Question

A conductor wire having $10^{29}$ free electrons/$m^{3}$ carries a current of $20\,A$. If the cross-section of the wire is $1\,mm^{2}$, then the drift velocity of electrons will be .......
$(e\, =\, 1.6 \times10^{-19}C).$

• A. $12.5 \times 10^{-3} ms^{-1}$
• B. $1.25 \times 10^{-4} ms^{-1}$
• C. $1.25 \times 10^{-5} ms^{-1}$
• D. $1.25 \times10^{-3} ms^{-1}$

Answer 1

Answer: (D)

Current, $l=20 A$
Free electron density, $n =10^{29}$
Cross-sectional area of wire, $A=1 mm ^{2}=10^{-6} m ^{2}$
Magnitude of charge on an electron, $e=1.6 \times 10^{-19} C$
We know that, current,
$
I = neAv _{ d }
$
$
\Rightarrow v_{d}=\frac{1}{n e A}=\frac{20}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-6}}
$
$
\Rightarrow v _{ d }=1.25 \times 10^{-3} m / s
$