Question

A current of 5 A is passing through a metallic wire of cross-sectional area $4\times {10}^{-6}{m}^2$. If the density of charge carriers of the wire is $5\times {10}^{26}{m}^{-3}$, then the drift velocity of the electrons will be

• A. $1\times {10}^{2}m/s$
• B. $1.56\times {10}^{-2}m/s$
• C. $1.56\times {10}^{-3}m/s$
• D. $1\times {10}^{-2}m/s$

Answer 1

Answer: (B)

$\text{Given :}n=5\times {10}^{26}{m}^{-3},A=4\times {10}^{-6}{m}^2,I=5A,e=1.6\times {10}^{-19}C$

$\text{Step 1 : Use the formula for drift velocity}$

Drift velocity, $v=\frac{I}{nAe}$ $.......\left(1\right)$

$\text{Step 2 : Substitute the values in equation (1)}$

$v=\frac{5A}{(4\times {10}^{-6}{m}^2)\times (5\times {10}^{26}{m}^{-3})\times (1.6\times {10}^{-19}C)}=1.56\times {10}^{-2}m/s$