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)

$6.4 \times 10^3$s

Answer 2

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, $t=\frac{l}{v_{d}}$ $=8 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-7}$ $=64 \times 10^{2}$ $=6.4 \times 10^{3} s$