Question

The number density of free electrons in a copper conductor is $8.5 \times 10^{28} \cdot m ^{-3}$. How long does an electron take to drift from one end of a wire $3.0\, m$ long to its other end? The area of crosssection of the wire is $2.0 \times 10^{-6}\, m ^{2}$ and it is carrying a current of $3.0\,A$.

• A. $ 8.1\times 10^{4}\,s $
• B. $ 2.7\times 10^{4}\,s $
• C. $ 9\times 10^{3}\,s $
• D. $ 3\times 10^{3}\,s $

Answer 1

The Correct Option is B

Given, $l=3.0$
$A=2.0 \times 10^{-6} m ^{2}$
$i=30 A$
$v_{d}=\frac{i}{n e A}$
$v_{d}=\frac{3}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2 \times 10^{-6}}$
$v_{d}=1.10 \times 10^{-4}$
Time $t=\frac{l}{v_{d}}$
$=\frac{3}{1.10 \times 10^{-4}}$
$=2.7 \times 10^{4} \,s$