Question

Define ‘current density’. Is it a scalar or a vector? An electric field $\vec{E$ is maintained in a metallic conductor. If $n$ be the number of electrons (mass $m$, charge $-e$) per unit volume in the conductor and $\tau$ its relaxation time, show that the current density $\vec{j} = \alpha \vec{E}$, where $\alpha = \left( \frac{ne^2}{m} \right) \tau$.}

Hint:

Current density $\vec{j}$ is always a vector quantity. It is proportional to the electric field $\vec{E}$, with proportionality constant $\alpha = \frac{ne^2 \tau}{m}$.

Answer 1

Definition of Current Density:
The current density $\vec{j}$ is defined as the current per unit cross-sectional area. Mathematically: \[ \vec{j} = \frac{\vec{I}}{A}, \] where $\vec{I}$ is the current and $A$ is the cross-sectional area. Current density is a vector quantity because it has both magnitude and direction. Derivation:
The current $I$ through a conductor is given by: \[ I = nqA v_d, \] where: \begin{itemize} \item $n$ is the number of charge carriers per unit volume, \item $q$ is the charge of each carrier, \item $A$ is the cross-sectional area, \item $v_d$ is the drift velocity of the charge carriers. \end{itemize} Substitute $v_d$ using $v_d = \mu E$, where $\mu$ is the mobility of charge carriers and $E$ is the electric field: \[ I = nqA (\mu E). \] The current density $\vec{j}$ is: \[ \vec{j} = \frac{I}{A} = nq (\mu E). \] Using $\mu = \frac{e \tau}{m}$, where $\tau$ is the relaxation time and $m$ is the mass of the charge carriers, substitute $\mu$: \[ \vec{j} = nq \left( \frac{e \tau}{m} \right) \vec{E}. \] Simplify: \[ \vec{j} = \left( \frac{ne^2 \tau}{m} \right) \vec{E}. \] Thus, $\alpha = \frac{ne^2 \tau}{m}$, and: \[ \boxed{\vec{j} = \alpha \vec{E}}. \]