Question

If current in a current carrying wire is $1.5A$, number of free electrons per unit volume is $8\times {10}^{28}{m}^3$ and area of cross section is $5m{m}^2$. Drift velocity of electrons will be:

• A. $0.02mm{s}^{-1}$
• B. $2mm{s}^{-1}$
• C. $0.2mm{s}^{-1}$
• D. None of these

Answer 1

The correct option is A

$0.02mm{s}^{-1}$

Step 1: Given data

Current, $I=1.5A$

Number of free electrons per unit volume, $v=8\times {10}^{28}{m}^3$

Area of cross section, $A=5m{m}^2$

Step 2: Formula and definition

Drift velocity

It is the average velocity gained by a charged particle in a body due to an electric field.

We know,

current, $I=nAe{v}_d$

where

$n\to$the number of electrons per unit volume.

$A\to$the cross-sectional area of the wire.

$e\to$electronic charge$=1.6\times {10}^{-19}C$

$v_d\to$drift velocity

Step 3: Substitution and final answer

$$\begin{gathered} I=nAe{v}_d \\ {v}_d=\frac{I}{nAe} \\ {v}_d=\frac{1.5}{8\times {10}^{28}\times 5\times {10}^{-6}\times 1.6\times {10}^{-19}} \\ {v}_d=2\times {10}^{-5}m/s \\ {v}_d=2\times {10}^{-2}mm/s \\ {v}_d=0.02mm/s \end{gathered}$$

Therefore, the drift velocity of electrons are $0.02mm{s}^{-1}$.

Hence, option A is correct.