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For a long cylindrical wire carrying axial current $I$, Ampère's law gives $B_s={\mu }_{0}I/(2\pi a)$ at its surface. With $a\gg \lambda$, the current resides mainly in a shell of thickness $\lambda$. If “critical” denotes the first equilibrium entry of vortices in an otherwise Meissner wire, set $B_s=B_{c1}$:The dependence on $a$ is necessary because no numerical radius is supplied. The local surface current density at this entry scale is different:Its conversion to the cross-section average is $\overline{J}\simeq (2\lambda /a)j_s$. Treating the shell current density as uniform over the entire thick wire would overestimate its total current by $a/(2\lambda )$.

Meaning of a maximum: in a type-II material, first vortex entry need not be the measured dissipative transport limit. Pinned vortices can sustain greater current, whereas freely moving vortices generate voltage. The hysteresis loop indicates pinning, but no geometry-dependent pinning force or transport criterion is given, so a unique practical maximum cannot be deduced solely from $\lambda ,\xi$ and $a$.

For comparison, the intrinsic homogeneous Ginzburg–Landau depairing scale follows from $j(q)={\Phi }_{0}q(1-{\xi }^2{q}^2)/(2\pi {\mu }_0{\lambda }^2)$. Maximizing at $q=1/(\sqrt{3}\xi )$ givesThis is a local, approximately uniform-flow theoretical scale, not the average current density throughout a thick self-field wire. Surface barriers, nonlinear screening, vortex motion and temperature determine how that scale relates to its actual critical current. Thus the entry estimate, local shell density and intrinsic depairing limit answer distinct physical criteria; the source's word “maximum” does not supply the missing pinning information.

In a contractible region, the Maxwell equations $\nabla \cdot \mathbf{B}=0$ and the Poincaré lemma imply the existence of a magnetic vector potential $\mathbf{A}$ such that $\mathbf{B}=\nabla \times \mathbf{A}$. Faraday's law of induction then givesThe expression in parentheses is a gradient there, so a scalar electric potential $\varphi$ can be chosen withThese statements are local, or global when the domain has the required topology and the electromagnetic fields are regular. A nonzero flux through a topologically nontrivial closed surface can obstruct a single globally defined vector potential, even though local expressions remain valid. The electric potential and magnetic vector potential are not unique: $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ and $\varphi \mapsto \varphi -{\partial }_t\chi$ leave both electromagnetic fields unchanged.

For the Lorenz gauge condition $\nabla \cdot \mathbf{A}=-c^{-2}{\partial }_t\varphi$, Gauss's law becomesThe Ampere-Maxwell law givesCancel the two gradient terms using the Lorenz gauge condition. HenceBoth potential equations have the same d'Alembert operator, with their respective charge density and current density sources. In a source-free region these are homogeneous wave equations with propagation speed $c$.

Let ${\rho }_q=3q/(4\pi a^3)$ be the uniform charge density and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. Rigid rotation produces current density $\mathbf{J}={\rho }_q\mathbf{\omega }\times \mathbf{r}$. The magnetic dipole moment isSpherical symmetry gives $\int r_i{r}_j{d}^{3}r={\delta }_{ij}\int r^2{d}^{3}r/3$, while $\int r^2{d}^{3}r=4\pi a^5/5$. ThereforeIt is parallel to the angular velocity for positive $q$ and antiparallel for negative $q$. The fully signed magnitude statement is $|{\mathbf{m}}_s|=|q||\omega |a^2/5$; the printed magnitude assumes positive $q$ and positive angular speed.

