Step 1: What current density means. Imagine slicing the conductor perpendicular to the flow. The current density \(\vec{J}\) tells how densely the current is packed through that slice: \(J = I/A\), with SI unit \(\text{A m}^{-2}\). It points along the flow direction.
Step 2: Start from Ohm's law in the wire. For a uniform conductor of length \(L\) and area \(A\), the macroscopic Ohm's law is \(V = IR\) with resistance \(R = \rho \dfrac{L}{A}\), where \(\rho\) is resistivity and \(\sigma = 1/\rho\) is conductivity.
Step 3: Express the field and the current density. A uniform field along the wire gives \(V = EL\), so \(E = V/L\). Also \(J = I/A\).
Step 4: Combine. From \(V = IR\): \(EL = I\rho\dfrac{L}{A}\). Cancel \(L\): \(E = \rho \dfrac{I}{A} = \rho J\). Hence \(J = \dfrac{E}{\rho} = \sigma E\).
Step 5: Cross-check with the free-electron model. The drift speed is \(v_d = eE\tau/m\) and \(J = ne v_d = \dfrac{ne^{2}\tau}{m}E\), which identifies \(\sigma = ne^{2}\tau/m\), the same result. \[ \boxed{\,\vec{J} = \sigma\vec{E}\,} \] Both the field description and the electron-drift description agree.