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 \(v_d\). Show that the conductivity \(\sigma\) of the material of the conductor is given by \[ \sigma=\frac{ne^2\tau}{m}. \]

Hint:

Important microscopic relation: \[ \sigma=\frac{ne^2\tau}{m} \] and \[ \rho=\frac{1}{\sigma}. \] For metals, resistance increases approximately linearly with temperature.

Answer 1

Derivation of Conductivity

Step 1: Write the expression for drift velocity. According to the electron theory, \[ v_d=\frac{eE\tau}{m}. \] where
• \(e\) = charge of electron,
• \(m\) = mass of electron,
• \(\tau\) = relaxation time,
• \(E\) = electric field.

Step 2: Write the expression for current. Current through a conductor is \[ I=neAv_d. \] Substituting drift velocity, \[ I = neA\left(\frac{eE\tau}{m}\right). \] \[ I = \frac{ne^2A\tau E}{m}. \]

Step 3: Find current density. Current density \[ J=\frac{I}{A}. \] Thus, \[ J = \frac{ne^2\tau}{m}E. \]

Step 4: Compare with microscopic Ohm's law. Microscopic Ohm's law is \[ J=\sigma E. \] Comparing, \[ \boxed{ \sigma = \frac{ne^2\tau}{m} } \] which is the required result.