Question

A conducting wire of cross-sectional area 1 $ c{{m}^{2}} $ has $ 3\times {{10}^{23}} $ charge carriers per $ metr{{e}^{3}} $ . If wire carries a current $24 \,mA$, then rift velocity of carriers is

• A. $ 5\times {{10}^{-2}}m/s $
• B. $ 0.5\text{ }m/s $
• C. $ 5\times {{10}^{-3}}\text{ }m/s $
• D. $ 5\times {{10}^{-6}}\text{ }m/s $

Answer 1

The Correct Option is C

The current $i$ crossing area of cross-section $A$ can be expressed in terms of drift velocity $v_{d}$ and the moving charges as $i=n e v_{d} A$ where, $n$ is number of charge carriers per unit volume and $e$ the charge on the carrier. $\therefore v_{d} =\frac{i}{n e A}=\frac{24 \times 10^{-3}}{\left(3 \times 10^{23}\right)\left(1.6 \times 10^{-19}\right)\left(10^{-4}\right)} $ $=5 \times 10^{-3} \,m / s$