Question

Drift speed of electrons, when $1.5\, A$ of current flows in a copper wire of cross section $5\, mm^2$, is $v$. If the electron density in copper is $9 \times 10^{28} /m^3$ the value of $v$ in mm/s is close to (Take charge of electron to be = $1.6 \times 10^{-19}C$)

• A. 0.2
• B. 3
• C. 2
• D. 0.02

Answer 1

Answer: (D)

0.02

Answer 2

$I = neAv_{d}$
$\Rightarrow v_{d} = \frac{I}{neA} = \frac{1.5}{9 \times10^{28} \times1.6 \times 10^{-19} \times 5\times 10^{-6}}$
= $0.02\, m/s$