CBSE
2023
Class Class 12 · Physics
5 Marks · Short
✅ Verified
Define current density $J$ of a metallic conductor. Deduce the relation connecting $J$ and conductivity $\sigma$ when electric field $E$ is applied.
✅ Answer & Solution
Current density: $J = I/A$ (current per unit cross-sectional area). Unit: A m$^{-2}$
Derivation:
Force on electron: $F = eE$
Acceleration: $a = eE/m$
Drift velocity: $v_d = a\tau = eE\tau/m$
Current density:
$$J = nev_d = \frac{ne^2\tau}{m} \cdot E$$
Comparing with $J = \sigma E$:
$$\sigma = \frac{ne^2\tau}{m}$$
$$\therefore J = \sigma E \quad \text{(Microscopic form of Ohm's law)}$$
✅ Verified by Super Admin