Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons.
CBSE DELHI - SET 1 2008
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Total number of free electrons, $N = n A.$
$\therefore$ Total Charge $Q = N e$
$= nAel$
$\tau = \frac{1}{\text{v}_{d}}$
$\therefore$ $\text{I} = \frac{Q}{\tau} = \frac{\text{nAel}}{l}\text{v}_{d} $
$= nAev_d$
$J =$ current dencity $= \frac{\text{I}}{\text{A}}$
$\therefore$ $\text{J} = \frac{nAev_d}{A} = \text{nev}_{d}$
$\therefore$ $\text{J} \propto \text{V}_{d}$
$\therefore$ Total Charge $Q = N e$
$= nAel$
$\tau = \frac{1}{\text{v}_{d}}$
$\therefore$ $\text{I} = \frac{Q}{\tau} = \frac{\text{nAel}}{l}\text{v}_{d} $
$= nAev_d$
$J =$ current dencity $= \frac{\text{I}}{\text{A}}$
$\therefore$ $\text{J} = \frac{nAev_d}{A} = \text{nev}_{d}$
$\therefore$ $\text{J} \propto \text{V}_{d}$
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