Question

There are $8.4\times {10}^{22}$ free electrons per ${\text{cm}}^3$ in copper. The current in the wire is $0.21$$\text{A}$ $(e=1.6\times {10}^{-19}C$). Then the drift velocity of electrons in a copper wire of $1$ ${\text{mm}}^2$ cross section will be

• A. $2.12\times {10}^{-5}$$\text{m/s}$
• B. $0.78\times {10}^{-5}$$\text{m/s}$
• C. $1.56\times {10}^{-5}$$\text{m/s}$
• D. none of these

Answer 1

The correct option is C $1.56\times {10}^{-5}$$\text{m/s}$

Given, electron density,

$n=8.4\times {10}^{22}$ electrons/${\text{cm}}^3$

$n=8.4\times {10}^{28}$ electrons/${\text{m}}^3$

$i=0.21$ $\text{A}$

$A=1{\text{mm}}^2$ $=$${10}^{-6}$ ${\text{m}}^2$

Now, drift velocity $=$ $v_d$ $=$ $\frac{I}{neA}$

$v_d$ $=\frac{0.21}{8.4\times {10}^{28}\times 1.6\times {10}^{-19}\times {10}^{-6}}$

$=\frac{0.21\times {10}^{-3}}{8.4\times 1.6}$

$=0.0156\times {10}^{-3}$

$=1.56\times {10}^{-5}$$\text{m/s}$