Question

There is a current of $40\, A$ in a wire of $10^{-6} \,m ^{2}$ area of cross-section. If the number of free electrons per cubic metre is $10^{29}$, then the drift velocity is

• A. $250 \times 10 ^{-3} \, ms^{-1}$
• B. $25.0 \times 10 ^{-3} \, ms^{-1}$
• C. $2.50 \times 10 ^{-3} \, ms^{-1}$
• D. $1.25 \times 10 ^{-3} \, ms^{-1}$

Answer 1

Answer: (C)

Drift velocity
$v_{d}=\frac{i}{n e A} $
$=\frac{40}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-6}}$
$=2.5 \times 10^{-3} \,ms ^{-1}$