses $F_x=2y/a$ and $F_y=-2x/a$, hence $(\nabla \times \mathbf{F}{)}_z=-4/a$, and multiplies by the ellipse area $\pi a^2\sinh 1\cosh 1$. The third component cannot affect either calculation because the surface and its boundary lie at fixed $z$. Equivalently one may apply Stokes to the smooth planar field $(2y/a,-2x/a,0)$, avoiding any need to differentiate the elliptic-coordinate expression for the third component at its foci.

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.

Stokes theorem converts circulation around an oriented boundary into curl flux through a spanning surface. For a sufficiently smooth vector field on an oriented piecewise smooth surface $S$,The boundary orientation follows the right-hand rule relative to $\mathbf{n}$. The field must be smooth on the surface, not merely on its boundary; a singularity crossing the surface cannot be ignored. For a fixed loop, applying the theorem to Faraday's differential law gives the integral flux law of induction.

The planar form is Green theorem. For a region $D$ with positively oriented boundary and sufficiently smooth $P,Q$,The outer boundary is counterclockwise and any holes have clockwise boundaries. For instance $P=-y/2,Q=x/2$ makes the area of $D$ equal to half the line integral of $xdy-ydx$; the unit circle gives area $\pi$.

A closely related use of “Green's theorems” in field theory is Green's identities, obtained by applying the divergence theorem to $\varphi \nabla \psi$:and, after subtracting the expression with $\varphi ,\psi$ exchanged,These relate interior sources to boundary data in electrostatics and other boundary-value problems. For example, a harmonic potential difference vanishing on a Dirichlet boundary condition has zero integral of its squared gradient by the first identity, establishing uniqueness. Both the planar circulation theorem and the field identities are included here to make the convention clear.

The differential form of Faraday's law of induction isBy Stokes theorem, for a stationary oriented circuit it becomes $\mathcal{E}=\oint \mathbf{E}\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$, where ${\Phi }_B=\int \mathbf{B}\cdot d\mathbf{S}$ is magnetic flux and $\mathcal{E}$ is electromotive force. Circuit orientation and surface normal obey the right-hand rule. Lenz's law is the physical meaning of the minus sign: the induced current's magnetic field opposes the change in linked magnetic flux. For a moving circuit, the force per unit electric charge is $\mathbf{E}+\mathbf{v}\times \mathbf{B}$, and its line integral is the corresponding electromotive force. Motion through a static magnetic field can therefore induce a voltage even when ${\partial }_t\mathbf{B}=0$.

Let $\theta$ be the wire's angle from the downward vertical, $r=L\sin \theta$, and choose $\mathbf{\omega }$ and $\mathbf{B}$ upward, as in the figure. The motional term along the wire has radial component $B\omega s$ at cylindrical radius $s$. In steady electrical equilibrium, integrating from the grounded axis to the bob givesThe Lorentz force on this positive electric charge is radially outward. Its ratio to the required centripetal term is $QB/(m\omega )\lesssim C{B}^2{L}^2/(2m)$, so the stated heavy-bob condition makes its effect on the zero-field conical pendulum geometry negligible. With wire tension $T_w$, $T_w\cos \theta =mg$ and $T_w\sin \theta =m{\omega }^{2}r$, giving $\cos \theta =g/({\omega }^{2}L)$. Hence the magnetically charged conical pendulum hasA nonzero-radius orbit requires ${\omega }^{2}L>g$. Reversing the sense of rotation or the magnetic field reverses the sign of $Q_0$.

The response to the added electric field requires an electrical convention. In the usual mechanical approximation, treat the pre-existing electric charge as $Q_0$, include its upward electric force $Q_{0}E$, and neglect all magnetic-force feedback. ThenThis predicts an increased radius for $Q_{0}E>0$. It is the result when the newly applied field is used only as a probe force on the original electric charge.

