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A uniform copper wire of length $1 \mathrm{~m}$ and crosssectional area $5 \times 10^{-7} \mathrm{~m}^{2}$ carries a current of $1 \mathrm{~A}$. Assuming that there are $8 \times 10^{28}$ free electrons/ $\mathrm{m}^{3}$ in copper, how long will an electron take to drift from one end of the wire to the other
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The correct answer is:
$6.4 \times 10^{3} \mathrm{~S}$
Time taken by free electrons to cross the conductor
$t=\frac{\ell}{v_{d}}$ where drift velocity $v_{d}=\frac{I}{n e A}$
$\Rightarrow \mathrm{v}_{\mathrm{d}}=\frac{1}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-7}}$
$=\frac{10^{-2}}{64} \mathrm{~m} / \mathrm{s}$
$\Rightarrow \mathrm{t}=\frac{1}{\mathrm{v}_{\mathrm{d}}}=\frac{64}{10^{-2}}=64 \times 10^{2} \mathrm{~s}=6.4 \times 10^{3} \mathrm{sec}$
$t=\frac{\ell}{v_{d}}$ where drift velocity $v_{d}=\frac{I}{n e A}$
$\Rightarrow \mathrm{v}_{\mathrm{d}}=\frac{1}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-7}}$
$=\frac{10^{-2}}{64} \mathrm{~m} / \mathrm{s}$
$\Rightarrow \mathrm{t}=\frac{1}{\mathrm{v}_{\mathrm{d}}}=\frac{64}{10^{-2}}=64 \times 10^{2} \mathrm{~s}=6.4 \times 10^{3} \mathrm{sec}$
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