Question

A current of 5A is passing through a metallic wire of cross-sectional area $4\times10^{-6}m^2$ If the density of the charge carriers in the wire is 5 × $10^{26}m^{-3}$, the drift speed of the electrons will be [e = 1.602 × $10^{-19}C$]

• A. $1.56 \times 10^{-2}ms^{-1}$
• B. $1.98 ×10^{-2}ms^{-1}$
• C. $2.42 ×10^{-2}ms^{-1}$
• D. $2.84 ×10^{-2}ms^{-1}$

Answer 1

Answer: (A)

In a metal, conduction current is due to electrons given by
I=nAev
$\Rightarrow $ drift velocity, $V=\frac{I}{nAe}$
$\Rightarrow v = \frac{5}{5 × 10^{26}× 4×10^{-6}× 1.602× 10^{-19}}$
$=\frac{1}{4 × 1.602× 10^1}$
$=\frac{10^{-1}}{6.408}=1.56× 10^{-2}m/s$