Question

Define current density \( (\vec{J}) \) and write its unit. Prove that \( \vec{J} = \sigma\vec{E} \), where \( \sigma \) is the specific conductivity (conductivity) of the conductor and \( \vec{E} \) the electric field inside the conductor.

Hint:

Current density \( J=I/A \) (unit A m\(^{-2}\)). Use \( v_d=eE\tau/m \) and \( I=nAev_d \) to get \( J=(ne^2\tau/m)E=\sigma E \).

Answer 1

Step 1: Definition of current density. Current density at a point is the electric current flowing per unit area of cross-section held perpendicular to the direction of flow. If a current \( I \) flows uniformly through an area \( A \), \[ \vec{J} = \frac{I}{A}\,\hat{n} \] It is a vector directed along the direction of current (conventional) flow.
Step 2: Unit. Ampere per square metre, \( \text{A m}^{-2} \).
Step 3: Set up drift motion. Consider a conductor of cross-section \( A \) with \( n \) free electrons per unit volume, each of charge \( e \) and mass \( m \). An electric field \( \vec{E} \) inside the conductor exerts a force \( \vec{F} = e\vec{E} \) on each electron, giving an acceleration \( a = \dfrac{eE}{m} \).
Step 4: Drift velocity. Between successive collisions the average (relaxation) time is \( \tau \). The steady drift velocity is \[ v_d = a\,\tau = \frac{eE\tau}{m} \]
Step 5: Relate current to drift velocity. In time \( t \) the electrons sweep a volume \( A v_d t \), containing \( nA v_d t \) electrons and charge \( nA v_d t\,e \). Hence \[ I = \frac{\text{charge}}{t} = nAe\,v_d \]
Step 6: Substitute the drift velocity. \[ I = nAe\left(\frac{eE\tau}{m}\right) = \frac{ne^{2}\tau}{m}\,A\,E \]
Step 7: Form the current density. Dividing by \( A \), \[ J = \frac{I}{A} = \frac{ne^{2}\tau}{m}\,E \] The quantity \( \dfrac{ne^{2}\tau}{m} \) depends only on the material and is called its conductivity \( \sigma \). Therefore \[ \boxed{\,\vec{J} = \sigma\vec{E}\,},\qquad \sigma = \frac{ne^{2}\tau}{m} \] This is the microscopic (point) form of Ohm's law.