Question

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

• A. 2
• B. 0.2
• C. 0.02
• D. 3

Answer 1

The correct option is C 0.02

Calculation of drift velocity:

Given, $i=1.5A,n=9{\times }^{28}/m^3$

$e=1.6\times {10}^{-19}C$ drift velocity $=v_d,A=5m{m}^2$

The current in relation with drift velocity is given by

$l=neA{v}_d$

$v_d=\frac{i}{neA}$

$=\frac{15}{9\times {10}^{28}\times 1.6\times {10}^{-19}\times 5\times {10}^{-6}}$

$=0.02mm/s$

Hence, option (D) is correct.