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

Derivation of Conductivity of a Material
Consider a cylindrical conductor of length \( l \) and cross-sectional area \( A \). Let the current \( I \) be maintained in the conductor and electrons drift with velocity \( \vec{v}_d \), where the magnitude of drift velocity is given by:
\[ |\vec{v}_d| = \frac{eE}{m} \tau \]
Here, \( n \) is the number density of electrons, \( e \) is the charge of an electron, \( m \) is the electron mass, \( E \) is the applied electric field, and \( \tau \) is the mean free time between collisions.
Step 1: Express Current in Terms of Drift Velocity
The current \( I \) through the conductor is:
\[ I = n e A |\vec{v}_d| \]
Substitute the given drift velocity:
\[ I = n e A \left(\frac{e E \tau}{m}\right) = \frac{n e^2 \tau}{m} A E \]
Step 2: Relate Current Density to Electric Field
Current density \( J \) is defined as:
\[ J = \frac{I}{A} = \frac{n e^2 \tau}{m} E \]
By definition, conductivity \( \sigma \) relates current density and electric field:
\[ J = \sigma E \]
Step 3: Expression for Conductivity
Comparing the two expressions for \( J \):
\[ \sigma = \frac{n e^2 \tau}{m} \]
Conclusion:
The conductivity of the material is given by:
\[ \boxed{\sigma = \frac{n e^2 \tau}{m}} \]
This shows that conductivity increases with the number of free electrons and their mean free time, and decreases with the mass of the electron.