Question

The diameter of a copper wire is $2mm$, a steady current of $6.25{\text{A}}$ is generated by $8.5 \times {10^{28}}/{m^3}$ electrons flowing through it. Calculate drift velocity of conduction electrons.

Answer 1

Hint
To solve this question, we need to use the expression of current in terms of the drift velocity, the number density of the electrons and area of the cross section of the wire. Putting the given values into that expression will give the required answer.
The formula used to solve this question is
$\Rightarrow I = neA{v_d}$, where $I$ is the current, $n$ is the number density of the electrons, $A$ is the area of cross section of the conductor, and ${v_d}$ is the drift velocity of the electrons.

Complete step by step answer
We know that the electric current conducted by the conduction electrons is given by
$\Rightarrow I = neA{v_d}$
So the drift velocity of the conduction electrons is given by
$\Rightarrow {v_d} = \dfrac{I}{{neA}}$ (1)
We know that the area of cross section is equal to
$\Rightarrow A = \pi {r^2}$
Substituting$r = \dfrac{d}{2}$ , we have
$\Rightarrow A = \pi {\left( {\dfrac{d}{2}} \right)^2}$
$\Rightarrow A = \dfrac{{\pi {d^2}}}{4}$
Substituting in (1), we get the drift velocity as
$\Rightarrow {v_d} = \dfrac{I}{{ne\left( {\dfrac{{\pi {d^2}}}{4}} \right)}}$
Multiplying numerator and denominator by $4$
$\Rightarrow {v_d} = \dfrac{{4I}}{{ne\pi {d^2}}}$ (2)
According to the question, $I = 6.25A$ $n = 8.5 \times {10^{28}}/{m^3}$$d = 2mm = 2 \times {10^{ - 3}}m$
Also, we know that the value of $e = 1.6 \times {10^{ - 19}}C$
Substituting these in (2), we get
$\Rightarrow {v_d} = \dfrac{{4(6.25)}}{{(8.5 \times {{10}^{28}})(1.6 \times {{10}^{ - 19}})\pi {{\left( {2 \times {{10}^{ - 3}}} \right)}^2}}}$
On solving, we get
$\Rightarrow {v_d} = 0.146 \times {10^{ - 3}}m/s$
Hence, the drift velocity of the conduction electron is equal to $0.146 \times {10^{ - 3}}m/s$

Note
Always remember to convert all the units of the quantities given in the problem into SI units. If any one of them is of a different system of units, our result will get incorrect. Like in this problem, the diameter was given in millimetres, which is not in SI units. It was supposed to be converted into meters. In these types of simple calculation based problems, the units are intentionally given from different unit systems.