Current Electricity 1 Question 11
11. Drift speed of electrons, when $1.5 A$ of current flows in a copper wire of cross-section $5 mm^{2}$ is $v$. If the electron density in copper is $9 \times 10^{28} / m^{3}$, the value of $v$ in $mm / s$ is close to (Take, charge of electron to be $=1.6 \times 10^{-19} C$ )
(201 9Main, 9 Jan I)
(a) 0.02
(b) 0.2
(c) 2
(d) 3
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Solution:
- Relation between current $(I)$ flowing through a conducting wire and drift velocity of electrons $\left(v _d\right)$ is given as
$$ I=n e A v _d $$
where, $n$ is the electron density and $A$ is the area of cross-section of wire.
$$ \Rightarrow \quad v _d=\frac{I}{n e A} $$
Substituting the given values, we get
or
$$ \begin{aligned} & v=\frac{1.5}{9 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-6}} \\ & v=\frac{1.5 \times 10^{-3}}{72} m / s=0.2 \times 10^{-4} m / s \\ & v=0.02 mm / s \end{aligned} $$
