Question

A uniform copper wire of length $1m$ and cross-sectional area $5\times 10^{-7}{m}^2$ carries a current of $1,A$. Assuming that there are $8\times 10^{28}$ free electron $/m^3$ 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)

Consider a conductor of length $l$ and of uniform area of cross-section $A$.
$\therefore$ Volume of the conductor $=Al$ If $n$ is the number of free electrons per unit volume of the conductor, then total number of free electrons in the conductor = Aln. If $e$ is the charge on each electron, then total charge on all the free electrons in the conductor, $q=$ Alne. Let a constant potential difference $V$ is applied across the ends of the conductor with the help of a battery.
The electric field set up across the conductor is given by
$E=V/l$
Due to this field, the free electrons present in the conductor will begin to move with a drift velocity $v_d$ towards the left hand side as shown in figure.

Therefore, time taken by the free electrons to cross the conductor,
$t=\frac{l}{v_d}$
Hence, current $i=\frac{q}{t}=\frac{\text{ Alne }}{l/v_d}$
or $i=$ Anev ${}_d$
Here, $i=1A,n=8\times 10^{28}$ electron $/m^3$
$A=5\times 10^{-7}{m}^2$
$⟹1=8\times 10^{28}\times 1.6\times 10^{-19}\times 5\times 10^{-7}\times v_d$
or $v_d=\frac{1}{8\times 10^{28}\times 1.6\times 10^{-19}\times 5\times 10^{-7}}$
Now, $t=\frac{l}{v_d}$
$=8\times 10^{28}\times 1.6\times 10^{-19}\times 5\times 10^{-7}$
$=64\times 10^2$
$=6.4\times 10^{3}s$