Axisymmetry permits a purely azimuthal magnetic vector potential, $\mathbf{A}=f(r)\sin \theta \hat{\mathbf{\phi }}$, with $A_r=A_{\theta }=0$. For the source's associated Legendre function convention, $P_1^1(\cos \theta )=+\sin \theta$, since its definition omits the Condon–Shortley phase. The given differential operator reduces toInside the sphere, $J_{\phi }={\rho }_q\omega r\sin \theta$. Since the radial operator maps $r^3$ to $10r$, the particular coefficient is ${\alpha }_0=-{\mu }_0{\rho }_q\omega /10$. Regularity at the origin and decay at infinity giveThere is no surface sheet electric current, so both normal and tangential magnetic field components are continuous at $r=a$. Usingthis is equivalent to continuity of $f$ and $f^{\prime }$. The two equations give $A={\mu }_0{\rho }_q\omega a^2/6$ and $C={\mu }_0{\rho }_q\omega a^5/15$. ThusIn the homogeneous expansion, ${\alpha }_1=A$, ${\beta }_1=C$ in their respective interior and exterior regions; all higher angular orders vanish. Indeed, matching the regular $r^{\ell }$ and decaying $r^{-\ell -1}$ terms with zero source at $\ell >1$ forces both coefficients to zero. The outside magnetic field is exactly the magnetic dipole field of ${\mathbf{m}}_s$, not merely its leading far-field approximation. For a positive moment along $z$, $A_{\phi }={\mu }_0{m}_s\sin \theta /(4\pi r^2)$, which is the printed cross-product magnitude expression.

For the flux coupling of a rotating charged sphere to a circular loop, take a loop of radius $b>a$ in the equatorial plane and use Stokes theorem on the regular full vector potential, evaluated around the rim. Its linked magnetic flux isWith $m_s=\pi a^2{I}_0$, the mutual inductance defined using this effective electric current isIntegrating an ideal magnetic dipole field through the entire disk would incorrectly use its singular expression inside the actual sphere; the rim calculation avoids that error.

For the prescribed time dependence, the quasistatic Faraday's law of induction result and the purely resistive loop approximation givePositive loop electric current is defined to produce a positive $z$ magnetic dipole moment, so the minus sign is Lenz's law. This assumes negligible retardation and $\nu L_b\ll R$, where $L_b$ is the loop's self-inductance. If that is not negligible, the governing equation is instead $L_b{\dot{I}}_b+R{I}_b=-{\dot{\Phi }}_b$; $R$ alone is insufficient for the exact dynamic response.

Use the constitutive relations $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{H}=\mathbf{B}/\mu$, where $\mathbf{D}$ is the electric displacement field, $\mathbf{H}$ the magnetic field strength, $ϵ$ the permittivity, and $\mu$ the magnetic permeability. Let ${\rho }_f$ be the charge density of free charge and ${\mathbf{J}}_f$ its current density. In this homogeneous medium, Gauss's law and the magnetostatic Ampère law are, respectively,Here $W$ is a volume with closed boundary $\partial W$, and $S$ is an oriented spanning surface whose boundary orientation follows the right-hand rule. Thus the electric field form is $\nabla \cdot \mathbf{E}={\rho }_f/ϵ$ and ${\oint }_{\partial W}\mathbf{E}\cdot \mathbf{n}dS=Q_f/ϵ$; the magnetic field form is $\nabla \times \mathbf{B}=\mu {\mathbf{J}}_f$ and ${\oint }_{\partial S}\mathbf{B}\cdot d\mathbf{\ell }=\mu I_f$.

The remaining static Maxwell equations areThese express Gauss's law for magnetism and the static limit of Faraday's law. The volume and circulation forms follow from the divergence theorem and Stokes theorem. The electric charge and electric current on the right of the macroscopic laws are free sources; material electric polarization and magnetization are already included in $\mathbf{D}$ and $\mathbf{H}$.

