Question

Consider a cylindrical conductor of length \( l \) and area of cross-section \( A \). Current \( I \) is maintained in the conductor and electrons drift with velocity \( \vec{v}_d \, (|\vec{v}_d| = \frac{eE}{m} \tau) \), where symbols have their usual meanings. Show that the conductivity of the material of the conductor is given by \[ \sigma = \frac{n e^2 \tau}{m}. \]

Hint:

Microscopic Ohm’s law: \[ J = \sigma E \] Use drift velocity and current density to derive conductivity formulas.

Answer 1

Concept: Current in a conductor is due to drift of free electrons under an electric field. Key relations:

Drift velocity: \[ v_d = \frac{eE}{m} \tau \]
Current density: \[ J = nqv_d \]

Step 1: Expression for current. If \( n \) = number of free electrons per unit volume, Charge passing per second through area \( A \): \[ I = nqAv_d \] Since electron charge magnitude = \( e \): \[ I = neAv_d \]
Step 2: Current density. \[ J = \frac{I}{A} = nev_d \]
Step 3: Substitute drift velocity. \[ v_d = \frac{eE}{m} \tau \] \[ J = ne \left(\frac{eE}{m} \tau\right) \] \[ J = \frac{ne^2 \tau}{m} E \]
Step 4: Compare with Ohm’s law (microscopic form). \[ J = \sigma E \] Comparing: \[ \sigma = \frac{ne^2 \tau}{m} \] Final Result: \[ \boxed{\sigma = \frac{ne^2 \tau}{m}} \] Conclusion: Conductivity depends on:

Number of charge carriers \( n \)
Relaxation time \( \tau \)
Electron charge and mass