Current Electricity Ques 5
- 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) $3$
Show Answer
Answer:
Correct Answer: 5.(a)
Solution:
Formula:
Electric Current in a Conductor:
- 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
$ \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} $
BETA
