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Electrical conductivity $\sigma$ measures the local response of current density to electric field. In the rest frame of an isotropic Ohmic conductor, the local form of Ohm's law isLarger positive electrical conductivity gives larger current density at the same electric field. It is a bulk material coefficient, distinct from the geometry-dependent electrical conductance of a whole component and from a sheet's surface conductivity. Electrical resistivity is ${\rho }_{\mathrm{el}}=1/\sigma$ in this scalar model. ^[ref].

The local and whole-wire forms fit together particularly neatly. For uniform current in a straight homogeneous wire, with length $L$, wire cross-sectional area $A$, constant electrical conductivity and terminal electric potential difference $V$, the axial electric field has magnitude $E=V/L$. ThusFor the defined electrical resistance example $\sigma =5\times 10^5(\Omega \mathrm{m}{)}^{-1}$, $L=1\mathrm{m}$, $A=10^{-6}{\mathrm{m}}^2$ and $V=5\mathrm{V}$, the local values are $E=5V/m$ and $j=2.5\times 10^{6}A/{\mathrm{m}}^2$; integrating gives $I=2.5\mathrm{A}$. One material law therefore supplies both a spatial description and a terminal relation.

The scalar form requires isotropy. In an anisotropic material the response can instead be a tensor, $J_i={\sum }_j{\sigma }_{ij}{E}_j$, and current density need not be parallel to electric field; uniaxial conductivity tensor is one existing example. These statements concern the material's rest frame. A moving electrical conductor instead uses the moving-conductor Ohm law.

Electrical resistance measures the electric potential difference needed to drive a given electric current through a component. For an Ohmic conductor under fixed material conditions, $R=V/I$ is constant and Ohm's law gives $V=IR$. The SI unit is the ohm, $1\Omega =1V/A$. At a fixed positive electric potential difference, increasing electrical resistance reduces electric current. Electrical conductance $G=1/R$ expresses the same response reciprocally. ^[ref].

For a straight uniform wire at rest, with constant electrical conductivity $\sigma >0$, length $L$ and wire cross-sectional area $A$, the uniform Ohmic conductor model givesThe electrical resistivity ${\rho }_{\mathrm{el}}$ characterizes the material; electrical resistance also includes geometry. Doubling length doubles $R$, while doubling wire cross-sectional area halves $R$, with material and temperature fixed. The uniform-field assumption includes insulated sides and equipotential end faces, as explained in uniform current in a straight homogeneous wire. ^[ref].

As a defined wire example, take $L=1\mathrm{m}$, $A=1\mathrm{m}{\mathrm{m}}^2$ and $\sigma =5\times 10^5(\Omega \mathrm{m}{)}^{-1}$. Then $R=2\Omega$; applying electric potential difference $5\mathrm{V}$ drives electric current $2.5\mathrm{A}$. With the same length and material, a second wire of wire cross-sectional area $2\mathrm{m}{\mathrm{m}}^2$ has $R=1\Omega$ and carries $5\mathrm{A}$ at that same electric potential difference. Fixing electric potential difference here differs from fixing electric current in the current density comparison: stating what is held fixed is essential.

The electric field does work on matter at local power per volumeHere $\mathbf{J}$ is current density and the dot product includes the relative direction. Positive $\mathbf{J}\cdot \mathbf{E}$ means energy passes from the field to matter; negative means matter supplies energy to the field. In general this is a work rate, not automatically an irreversible heating rate.^[ref]

For a stationary isotropic Ohmic conductor with real scalar electrical conductivity $\sigma >0$, local Ohm's law gives $\mathbf{J}=\sigma \mathbf{E}$. Thus the local Joule heating rate isThe electrical resistivity ${\rho }_{\mathrm{el}}$ has units $\Omega \mathrm{m}$, and electrical conductivity has units $(\Omega \mathrm{m}{)}^{-1}$. These formulas use the material's rest frame and dissipative conduction current density. Moving media, complex conductivity, and additional storage or transport processes require their appropriate energy balance rather than this scalar substitution.

