Question

There are $8.4 \times {10^{22}}$ free electrons per $cm^3$ in copper, the current in the wire is $0.21A$. Then the drift velocity of the electron in a copper wire of $1 mm^2$ cross section will be:
(A) $2.12 \times {10^{ - 5}}m/s$
(B) $0.78 \times {10^{ - 5}}m/s$
(C) $1.56 \times {10^{ - 5}}m/s$
(D) $\text{None of these}$

Answer 1

In this question, the concept of the total charge flow through the wire is used, that is, it depends on the total number of electrons, the area of cross-section of the wire, the drift velocity of the electrons, the charge of the single electro, and the time taken.

Complete step by step solution:
So, what current has flowed. We need work out how much charge has passed point $A$
We know that the total charge carried by electron in wire can be written as,
$Q = nAvte$
Here, the total number of electrons is $n$, the area of cross-section of the wire is $A$, the drift velocity is $v$ and the time taken is $t$, and the charge of the electron is $q$.
As we know that the charge per unit time is the current which can be written as,
$I = \dfrac{Q}{t}$
Now, we substitute the value of the total charge in the above equation as,
$\Rightarrow I = \dfrac{{nAvte}}{t}$
$\Rightarrow I = nAvq$
By rearranging we get,
$v = \dfrac{I}{{neA}}......\left( 1 \right)$
Where, the Number of free electrons, $n = 8.4 \times {10^{28}}m^3$, the current in the wire is $I = 0.21A$, and the cross section of the wire is $A = 1mm^2$.
Convert the area from millimeter to meter.
$\Rightarrow A = {10^{ - 6}}m^2$
Putting the value of $n,A$, and $I$ in equation (1)
$v = \dfrac{{0.21}}{{\left( {8.4 \times {{10}^{28}}} \right) \times \left( {1.6 \times {{10}^{ - 19}}} \right) \times \left( {{{10}^{ - 6}}} \right)}}$
Solve the above expression we get,
$\Rightarrow v = 1.56 \times {10^{ - 5}} m/s$
Therefore, the drift velocity of electrons in a copper wire of $1 mm^2$ cross section will be $v = 1.56 \times {10^{ - 5}}m/s$.

Hence, the correct option is (C).

Note: Do not forget to convert the area from millimeter to meter. if you forget to convert the area then the final answer of the drift velocity is not calculated correctly. The final answer is that the drift velocity should be in meter per second as the drift velocity in the given choices is in meter per second.