Question

A wire of diameter 0.02 metre contains \[{10^{28}}\] free electrons per cubic metre. For an electrical current of 100 A, the drift velocity of the free electrons in the wire is nearly.
A. \[1 \times {10^{ - 19}}\] m/s
B. \[5 \times {10^{ - 10}}\] m/s
C. \[2 \times {10^{ - 4}}\] m/s
D. \[8 \times {10^3}\] m/s

Answer 1

Hint:The drift velocity is the rate of distance covered by the charge particle between two consecutive collisions from neighbouring atoms in an atom. In this problem we are going to use the expression of drift velocity of the free electrons.

Formula used:
\[{v_d} = \dfrac{J}{{ne}}\]
where \[{v_d}\] is the drift velocity, \[J\] is the current density, n is the number of free electrons per unit volume and e is the charge on the electron.

Complete step by step solution:
When potential is applied across the wire then there is an electric field generated inside the wire which applies force on the free electrons. The applied force on the free electrons makes them move inside the wire. The average velocity gained by the free electron is called the drift velocity of the electron.

The time taken to cover the distance between two consecutive collisions is called the relaxation time period for the charge in motion. The current in the wire is given as 40 amperes.
\[i = 100A\]
The diameter of the wire is 0.02 metre, so the area of cross-section of the wire is,
\[A = \dfrac{{\pi \times {{\left( {0.02} \right)}^2}}}{4}{m^2}\]
\[\Rightarrow A = 3.14 \times {10^{ - 4}}{m^2}\]

The density of the free electron is given as \[{10^{28}}\].
\[n = {10^{28}}\]
Using the formula of the drift velocity,
\[{v_d} = \dfrac{J}{{ne}}\]
As the J is the current density of the wire, so it is equal to the electric current flowing per unit cross-sectional area of the wire,
\[J = \dfrac{i}{A}\]

On substituting the expression for the current density in the formula of drift velocity, we get
\[{v_d} = \dfrac{i}{{neA}}\]
Putting the values, we get
\[{v_d} = \dfrac{{100}}{{{{10}^{28}} \times 1.6 \times {{10}^{ - 19}} \times 3.14 \times {{10}^{ - 4}}}}m/s \\ \]
\[\Rightarrow {v_d} = 1.99 \times {10^{ - 4}}m/s \\ \]
\[\therefore {v_d} \approx 2 \times {10^{ - 4}}m/s\]
Hence, the drift velocity of the electron in the given wire is approximately \[2 \times {10^{ - 4}}\] m/s.

Therefore, the correct option is C.

Note: The drift velocity depends on the cross-section of the wire (A), the number of free electrons per unit volume (n) and the magnitude of the electric current. It does not depend upon the length of the wire.