Question

A current of $5\text{A}$ is passing through a metallic wire of cross sectional area $14\times {10}^{-6}{\text{m}}^2$. If the density of the charge carries in the wire is $5\times {10}^{26}{\text{m}}^3$, the drift speed of the electrons is-

• A. $3.46\times {10}^{-3}{\text{ms}}^{-1}$
• B. $4.46\times {10}^{-3}{\text{ms}}^{-1}$
• C. $5.46\times {10}^{-3}{\text{ms}}^{-1}$
• D. $6.46\times {10}^{-3}{\text{ms}}^{-1}$

Answer 1

The correct option is B $4.46\times {10}^{-3}{\text{ms}}^{-1}$

Given, $I=5\text{A};A=14\times {10}^{-6}{\text{m}}^2$

$n=5\times {10}^{26}{\text{m}}^3;e=1.6\times {10}^{-19}\text{C}$

The drift speed , $v_d=\frac{I}{neA}$

$=\frac{5}{5\times {10}^{26}\times 1.6\times {10}^{-19}\times 14\times {10}^{-6}}$

$=\frac{1}{16\times 14}=\frac{1}{224}$

$=4.46\times {10}^{-3}{\text{ms}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.