Question

Consider a wire of length $4\,m$ and cross sectional area $1\,m{m^2}$carrying a current of $2\,A$. If each cubic meter of the material contains ${10^{29}}$ free electrons, find the average time taken by an electron to cross the length of the wire.

Answer 1

Hint:Average velocity attained by random moving electrons, when an external electric is applied, which causes the electrons to move in only one direction is called Drift velocity. Current produced by the electrons towards high potential is called drift current.

Formula used:
$u = \mu E$
Where $u$ is the drift velocity, $\mu $ is the electron mobility and $E$ is the electric field.
Drift is calculated from the formula,
$i = neAvd$
Here $i$ is current, $n$ is number of electrons, $A$ is area of cross sectional and $vd$ is drift velocity.

Complete step by step answer:
Drift velocity occurs due the fact that when an electric field is applied to the randomly moving electron, they slightly tend to shift to the higher potential end, thus they gain a velocity while in random motion, to shift to higher potential and this is known as drift velocity.

Given measurements are,
$i = 2A \\
\Rightarrow n = {10^{29}}{m^3} \\
\Rightarrow e = 1.6 \times {10^{ - 19}}C \\
\Rightarrow A = 1m{m^2} = {10^{ - 6}}{m^2} $
Substituting all the values in drift formula we get,
$v_d = \dfrac{2}{{{{10}^{29}} \times 1.6 \times {{10}^{ - 19}} \times {{10}^{ - 6}}}} \\
\Rightarrow v_d = \dfrac{2}{{1.6 \times {{10}^4}}} \\
\Rightarrow v_d = \dfrac{1}{{8000}} $
We have calculated the drift value, now we have to calculate the time.As mentioned in the question, time taken by an electron to cross the length of the wire.Given length of the wire is 4m and time taken cross the length of the wire is,
$t = \dfrac{l}{{vd}}$
$l$ length of the wire is 4m and $v_d$ is drift velocity.
$t = \dfrac{4}{{\dfrac{1}{{8000}}}} \\
\therefore t = 32000\,s $
We are converting into seconds then we get, $t = 3.2 \times {10^4}$ seconds. Each electron time taken to cross the length of the wire is $3.2 \times {10^4}$ seconds.

Note:Drift velocity occurs in a current carrying body in the direction opposite to the direction of the electric field. The time has been calculated just using the normal concept of distance equal to product of time and velocity where distance here represents the length of wire that the electron has to travel and the speed as drift velocity.