Question

A uniform copper wire of length $1 \, m$ and cross-section area $5\times 10^{- 7} \, m^{2}$ carries a current of $1 \, A$ . 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 $i=Anev_{d}$
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$