Question

A uniform copper wire of length $1 m$ 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 $=A l$ 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, $i=1 A , n=8 \times 10^{28}$ electron $/ m ^{3}$
$A=5 \times 10^{-7} m ^{2}$
$\Longrightarrow 1=8 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-7} \times v_{d}$
or $v_{d}=\frac{1}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-7}}$
Now, $t=\frac{l}{v_{d}}$
$=8 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-7}$
$=64 \times 10^{2}$
$=6.4 \times 10^{3} s$