Question

When $5\,V$ potential difference is applied across a wire of length $0.1\,m$, the drift speed of electrons is $2.5\times10^{-4}$ $ms^{-1}$. If the electron density in the wire is $8\times10^{28}\, m^{-3}$ the resistivity of the material is close to

• A. $1.6\times10^{-8}\Omega\,m$
• B. $1.6\times10^{-7}\Omega\,m$
• C. $1.6\times10^{-5}\Omega\,m$
• D. $1.6\times10^{-6}\Omega\,m$

Answer 1

Answer: (C)

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