Question

When $5V$ potential difference is applied across a wire of length $0.1m$, the drift speed of electrons is $2.5\times 10^{-4}$ $m{s}^{-1}$. If the electron density in the wire is $8\times 10^{28}{m}^{-3}$ the resistivity of the material is close to

• A. $1.6\times 10^{-8}\Omega m$
• B. $1.6\times 10^{-7}\Omega m$
• C. $1.6\times 10^{-5}\Omega m$
• D. $1.6\times 10^{-6}\Omega m$

Answer 1

Answer: (C)

$i=$neAV${}_d$
$\Rightarrow \frac{V}{R}=$ neAV ${}_d\left\{R=\frac{\rho l}{A}\right\}$
$\Rightarrow \frac{V\times A}{\rho \ell }=neA{V}_d$
$\Rightarrow \frac{5}{\rho \times 0.1}=8\times 10^{28}\times 1.6\times 10^{-19}\times 2.5\times 10^{-4}$
$\Rightarrow \rho =1.56\times 10^{-5}\Omega m$
$\Rightarrow \rho \simeq 1.6\times 10^{-5}\Omega m$