Question

In $N,e,\tau$ and $m$ are representing electron density, charge, relaxation time and mass of an electron respectively, then the resistance of wire of length $l$ and cross-sectional area $A$ is given by

• A. $\frac{2ml}{N{e}^{2}A\tau }$
• B. $\frac{2m\tau A}{N{e}^{2}l}$
• C. $\frac{N{e}^2\tau A}{2ml}$
• D. $\frac{N{e}^{2}A}{2m\tau l}$

Answer 1

The correct option is A $\frac{2ml}{N{e}^{2}A\tau }$

Since average drift velocity $=\frac{1}{2}\frac{eE}{m}\times \left(\tau \right)$

Now $I=NeA\times \left(Averagedriftvelocity\right)$

$I=\frac{N{e}^{2}AE}{2ml}\times \tau =\frac{N{e}^{2}AV}{2ml}\times \tau$

$R=\frac{V}{I}=\frac{2ml}{N{e}^2\tau A}$, where $N$ is electron density.