Question

When $5V$ potential difference is applied across a wire length $0.1m$, the drift speed of electron 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. $1.6\times {10}^{-6}\Omega$m
• D. $1.6\times {10}^{-5}\Omega$m

Answer 1

The correct option is D $1.6\times {10}^{-5}\Omega$m

$\text{Step 1: Current flowing through the wire}$

We know that

$I=nAe{v}_d$
We also know by ohm's law , $I=\frac{V}{R}$

$\Rightarrow \frac{V}{R}=Ane{v}_d$ $.....\left(1\right)$

$\text{Step 2: Resistance of the wire}$

$R=\frac{\rho l}{A}$ $.....\left(2\right)$

$\text{Step 3: Solving above equations}$

From $\left(1\right)$ and $\left(2\right)$

$\frac{V}{\left(\rho \frac{l}{A}\right)}=Ane{v}_d\Rightarrow \frac{VA}{\rho l}=Ane{v}_d\Rightarrow \rho =\frac{V}{lne{v}_d}=\frac{5V}{0.1m\times (8\times {10}^{28}{m}^{-3})\times (1.6\times {10}^{-19}C)\times (2.5\times {10}^{-4}m{s}^{-1})}=1.5625\times {10}^{-5}\Omega m$

$\Rightarrow \rho \approx 1.6\times {10}^{-5}\Omega m$

Hence, option (D) is correct.