Question

The number density of free electrons in a copper conductor as estimated is $8.5 \times {10^{28}}{m^{ - 3}}$. How long does an electron take to drift form one end of a wire 3.0 m long to its other end? The area of cross- section of the wire is $2.0 \times {10^{ - 6}}{m^2}$ and it is carrying a current of 3.0A.

Answer 1

Hint: In this problem we are going to write first all the forces which are acting on the rod. These include gravitational force, normal force and frictional force. And then by applying force balance we are going to get normal and friction force that the floor exerts on the rod.
When we give potential difference across some metal, an electric field is created within the metal piece. Due to the presence of the electric field, the free electrons of the metals start moving from lower to higher potential and a current flow is created in the opposite direction. The average uniform velocity of the free electrons is known as the drift velocity for that metal.

Formula Used:
$i = nAe{v_d}$

Complete answer:
Dependence of produced current on the cross-sectional area, drift velocity and number density of electron is shown by:
$i = nAe{v_d}$----(1)
Where,
i is the current through the conductor,
n is the free electron density within the conductor,
e is electron charge magnitude, $e = 1.6 \times {10^{ - 19}}C$
${v_d}$ is the drift velocity

First, rewrite the eq. (1) to find an expression of ${v_d}$in terms of other variables
${v_d} = \dfrac{i}{{nAe}}$
$ \Rightarrow {v_d} = \dfrac{3}{{8.5 \times {{10}^{28}}{m^{ - 3}} \times 2.0 \times {{10}^{ - 6}}{m^2} \times 1.6 \times {{10}^{ - 19}}}}$
$ \Rightarrow {v_d} = 1.103 \times {10^{ - 4}}$

Now the time an electron takes to drift form one end of a wire 3.0 m long to its other end is given by
$t = \dfrac{l}{{{v_d}}}$

Substituting the values, we get
$t = 27200$sec

Note: Many students have mistaken about the direction of drift velocity of the electrons. They think that the current direction is the same as the direction of drift velocity. But its totally wrong. Since, electrons are negatively charged particles so they always move from lower potential to higher potential but the direction of current is to be from higher to lower potential. Hence, their directions are completely opposite.