Question

A conductor of length l and area of cross-section A is connected across an ideal battery of emf V . Derive the formula for the current density in terms of relaxation time τ

Answer 1

We are given a conductor with length \( l \) and cross-sectional area \( A \) connected to an ideal battery with emf \( V \). The task is to derive an expression for the current density \( J \) in terms of the relaxation time \( \tau \).

1. **Ohm’s Law:** According to Ohm’s law, the current \( I \) through a conductor is given by:

\[ I = \frac{V}{R} \]

Where \( V \) is the potential difference (emf), and \( R \) is the resistance of the conductor. The resistance \( R \) of the conductor is given by:

\[ R = \rho \frac{l}{A} \]

Where \( \rho \) is the resistivity of the material, \( l \) is the length of the conductor, and \( A \) is the cross-sectional area of the conductor. Substituting the expression for \( R \) in Ohm's law:

\[ I = \frac{V}{\rho \frac{l}{A}} = \frac{V A}{\rho l} \]

2. **Current Density:** The current density \( J \) is defined as the current per unit area of cross-section of the conductor:

\[ J = \frac{I}{A} \]

Substituting the expression for \( I \) into this equation:

\[ J = \frac{V A}{\rho l A} = \frac{V}{\rho l} \]

3. **Relation to Relaxation Time \( \tau \):** The current density can also be expressed in terms of the charge carrier density \( n \), the charge of the carriers \( e \), the relaxation time \( \tau \), and the drift velocity \( v_d \) of the charge carriers:

\[ J = n e v_d \]

Where \( n \) is the number of charge carriers per unit volume, and \( v_d \) is the drift velocity of the carriers. The drift velocity is related to the electric field \( E \) and the relaxation time \( \tau \) by:

\[ v_d = \mu E = \frac{e \tau}{m} E \]

Where \( \mu = \frac{e \tau}{m} \) is the mobility of the charge carriers, and \( m \) is the mass of the charge carriers. The electric field \( E \) is related to the potential difference \( V \) and the length \( l \) of the conductor:

\[ E = \frac{V}{l} \]

Substituting this into the expression for \( v_d \):

\[ v_d = \frac{e \tau}{m} \frac{V}{l} \]

Now, substituting this into the equation for current density \( J \):

\[ J = n e \frac{e \tau}{m} \frac{V}{l} = \frac{n e^2 \tau}{m} \frac{V}{l} \]

Thus, the current density in terms of the relaxation time \( \tau \) is:

\[ J = \frac{n e^2 \tau}{m} \frac{V}{l} \]

4. **Final Expression for Current Density:** Therefore, the formula for the current density \( J \) in terms of the relaxation time \( \tau \) is:

\[ J = \frac{V}{\rho l} \]

This equation relates the current density \( J \) to the emf \( V \), the resistivity \( \rho \), and the length \( l \) of the conductor.