The macroscopic relation $V=IR$ describes an ohmic component as a whole. A conducting material can also be described locally by relating the current density to the electric field inside it.

For an isotropic ohmic material,

$$\boxed{\mathbf J=\sigma\mathbf E},$$

where $\sigma$ is the electrical conductivity. The current density points along the electric field when the charge transport has positive scalar conductivity.

The reciprocal quantity

$$\boxed{\rho=\frac1\sigma}$$

is the electrical resistivity. Its SI unit is

$$\Omega,\mathrm m.$$

Conductivity and resistivity are material properties within a specified range of temperature and other operating conditions. Resistance, by contrast, also depends on geometry.

Deriving the resistance of a uniform wire

Consider a uniform wire of length $L$ and cross-sectional area $A$, made from material with resistivity $\rho$.

For uniform current density,

$$J=\frac{I}{A}.$$

For an approximately uniform electric field along the wire,

$$E=\frac{V}{L}$$

in magnitude.

Using

$$J=\sigma E=\frac{E}{\rho},$$

we get

$$\frac{I}{A}=\frac{1}{\rho}\frac{V}{L}.$$

Rearranging,

$$V=I\rho\frac{L}{A}.$$

Comparing with $V=IR$ gives

$$\boxed{R=\rho\frac{L}{A}}.$$

Thus a longer conductor has greater resistance because charge carriers must sustain drift through more material, while a larger cross-sectional area provides more parallel conducting area and reduces resistance.

Microscopic interpretation

An electric field exerts forces on mobile charge carriers and creates a net drift. Collisions and scattering in the material prevent indefinite acceleration and produce a steady average drift under a steady field.

In the linear regime, doubling $\mathbf E$ doubles the drift contribution to $\mathbf J$, giving the proportionality $\mathbf J=\sigma\mathbf E$.

The value of $\sigma$ encodes microscopic properties such as carrier density, carrier charge, and how readily their momentum is disrupted by scattering. The simple local Ohm law does not by itself specify those microscopic mechanisms.

Worked example

A wire has

$$L=2.0,\mathrm m,$$

$$A=1.0,\mathrm{mm^2}=1.0\times10^{-6},\mathrm{m^2},$$

and resistivity

$$\rho=1.7\times10^{-8},\Omega,\mathrm m.$$

Its resistance is

$$R=\rho\frac{L}{A} =(1.7\times10^{-8})\frac{2.0}{1.0\times10^{-6}} =3.4\times10^{-2},\Omega.$$

So

$$\boxed{R=0.034,\Omega}.$$

If it carries $5.0,\mathrm A$, the voltage magnitude across it is

$$V=IR=(5.0)(0.034)=0.17,\mathrm V.$$

The corresponding average field magnitude is

$$E=\frac{V}{L}=0.085,\mathrm{V/m}.$$

Limits of the linear law

Not every material satisfies $\mathbf J=\sigma\mathbf E$ with constant $\sigma$. Conductivity can depend on temperature, field strength, carrier concentration, direction in anisotropic materials, and other state variables.

Microscopic Ohm's law is therefore a constitutive model: it connects local electric field to local current density for materials and operating regimes where the response is approximately linear.