Question

There is a current of \[20\] amperes in a copper wire of \[{10^{ - 6}}\] square meter area of cross-section. If the number of free electrons per cubic meter is \[{10^{29}}\] , then the drift velocity is
A. $125 \times {10^{ - 3}}m/\sec $
B. $12.5 \times {10^{ - 3}}m/\sec $
C. $1.25 \times {10^{ - 3}}m/\sec $
D. $125 \times {10^{ - 4}}m/\sec $

Answer 1

Hint: The drift velocity is the average speed that charged particles like electrons achieve within a material as a result of an electric field. To find the value of drift velocity, we will use the relationship between drift velocity, current, area of cross section, number of electrons and charge on each electron.

Formula Used:
Drift velocity, $v = \dfrac{I}{{nAe}}$
where $v$ is the drift velocity of free electrons, $I$ is the current, $n$ is the number of electrons and $e$ is the charge on each electron $\left( {e = 1.6 \times {{10}^{ - 19}}C} \right)$.

Complete step by step solution:
Given: Current through the copper wire, $I = 20A$
Area of cross section, $A = {10^{ - 6}}{m^2}$
Number of electrons, $n = {10^{29}}$
And charge on each electron, \[e = 1.6 \times {10^{ - 19}}C\]

Drift is the slow progression to an object. The drift velocity is the average speed that charged particles—like electrons—achieve within a material as a result of an electric field. An electron will typically move randomly across a conductor at the Fermi velocity, producing an average velocity of zero. It is the average speed at which electrons "drift" in the presence of an electric field. The electric current is a result of the drift velocity (or drift speed).

We know that, $v = \dfrac{I}{{nAe}}$ . . .(1)
Substituting all the known values in equation (1), we get,
$v = \dfrac{{20}}{{{{10}^{29}} \times {{10}^{ - 6}} \times 1.6 \times {{10}^{ - 19}}}} \\ $
Solving further we get,
$v = \dfrac{{20}}{{1.6 \times {{10}^4}}} \\ $
$\therefore v = 1.25 \times {10^{ - 3}}\,m/\sec $

Hence, option C is the correct answer.

Note: While solving such numerical problems we should always remember to convert all the physical quantities into the same system of units, that is, CGS, SI etc. Also the area taken in the formula is not the complete area of the wire rather just the area of the cross section of the wire. The final answer of a numerical should always be written with its unit.