Question

A copper wire has a square cross-section 2.0 mm on a side. It carries a current of 8 A and the density of the electrons is \[8 \times {10^{28}}{m^{ - 3}}\]. The drift speed of electrons is equal to
A. \[0.156 \times {10^{ - 3}}\]m/s
B. \[0.156 \times {10^{ - 2}}\]m/s
B. \[3.12 \times {10^{ - 3}}\]m/s
D. \[3.12 \times {10^{ - 2}}\]m/s

Answer 1

Hint: The velocity of drift is exactly proportional to the current. When an electric field is applied to a conductor, electrons leap towards the wire's high potential terminal. Unless the conductor is subjected to an electric field, the internal electrons flow at random speeds and trajectories. Here we use the expression of drift velocity to find the accurate answer.

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.
\[i = 8A\]

The square cross-sectional of the wire is of side length 2.0 mm. So, the area of cross-section of the wire is,
\[A = {\left( {2 \times {{10}^{ - 3}}} \right)^2}{m^2}\]
\[\Rightarrow A = 4 \times {10^{ - 4}}{m^2}\]
The density of the free electron is given as \[8 \times {10^{28}}\]
\[n = 8 \times {10^{28}}\]
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
\[{v_d} = \dfrac{8}{{8 \times {{10}^{28}} \times 1.6 \times {{10}^{ - 19}} \times 4 \times {{10}^{ - 6}}}}\,m/s \\ \]
\[\therefore {v_d} = 0.156 \times {10^{ - 3}}m/s\]
Hence, the drift velocity of the electron in the given wire is \[0.156 \times {10^{ - 3}}\] m/s.

Therefore, the correct option is A.

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.