KCET · Physics · Current Electricity
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?
- A \(0.8 \times 10^{3} \mathrm{~s}\)
- B \(1.6 \times 10^{3} \mathrm{~s}\)
- C \(3.2 \times 10^{3} \mathrm{~s}\)
- D \(6.4 \times 10^{3} \mathrm{~s}\)
Answer & Solution
Correct Answer
(D) \(6.4 \times 10^{3} \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
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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