For uniform current in a straight homogeneous wire of length $L$, wire cross-sectional area $A$ and constant electrical conductivity, $E=V/L$, $J=I/A$ and $R=L/(\sigma A)$. The volume integral recovers terminal power:As a numerical check, the wire already used in electrical resistance has $L=1\mathrm{m}$, $A=10^{-6}{\mathrm{m}}^2$, $\sigma =5\times 10^5(\Omega \mathrm{m}{)}^{-1}$ and $V=5\mathrm{V}$. It gives $J=2.5\times 10^{6}A/{\mathrm{m}}^2$, $p_{\mathrm{vol}}=1.25\times 10^{7}W/{\mathrm{m}}^3$ and $P=12.5\mathrm{W}$, agreeing with $I=2.5\mathrm{A}$ and $R=2\Omega$. Concentrating the same electric current in a smaller wire cross-sectional area therefore raises the local heating rate if electrical resistivity stays fixed.

In a steady state, Maxwell's equations imply $\nabla \times \mathbf{E}=0$, since the magnetic field is time independent. The divergence of the steady Ampère equation gives $\nabla \cdot \mathbf{j}=0$. With uniform electrical conductivity, Ohm's law is $\mathbf{j}=\sigma \mathbf{E}$, soFor the usual straight-wire model, the cylindrical side is insulating and the two flat end faces are equipotentials. Put $\mathbf{E}=-\nabla V$. The electric potential solves Laplace's equation, with constant end values and zero normal derivative on the side. The linear axial potential $V(z)=V(0)-\Delta Vz/l$ satisfies all these conditions. It is unique: the difference $h$ of two solutions has zero end values and zero side normal derivative, and integration by parts gives $\int |\nabla h{|}^{2}dV=0$. ThusThis proves uniform current in a straight homogeneous wire with the usual contact conditions. Integrating the electric current density over a cross-section and using Ohm's law then givesThe second expression is the rate of Ohmic heating.

Strictly, steady current and uniform conductivity alone do not force uniformity; the contact conditions are implicit in the wire model. For example, let $\chi$ be a nonconstant transverse Neumann eigenfunction on the disk, satisfying ${\Delta }_{\perp }\chi =-{\lambda }^2\chi$ and zero normal derivative at its edge. Then $V=-E_{0}z+C\chi \cosh (\lambda (z-l/2))$ is harmonic and preserves an insulating cylindrical side, but its nonconstant end potentials produce nonuniform current. Its added axial current has zero cross-sectional mean. This supplies a counterexample if arbitrary contacts are permitted.

A bend generally redistributes the electric current density: the inner side of the bend carries more current than the outer side, as illustrated by $j_{\theta }\propto 1/r$ in an annular conductor. Exact resistance and local Ohmic heating can therefore change. For a slender, gently curved wire the straight-wire formulas remain the leading approximation; the identity $\dot{Q}=I^{2}R$ remains valid for the total steady dissipation.

Take the electric current along $+z$, and use cylindrical coordinates. A uniform current in a straight homogeneous wire hasThe local Ohm's law $\mathbf{J}=\sigma \mathbf{E}$ gives the longitudinal electric fieldFor a circular Amperian loop of radius $r\le a$, the enclosed electric current is $I_{\mathrm{enc}}=I{r}^2/a^2$. The magnetostatic Ampère law gives $2\pi r{H}_{\varphi }=I_{\mathrm{enc}}$, henceSince $\hat{\mathbf{z}}\times \hat{\mathbf{\varphi }}=-\hat{\mathbf{r}}$, the Poynting vector isIt is zero at the centre and points radially inward elsewhere. Its units are $\mathrm{W}{\mathrm{m}}^{-2}$. These are the interior fields in the long-wire approximation, neglecting end and contact effects.