Question:
Drift speed of electrons, when $1.5 \mathrm{~A}$ of current flows in a copper wire of cross section $5 \mathrm{~mm}^{2}$, is $v$. If the electron density in copper is $9 \times 10^{28} / \mathrm{m}^{3}$ the value of $v$ in $\mathrm{mm} / \mathrm{s}$ close to (Take charge of electron to be $=1.6 \times 10^{-19} \mathrm{C}$ )
Correct Option: 1
Solution:
(1) Using, $\mathrm{I}=\mathrm{neAv}_{\mathrm{d}}$
$\therefore$ Drift speed $\mathrm{v}_{\mathrm{d}}=\frac{1}{\text { neA }}$
$=\frac{1.5}{9 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-6}}=0.02 \mathrm{~mm} / \mathrm{s}^{-1}$
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