Current Electricity Ques 7
- 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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Answer:
Correct Answer: 7.(c)
Solution:
Formula:
Electric Current in a Conductor:
- $i=n e A v _d$ 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} $
BETA