Orient the unit normal $\mathbf{n}$ from $R_1$ to $R_2$ and write $\mathbf{H}$ for the magnetic field strength. For static electric fields, $\mathbf{E}=-\nabla \varphi$; the static Ampère's law gives $\nabla \times \mathbf{H}={\mathbf{J}}_f$. The curl product rule therefore gives the local Poynting flux and terminal power identityIntegrate over an oriented piece ${\Sigma }_R$ of the separating surface and apply Stokes theorem:Once all the wires are included, the first surface integral is ${\sum }_k{\varphi }_k{I}_k$: the electric potential is constant on each perfect conductor and the integral of its normal current density is its signed electric current. Equivalently, remove the wire cross-sections and apply Stokes theorem in the surrounding insulator. Each resulting hole has negative boundary orientation, and the magnetostatic Ampère law around it gives the positive contribution ${\varphi }_k{I}_k$.

For a closed separating surface the boundary term is absent. For a noncompact separating surface, take $R\to \infty$ with the usual isolated-circuit condition ${\oint }_{\partial {\Sigma }_R}\varphi \mathbf{H}\cdot d\mathbf{\ell }\to 0$. In either case,For a steady closed circuit the signed electric currents sum to zero, so shifting the reference of electric potential does not change the terminal power.

For a fully general noncompact surface, the hypotheses as printed need an additional condition at infinity. The exact result before imposing it isTime independence alone does not remove this boundary term or ensure a finite Poynting flux; for example, spatially constant crossed electric fields and magnetic fields carry a constant Poynting vector through an infinite plane. The stated terminal-power formula holds with the boundary condition specified above.

With charge density $\rho$, current density $\mathbf{J}$, and speed of light $c=({ϵ}_0{\mu }_0{)}^{-1/2}$, the four Maxwell equations in vacuum areThe electric potential and magnetic vector potential determine the electric field and magnetic field byThese definitions satisfy Gauss's law for magnetism and Faraday's law identically. The Lorenz gauge condition isSubstituting into Gauss's law and using the Lorenz gauge condition givesFor the Ampère-Maxwell equation, the double-curl identity givesThe two gradient terms cancel by the Lorenz gauge condition. Defining the d'Alembert operator with the sign convention $\square ={\nabla }^2-c^{-2}{\partial }_t^2$, the remaining equations areConversely, if the electric potential and magnetic vector potential satisfy these wave equations and the Lorenz gauge condition, the same substitutions in reverse give Gauss's law and the Ampère-Maxwell equation. The other two Maxwell equations follow identically from the definitions, proving equivalence within the Lorenz gauge.

The moving point charge has distributional charge density and current densityLetThe given retarded fundamental solution identity immediately implies that the retarded potentialssolve $\square \varphi =-\rho /{ϵ}_0$ and $\square \mathbf{A}=-{\mu }_0\mathbf{J}$ in the sense of distributions. Integrating the spatial Dirac delta function first givesand the analogous integral for $\mathbf{A}$ has numerator ${\mu }_{0}q{\mathbf{v}}_0(t^{\prime })/(4\pi )$.

Assume the physical trajectory has subluminal speed and the retarded time exists. At its root $t_r$, with ${\mathbf{n}}_r=\mathbf{R}(t_r)/R(t_r)$,The positive factor $\kappa =1-{\mathbf{n}}_r\cdot {\mathbf{v}}_0/c$ also proves uniqueness of retarded time for a subluminal source. The Dirac delta function substitution formula gives $\delta (g(t^{\prime }))=\delta (t^{\prime }-t_r)/\kappa$, soThese are the Lienard-Wiechert potentials. The denominator includes the time-substitution factor, rather than just the retarded distance. Away from the trajectory the electric potential and magnetic vector potential are ordinary smooth functions; on the trajectory their wave equations have the indicated distributional sources.

For completeness, the charge continuity equation ${\partial }_{t^{\prime }}\rho +{\nabla }^{\prime }\cdot \mathbf{J}=0$ also verifies the Lorenz gauge. Using $\nabla K=-{\nabla }^{\prime }K$ and ${\partial }_{t}K=-{\partial }_{t^{\prime }}K$, integration by parts in the retarded integrals giveswith the retarded boundary prescription. Thus the construction also has the required Lorenz gauge condition.

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.