There is a first-order qualification if the bob remains electrically connected to the grounded pivot and the uniform added field's potential is included consistently. Put $s=\cos \theta$ and set the external electric potential to zero at the pivot. The external potential at the bob is then $ELs$, so the charging relation and full force balance becomeThe second equation follows by eliminating the wire tension between vertical balance and radial balance, including the outward magnetic Lorentz force. Expanding both equations to first order in $E$, and to leading order in $C{B}^2{L}^2/m$, with $s_0=g/({\omega }^{2}L)$, givesThe extra factor arises from the added field changing $Q$ by $-CEL{s}_0$ and therefore changing the radial magnetic force. That term is linear in $E$ too; the heavy-bob assumption alone does not justify retaining $Q_{0}E$ while dropping it. The two expressions refer to the two electrical approximations just specified. The expansion requires a small change relative to the initial radius and is nonuniform as that radius tends to zero.

For an electrical measurement, route the grounded connection through a sensitive electrometer operated near zero input voltage, and integrate the charging electric current when $B$ or $\omega$ is changed. Starting with the field off, the transferred electric charge is $Q_0$ in this ideal capacitance model. Reversing $B$ separates the induced signal from offsets. Alternatively, isolate the charged bob briefly and measure its electric potential with a high-input-impedance electrometer, accounting for the instrument's extra capacitance and for the fact that isolation changes the electrical boundary condition.

The absence of magnetic monopoles is expressed by $\nabla \cdot \mathbf{B}=0$. In a regular simply connected region, a magnetic vector potential may be introduced so thatIts units are $\mathrm{T}\mathrm{m}=\mathrm{Wb}{\mathrm{m}}^{-1}$. It is not unique: adding a gradient leaves the magnetic field unchanged because $\nabla \times \nabla \chi =0$. In time-dependent electromagnetism,is a gauge transformation preserving both physical fields. Stokes theorem gives ${\oint }_C\mathbf{A}\cdot d\mathbf{\ell }={\int }_S\mathbf{B}\cdot d\mathbf{S}={\Phi }_B$ for a contour bounding an appropriate surface.

For localized steady currents, choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. Substitution into Ampère's law and the curl-of-curl identity givesBoundary conditions are needed in addition to the gauge choice. For a uniform magnetic field one convenient choice is $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$. For a magnetic dipole, $\mathbf{A}={\mu }_0\mathbf{m}\times \mathbf{r}/(4\pi r^3)$ outside its source. The magnetic vector potential simplifies field calculations and appears directly in the Lagrangian of a charged particle, $L=\frac{1}{2}m{v}^2+q\mathbf{v}\cdot \mathbf{A}-q\varphi$. In multiply connected regions, a field-free region can retain nontrivial circulation of $\mathbf{A}$, relevant to the Aharonov-Bohm effect; a local gauge argument cannot remove every global circulation.

Induction and its sign. The Maxwell-Faraday equation is $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$. For a fixed loop, Stokes theorem givesThis is Faraday's law of induction. The orientation of the surface normal and positive circulation are linked by the right-hand rule. The minus sign is Lenz's law: the induced current's linked flux opposes the imposed flux change, rather than reinforcing it.

The inductive current. Let the loop normal make angle $\theta$ with the applied field, with positive current defined by that normal. The external magnetic flux is ${\Phi }_{\mathrm{ext}}=A{B}_0\sin \omega t\cos \theta$. With negligible electrical resistance, the circuit equation is $L\dot{I}=-{\dot{\Phi }}_{\mathrm{ext}}$, so total linked flux $LI+{\Phi }_{\mathrm{ext}}$ is constant. For zero initial trapped flux,The printed result needs this initial-flux assumption; in general an additional constant current is allowed. The units are amperes because a weber divided by a henry is an ampere.

Current and forces at one instant. Take $B>0$ along $+x$ and $0<\theta <\pi /2$, so $I<0$ relative to the chosen normal. The loop magnetic dipole moment $\mathbf{\mu }=IA\mathbf{n}$ therefore has a component opposite the applied field. Each wire element experiences Lorentz force $d\mathbf{F}=Id\mathbf{\ell }\times \mathbf{B}$. Opposite sides have equal and opposite forces, giving zero net force in the uniform field but a nonzero couple.

The horizontal top and bottom wire segments also experience equal and opposite vertical forces. They cancel without adding torque about the vertical rotation axis; the displayed force pair supplies that torque.

