Question

A uniform copper wire of length $1m$ and cross-sectional area $5\times {10}^{-7}{m}^2$ carries current of $1A$. Assuming that there are $8\times {10}^{28}$ free electron per $m^3$ in copper, how long will an electron take to drift from one end of the wire to the other ?

• A. $0.8\times {10}^{3}s$
• B. $1.6\times {10}^{3}s$
• C. $3.2\times {10}^{3}s$
• D. $6.4\times {10}^{3}s$

Answer 1

The correct option is D $6.4\times {10}^{3}s$

$i=1A$
$n=8\times {10}^{28}$ electrons/$m^3$
$A=5\times {10}^{-7}{m}^2$
$i=Ane{v}_d$

$1=8\times {10}^{28}\times 1.6\times {10}^{-19}\times 5\times {10}^{-7}\times v_d$

$v_d=\frac{1}{8\times {10}^{28}\times 8\times {10}^{-26}}$

$t=\frac{1}{v_d}=8\times {10}^{28}\times 8\times {10}^{-26}$

$t=64\times {10}^2=6.4\times {10}^{3}s$