The identity divergence of a curl is zero follows by the symmetry of second partial derivatives and antisymmetry of the Levi-Civita symbol. ThereforeIf $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{A}$ is in Coulomb gauge, the curl of the curl identity gives $\mathbf{J}=-{\nabla }^2\mathbf{A}$. Apply the supplied scalar Poisson equation representation to each component of the decaying vector potential. In the units of this problem the Coulomb-gauge vector potential of a localized steady current isHere one needs $\mathbf{A}=O(|\mathbf{x}{|}^{-1})$, $\nabla \mathbf{A}=O(|\mathbf{x}{|}^{-2})$ and enough localization of $\mathbf{J}$ for the stated representation and differentiations. A smooth compactly supported current density is a sufficient convenient case. The local differential equations and gauge alone do not impose these conditions: $\mathbf{A}=(-y,x,0)$, $\mathbf{B}=(0,0,2)$, $\mathbf{J}=0$ satisfy them but do not satisfy the displayed integral formula. The formula selects the decaying solution, excluding nondecaying homogeneous harmonic functions.

With $\mathbf{r}=\mathbf{x}-\mathbf{y}$, the gradient of the kernel is ${\nabla }_x(1/|\mathbf{r}|)=-\mathbf{r}/|\mathbf{r}{|}^3$. Since $\mathbf{J}(\mathbf{y})$ is independent of $\mathbf{x}$,Thus taking the curl gives the Biot-Savart law in the problem's normalization:Finally, change variables $\mathbf{z}=\mathbf{x}-\mathbf{y}$ in the integral for $\mathbf{A}$. This keeps the singular kernel independent of the differentiation variable:For smooth localized $\mathbf{J}$, differentiation under the integral sign is legitimate: $1/|\mathbf{z}|$ is locally integrable in three dimensions, and on a bounded neighbourhood of $\mathbf{x}$ the translated sources and their derivatives have common compact support. The integral indeed satisfies Coulomb gauge.

In SI units the Ampère-Maxwell equation is $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}+{\mu }_0{\epsilon }_0{\partial }_t\mathbf{E}$. Since $\mathbf{E}=0$, the displacement-current term vanishes and $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}$. Apply Stokes theorem to an oriented surface $S$ with positively oriented boundary $C$:Hence the integral form of Ampère's law isHere ${\mu }_0$ is vacuum magnetic permeability, $\mathbf{J}$ is current density, and positive enclosed electric current is defined by the orientation of $S$.

Use the fully developed flow branch of Hartmann flow. Translation invariance along the walls makes the velocity and induced magnetic field functions of $z$ alone. Incompressible flow and zero normal velocity at the walls give $u_z^{\prime }=0$ and hence $u_z=0$. The zero divergence of the magnetic field makes $B_z$ constant, equal to the imposed $B_0$. There is no forcing in $y$; the homogeneous $y$-components obey the same coupled viscous-resistive equations as the $x$-components, with zero boundary data. Multiplying them by $u_y$ and $b_y/({\mu }_0\rho )$, integrating by parts and adding givesThus $u_y=b_y=0$, establishing the asserted forms on this fully developed flow branch. This is a symmetry reduction of the steady channel model, rather than a claim that every possible flow in a channel is translation invariant.

The current density and Lorentz force density areThe advective acceleration vanishes because $\mathbf{u}\cdot \nabla =u(z){\partial }_x$ acts on fields independent of $x$. The $x$-component of the magnetohydrodynamic momentum equation therefore givesThe $z$-component must also balance: it requires ${\partial }_z(p+b^2/(2{\mu }_0))=0$. Hence a compatible pressure is $p(x,z)=p_{\ast }-\rho Gx-b(z{)}^2/(2{\mu }_0)$; the specified streamwise pressure gradient does not require the ordinary pressure to be uniform in $z$. This accounts for the magnetic pressure of the induced field.

