Current Electricity 1 Question 13
13. When $5 V$ potential difference is applied across a wire of length $0.1 m$, the drift speed of electrons is $2.5 \times 10^{-4} ms^{-1}$. If the electron density in the wire is $8 \times 10^{28} m^{-3}$ the resistivity of the material is close to
(2015 Main)
(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) $1.6 \times 10^{-6} \Omega m$
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Solution:
- $i=n e A v _d$ or $\frac{V}{R}=n e A v _d$
$$ \begin{aligned} \text { or } & \frac{V}{\frac{\rho l}{A}} & =n e A v _d \\ \therefore & \rho & =\frac{V}{n e l v _d}=\text { resistivity of wire } \end{aligned} $$
Substituting the given values we have
$$ \begin{aligned} \rho & =\frac{5}{\left(8 \times 10^{28)}\left(1.6 \times 10^{-19}\right)(0.1)\left(2.5 \times 10^{-4}\right)\right.} \\ & \approx 1.6 \times 10^{-5} \Omega-m \end{aligned} $$
