Question

A uniform copper wire of length $1m$ and cross-section area $5\times 10^{-7}{m}^2$ carries a current of $1A$ . Assuming that there are $8\times 10^{28}$ free electrons 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

Answer: (D)

Current in wire $i=Ane{v}_d$
Here,
$A=5\times 10^{-7}{m}^2$
$\therefore 1=8\times 10^{28}\times 1.6\times 10^{-19}\times 5\times 10^{-7}\times v_d$
$or{v}_d=\frac{1}{8\times 10^{28}\times 1.6\times 10^{-19}\times 5\times 10^{-7}}$
Now,
$t=\frac{1}{v_d}=8\times 10^{28}\times 1.6\times 10^{-19}\times 5\times 10^{-7}$
$=64\times 10^2=6.4\times 10^{3}s$