The resistive induction equation with constant magnetic diffusivity gives ${\partial }_t\mathbf{B}=\nabla \times (\mathbf{u}\times \mathbf{B})+\eta {\nabla }^2\mathbf{B}$. Since $\mathbf{u}\times \mathbf{B}=(0,-B_{0}u,0)$, its curl has $x$-component $B_0{u}^{\prime }$. Steadiness consequently givesFinally the no-slip boundary condition gives $u(\pm L)=0$, and the normal magnetic field boundary condition gives $b(\pm L)=0$. The coupled equations and all wall conditions follow directly from momentum balance and magnetic induction.

The free-space Green function of the Laplacian for the Poisson equation gives the electric potentialFor $r=|\mathbf{x}|\gg R$, its electric multipole expansion begins withDefine the total electric charge and electric dipole moment byThen

The analogous solution for the magnetic vector potential isIts apparent $r^{-1}$ coefficient is ${\mu }_0(4\pi {)}^{-1}\int \mathbf{J}{d}^3{x}^{\prime }$. For each component, the supplied identity and the divergence-free vector field condition giveThe divergence theorem turns this into a surface integral outside the compact support of the current density, where $\mathbf{J}=0$. Hence $\int \mathbf{J}{d}^3{x}^{\prime }=0$, so the $1/r$ term vanishes.

The magnetic field is $B=\nabla \times A$. The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$, so the curl of the curl identity givesA gauge transformation $A\mapsto A+\nabla \chi$ does not change $B$. We may therefore choose the Coulomb gauge $\nabla \cdot A=0$ by solving the Poisson equation ${\nabla }^2\chi =-\nabla \cdot A$. It follows component by component in a Cartesian coordinate system thatThe free-space Green function of the Laplacian then givesFor a localized steady current density, conservation of electric charge gives ${\nabla }^{\prime }\cdot J=0$; integration by parts confirms directly that this integral has zero divergence and hence obeys the Coulomb gauge.

For the aligned magnetic dipole field,Since ${\mathbf{e}}_z=\cos \theta {\mathbf{e}}_r-\sin \theta {\mathbf{e}}_{\theta }$,The corotation velocity is $\mathbf{u}=\Omega r\sin \theta {\mathbf{e}}_{\varphi }$. The current density is therefore the advected charge density,

For a point charge, the charge density and current density areInsert these into the retarded electromagnetic potential. The argument of the Dirac delta function depends on $x^{\prime }$ throughAt its unique zero $x^{\prime }=y(t_{\mathrm{ret}})$, the Jacobian determinant contributes the factorSince the source is subluminal, this quantity is positive. Writing $\mathbf{R}=x-y(t_{\mathrm{ret}})$, $R=|\mathbf{R}|$, and evaluating $\mathbf{v}$ at $t_{\mathrm{ret}}$ gives the Lienard-Wiechert potentialsThe identity ${\mu }_0{ϵ}_0{c}^2=1$ equivalently gives $\mathbf{A}=\mathbf{v}\varphi /c^2$.

In a magnetostatic equilibrium, inertia is absent and all forces balance. A force-free magnetic field is the limiting case in which the Lorentz force density vanishes:Thus the current density is parallel to the magnetic field, so for some scalar force-free parameter $\alpha (\mathbf{r})$,Using Ampère's law in magnetostatics givesTaking the divergence and using both the divergence of a curl is zero and $\nabla \cdot \mathbf{B}=0$ yieldssoThe force-free parameter is therefore constant along every magnetic field line.

First hold the production rate fixed, so total electric current remains $200\mathrm{A}$. If only the impure anode receives a factor $10^5$ increase in accessible area, its local current density becomes $0.002\mathrm{A}{\mathrm{m}}^{-2}$, far below $j_0=1\mathrm{A}{\mathrm{m}}^{-2}$. Extrapolating the large-current Tafel equation would give a negative anodic overpotential, which is physically the wrong regime. Near equilibrium the Butler-Volmer equation instead gives, in the same conditional two-electron model,The anode activation loss nearly disappears; the 1.20 V ohmic loss does not disappear. The cathode still operates close to its original transport limit if it is unchanged. The conditional kinetic model gives $V_{\mathrm{app}}\simeq 1.3286$ V for anode-only enlargement. Its anodic potential returns close to 0.34 V, making retention of noble silver and gold more favorable, while less noble nickel dissolution and lead sulfate formation remain relevant.