Average torque. Since $\mathbf{\tau }=\mathbf{\mu }\times \mathbf{B}$, its generalized component for increasing $\theta$ is ${\tau }_{\theta }=-IAB\sin \theta$. Substitution givesand, using $⟨{\sin }^2\omega t⟩=1/2$,The orientation torque on a flux-conserving loop drives its normal away from the field for the stated initial angle. Its magnitude has units joules, or newton metres.

Free motion with large inertia. Write $J$ for the mechanical moment of inertia. If its orientation changes little during each field cycle, the slow equation is $J\ddot{\theta }=⟨{\tau }_{\theta }⟩$, with effective potentialThe stable orientation is $\theta =\pi /2$, where the loop plane contains the applied field. Released from rest between $0$ and $\pi /2$, the loop accelerates toward that orientation, overshoots and librates around it. In the leading averaged, undamped model its turning angles are ${\theta }_i$ and $\pi -{\theta }_i$. It does not settle without a dissipative mechanism. For small librations the squared angular frequency is $A^2{B}_0^2/(2LJ)$.

Negligible inductance and finite resistance. Now the current is proportional to the induced electromotive force, not to the flux. Including slow rotation,The rotational damping of a resistive loop in an oscillating field follows fromAt fixed orientation, the first term alternates at $2\omega$ and has zero mean. The loop can therefore rock slightly, but there is no leading mean aligning torque like that in the inductive case. Slow rotation giveswhich opposes motion and produces Joule heating. Initial-phase transients can initiate motion, and the instantaneous torque is not zero. This description retains the large-inertia, slow-motion approximation; it does not assert an exact stationary solution at every starting angle.

The integral field equation. The Ampere-Maxwell law is $\nabla \times \mathbf{H}=\mathbf{J}+{\partial }_t\mathbf{D}$. Apply Stokes theorem to any fixed oriented surface bounded by the chosen contour:The added displacement current is required for compatibility with electric charge conservation. Taking a divergence and using $\nabla \cdot \mathbf{D}={\rho }_f$ gives $\nabla \cdot \mathbf{J}+{\partial }_t{\rho }_f=0$.

Field geometry. Choose the upper plate positive and let $z$ increase upward. In the long-plate, slow-discharge approximation, $\mathbf{E}=-(V/d)\hat{\mathbf{z}}$. Conventional wire current runs downward from the positive plate. The magnetic field intensity circulates clockwise as viewed from above. The Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{H}$ points radially toward the resistive wire.

Wire and displacement currents. The capacitance and discharge law areBecause $V$ decreases, ${\partial }_t\mathbf{D}$ points upward, opposing the downward wire current. Through a disk of radius $r$ between the plates its magnitude isThe effective downward current enclosed by the circular contour is therefore $I-I_D(r)$. The Ampere-Maxwell law givesThis is the magnitude; with the stated coordinates, $\mathbf{H}=-H\hat{\mathbf{\varphi }}$. The factor $1-r^2/a^2$ expresses cancellation between conduction and displacement currents. The approximation requires distances outside the thin wire and away from detailed electrode-edge fields.

Energy flux and power. Since $\hat{\mathbf{z}}\times \hat{\mathbf{\varphi }}=-\hat{\mathbf{r}}$,Its units are watts per square metre. Integrating over the cylinder's side area $2\pi rd$, the signed outward power is negative:There is no leading axial flux through the cylinder's ends in this ideal field geometry.

Where the energy goes. The Poynting flux in a centrally discharged parallel-plate capacitor carries field energy through the dielectric space into the resistor, where it becomes Joule heating at rate $V^2/R$. Inside radius $r$ the stored electric energy isHenceThe resistor receives the inward flux from outside that cylinder plus the field energy released inside it. As $r$ approaches the wire radius, essentially all its power arrives as radial electromagnetic flux. Over the whole capacitor, $-d(C{V}^2/2)/dt=V^2/R$. The Poynting theorem thus accounts for the complete discharge without treating energy as transported only along the conducting wire. High resistance makes the quasistatic electric-field approximation and neglect of changing magnetic energy consistent.