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A silver wire of length $3 \mathrm{~m}$ and of cross-sectional area $6.14 \times 10^{-6} \mathrm{~m}^2$ carries a current of 6 A. Atomic weight and density of silver are 108 and $10500 \mathrm{kgm}^{-3}$, respectively. A silver atom contributes one free electron for conduction. Avogadro number is $6.023 \times 10^{23} / \mathrm{mol}$. Drift velocity of electrons in silver is, close to
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The correct answer is:
$0.1 \mathrm{~ms}^{-1}$
Given, length of wire, $l=3 \mathrm{~m}$
Cross-sectional area, $A=6.14 \times 10^{-6} \mathrm{~m}^2$
Current in the wire, $I=6 \mathrm{~A}$
Atomic weight of silver $=108 \mathrm{amu}$
Density of silver, $\rho=10500 \mathrm{kgm}^{-3}$
$$
\begin{aligned}
\text { Number of electrons per } \mathrm{kg} \text { of silver } & =N_A / 108 \\
& =\frac{6.023 \times 10^{23}}{108}
\end{aligned}
$$
Number of electrons per unit volume of silver.
$$
n=\frac{6.023 \times 10^{23}}{108} \times 10500
$$
Drift current, $I=$ Anev $_d$
where $v_d$ is drift velocity and $n$ is number of electrons per unit volume.
$$
\begin{aligned}
& \Rightarrow \quad v_d=\frac{I}{n e A} \\
& =\frac{6 \times 108}{6.023 \times 10^{23} \times 10500 \times 1.6 \times 10^{-19} \times 6.14 \times 10^{-6}} \\
& =0.1 \mathrm{~ms}^{-1}
\end{aligned}
$$
Cross-sectional area, $A=6.14 \times 10^{-6} \mathrm{~m}^2$
Current in the wire, $I=6 \mathrm{~A}$
Atomic weight of silver $=108 \mathrm{amu}$
Density of silver, $\rho=10500 \mathrm{kgm}^{-3}$
$$
\begin{aligned}
\text { Number of electrons per } \mathrm{kg} \text { of silver } & =N_A / 108 \\
& =\frac{6.023 \times 10^{23}}{108}
\end{aligned}
$$
Number of electrons per unit volume of silver.
$$
n=\frac{6.023 \times 10^{23}}{108} \times 10500
$$
Drift current, $I=$ Anev $_d$
where $v_d$ is drift velocity and $n$ is number of electrons per unit volume.
$$
\begin{aligned}
& \Rightarrow \quad v_d=\frac{I}{n e A} \\
& =\frac{6 \times 108}{6.023 \times 10^{23} \times 10500 \times 1.6 \times 10^{-19} \times 6.14 \times 10^{-6}} \\
& =0.1 \mathrm{~ms}^{-1}
\end{aligned}
$$
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