If both active areas increase by $10^5$ and their independent mass-transfer capacity scales by the same factor, both local kinetic losses become small and the conditional prediction is about $1.20005$ V. Porous internal area does not generally multiply an external diffusion-limited flux by $10^5$: pore access, ion depletion, tortuosity, electrical connectivity and slime blockage must be checked. If the specified 200 A/m² is instead held per true enlarged area, total current rises to $2\times 10^7$ A and the unchanged electrolyte resistance would give $1.2\times 10^5$ V, so this is an entirely different operating condition.

For the same refining rate and the same effective kinetic model, the anode-only estimate is therefore $1.3286\mathrm{V}$. The robust conclusion at fixed total current is reduced local activation polarization on the genuinely accessible enlarged surface, with an unchanged ohmic floor unless cell geometry or conductivity also changes. Increased geometric area alone does not prove a proportionate decrease in energy cost or improved purity. The distinct area and current conventions above specify how to interpret the prediction; they must not be mixed in one calculation.

The reversible voltage follows from the two-electron hydrogen reaction:At the specified geometric current density, the magnitude of each activation loss follows its Tafel equation, $\eta =b{\log }_{10}(j/j_0)$:Both losses reduce the delivered fuel cell voltage. Therefore $V=0.36048\mathrm{V}$ and $P/A=jV=3.60\mathrm{W}{\mathrm{m}}^{-2}$. The very small oxygen exchange current is responsible for most of the activation loss. The calculation uses current densities on a common area basis, unit water activity and the stated gas conditions, and neglects the losses the problem excludes.

At unchanged intrinsic catalyst kinetics, increasing the electrochemically accessible true area by $100$ multiplies exchange current per geometric area by $100$. Equivalently, the local current density at a fixed total geometric current falls by $100$. Each Tafel equation activation loss decreases by $b{\log }_{10}100=2b$:Thus $V_{\mathrm{new}}=0.59948\mathrm{V}$ and $P_{\mathrm{new}}/A=5.99\mathrm{W}{\mathrm{m}}^{-2}$, an increase of $2.39\mathrm{W}{\mathrm{m}}^{-2}$, or about $66.3\%$. Power does not increase a hundredfold, because the voltage improvement is logarithmic and the specified geometric current is unchanged. This assumes both electrodes gain accessible active area, the original exchange currents use the original area basis, and the porous structure does not introduce additional transport or Ohmic losses.

Four advantages are strong electrostatic gate control, rapid charge transport, high current-carrying capability and mechanical flexibility.

A semiconducting single-walled carbon nanotube has a nanometre-scale channel cross-section, allowing efficient gate control in a field-effect transistor. A clean short tube can have a long mean free path and correspondingly little channel scattering. Strong covalent bonding supports large current densities for interconnect applications, while high axial thermal conductivity can help distribute heat. Small mass, large aspect ratio and mechanical compliance also permit conducting or semiconducting networks on flexible substrates. Metallic interconnects and low-leakage transistors require different electronic tube populations.

Manufacturing must control diameter, carbon nanotube chiral vector, purity, alignment and placement. Metallic contaminants short-circuit transistor networks; low network density reduces current, whereas excessive density makes isolation difficult. Tube–metal contact resistance, schottky barriers, tube–tube junctions, catalyst residues and substrate charge traps can dominate the intrinsic channel response. Thermal boundary resistance can limit cooling despite high axial thermal conductivity. Compare sorted and unsorted networks at matched density and contact geometry, measure contact and channel contributions separately, and assess uniformity across many devices. High intrinsic mobility alone is not evidence of a reproducible integrated circuit.