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A copper wire of length $1 \mathrm{~m}$ and uniform cross-sectional 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 per $\mathrm{m}^{3}$ in copper, how long will an electron take to drift from one end of the wire to the other?
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Verified Answer
The correct answer is:
$6.4 \times 10^{3} \mathrm{~s}$
Given, $l=1 \mathrm{~m}$
$\begin{aligned}
&A=5 \times 10^{-7} \mathrm{~m}^{2} \\
&I=1 \mathrm{~A} \\
&n=8 \times 10^{28} \text { electrons } / \mathrm{m}^{3}
\end{aligned}$
Time taken by an electron to take the drift from one end to the other is given as
$T=\frac{l}{v_{d}}...(i)$
where, $v_{d}$ is the drift velocity.
As, $v_{d}=\frac{I}{n e A}$
$=\frac{1}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-7}}=\frac{1}{6.4 \times 10^{3}}$
So, substituting the given values in Eq. (i), we get
$T=\frac{1}{\frac{1}{6.4 \times 10^{3}}}=6.4 \times 10^{3} \mathrm{~s}$
$\begin{aligned}
&A=5 \times 10^{-7} \mathrm{~m}^{2} \\
&I=1 \mathrm{~A} \\
&n=8 \times 10^{28} \text { electrons } / \mathrm{m}^{3}
\end{aligned}$
Time taken by an electron to take the drift from one end to the other is given as
$T=\frac{l}{v_{d}}...(i)$
where, $v_{d}$ is the drift velocity.
As, $v_{d}=\frac{I}{n e A}$
$=\frac{1}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-7}}=\frac{1}{6.4 \times 10^{3}}$
So, substituting the given values in Eq. (i), we get
$T=\frac{1}{\frac{1}{6.4 \times 10^{3}}}=6.4 \times 10^{3} \mathrm{~s}$
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