Question

If $n$ , $e$ , $\tau$ and $m$ represent 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

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

Answer 1

Answer: (A)

$R=\rho \frac{l}{A}$
Since we know resistivity is given by $\rho =\frac{m}{n{e}^{2}r}$
Resistance $R=\frac{ml}{n{e}^2\tau A}$
where,
$m=$ Mass of electron
$l=$ length of conductor
$n=$ charge density
$\text{e}=$ charge of electron
$\tau =$ relaxation time
$A=$ cross-sectional area