Given a conductor of length \( l \) and cross-sectional area \( A \) connected to an ideal battery with emf \( V \), we aim to derive an expression for the current density \( J \) in terms of the relaxation time \( \tau \).
1. Ohm’s Law and Resistance: Ohm's law states that the current \( I \) is \( I = \frac{V}{R} \), where \( V \) is the potential difference and \( R \) is the resistance. The resistance is given by \( R = \rho \frac{l}{A} \), where \( \rho \) is the resistivity. Substituting \( R \) into Ohm's law yields \( I = \frac{V}{\rho \frac{l}{A}} = \frac{V A}{\rho l} \).
2. Current Density Definition: Current density \( J \) is defined as current per unit area: \( J = \frac{I}{A} \). Substituting the expression for \( I \) gives \( J = \frac{V A}{\rho l A} = \frac{V}{\rho l} \).
3. Relating to Relaxation Time \( \tau \): Current density can also be expressed as \( J = n e v_d \), where \( n \) is the charge carrier density, \( e \) is the charge, and \( v_d \) is the drift velocity. The drift velocity is given by \( v_d = \mu E \), where \( \mu \) is the mobility and \( E \) is the electric field. We know \( \mu = \frac{e \tau}{m} \), where \( \tau \) is the relaxation time and \( m \) is the carrier mass. The electric field is \( E = \frac{V}{l} \). Thus, \( v_d = \frac{e \tau}{m} \frac{V}{l} \). Substituting this into the expression for \( J \): \( J = n e \left( \frac{e \tau}{m} \frac{V}{l} \right) = \frac{n e^2 \tau}{m} \frac{V}{l} \).
4. Final Current Density Formula: The current density in terms of relaxation time is therefore \( J = \frac{n e^2 \tau}{m} \frac{V}{l} \). This equation relates \( J \) to \( V \), \( \rho \), and \( l \) through the relation \( \rho = \frac{m}{n e^2 \tau} \).