Question

If $n$, $e$, $\tau$ , $m$ 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. $\frac{ml}{n{e}^2\tau A}$
• B. $\frac{2ma}{n{e}^2\tau }$
• C. $n{e}^2\tau A$
• D. $\frac{n{e}^2\tau A}{2m}$

Answer 1

Answer: (A)

$\frac{ml}{n{e}^2\tau A}$

$n$, $e$, $\tau$, $m$ are representing electron density, charge, relaxation time and mass of an electron respectively of a wire.
Now, the resistance of the wire is given by:
$R=\frac{\rho l}{A}=\frac{l}{\sigma A}$, where $\sigma$ is the electrical conductivity of the wire and $\sigma =\frac{n{e}^2\tau }{m}$.
So the resistance will be: $R=\frac{l}{\sigma A}=\frac{ml}{n{e}^2\tau A}$