Question

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

• A. $\raisebox{1ex}{$ml$}\!\left/ \!\raisebox{-1ex}{$n{e}^2\tau A$}\right.$
• B. $\raisebox{1ex}{$2mA$}\!\left/ \!\raisebox{-1ex}{$n{e}^2\tau $}\right.$
• C. $n{e}^2\tau A$
• D. $\raisebox{1ex}{$n{e}^2\tau A$}\!\left/ \!\raisebox{-1ex}{$2m$}\right.$

Answer 1

The correct option is A

$\raisebox{1ex}{$ml$}\!\left/ \!\raisebox{-1ex}{$n{e}^2\tau A$}\right.$

Given that $n$, $e$, $\tau$, $m$ are representing electron density, charge, relaxation time, and mass of an electron respectively of a wire respectively.
The resistance of the wire is given by:

$R={\displaystyle \frac{\rho l}{A}}={\displaystyle \frac{l}{\sigma A}}$……………..$\left(1\right)$,

Where, $\sigma$ is the electrical conductivity of the wire, and $\rho$= resistivity of the material of the wire.

The electrical conductivity of the wire $\sigma$ is also given as $\sigma =\frac{n{e}^2\tau }{m}$ and its SI unit is siemens per meter.

Putting the value of $\sigma$ in $\left(1\right)$, we get

$R=\frac{l}{\sigma A}=\frac{ml}{n{e}^2\tau A}as\frac{1}{\sigma }=\frac{m}{n{e}^2\tau }$,

Thus, the correct option is option A.