Question

There is a current of 1.34 Amperes in a copper wire whose area of cross-section normal to the length of the wire is $1{\text{ m}}{{\text{m}}^2}$. If the number of the free electrons per $c{m^2}$ is $8.4 \times {10^{22}}$ then, the drift velocity would be
A. 1.0 mm/sec
B. 1.0 m/sec
C. 0.1 mm/sec
D. 0.01 mm/sec

Answer 1

Hint: During the transition of the electrons in a conductor for the transmission of the current, electrons flow could be deferred by protons and other electrons and due to which the velocity of the moving electrons starred. Again due to the presence of the electric field, electrons gain velocity and again collide with other electrons, and this process goes on. The average velocity id the total distance covered by the electrons in the meantime and is known as the Drift velocity of the electrons.
Here in the question, we need to determine the drift velocity of the electrons moving in the copper wire. For this, we will use the formula $i = neA{v_d}$, where, n is the number of free electrons per square units, e is the charge on an electron, A is the area of the cross-section of the conductor and ${v_d}$ is the drift velocity of the electrons.

Complete step by step answer:
From the question, it is given that:
$
  i = 1.34{\text{ Amperes}} \\
  A = 1{\text{ m}}{{\text{m}}^2} \\
  n = 8.4 \times {10^{22}}{\text{ electrons/c}}{{\text{m}}^2} \\
 $

Here, the area of the cross-section is given in square millimetres which should be converted to the centimetre square by dividing the given quantity by 100. So, we get $A = 0.01{\text{ c}}{{\text{m}}^2}$.

Also, the charge of an electron is given as, $e = 1.6 \times {10^{ - 19}}$

Now, following the formula for the current in the conductor as $i = neA{v_d}$.
So, substitute $A = 0.01{\text{ c}}{{\text{m}}^2}$, $e = 1.6 \times {10^{ - 19}}$, $i = 1.34{\text{ Amperes}}$ and $n = 8.4 \times {10^{22}}{\text{ electrons/c}}{{\text{m}}^2}$ in the formula $i = neA{v_d}$ to determine the value of the drift-velocity of the electrons as:

$
  i = neA{v_d} \\
  1.34 = 8.4 \times {10^{22}} \times 1.6 \times {10^{ - 19}} \times 0.01 \times {v_d} \\
  {v_d} = \dfrac{{1.34}}{{8.4 \times {{10}^{22}} \times 1.6 \times {{10}^{ - 19}} \times 0.01}} \\
   = \dfrac{{1.34}}{{13.44 \times {{10}^{(22 - 19 - 2)}}}} \\
   = \dfrac{{1.34}}{{13.44 \times 10}} \\
   \approx 0.01{\text{ cm/sec}} \\
   \approx {\text{0}}{\text{.1 mm/sec}} \\
 $

Hence, the drift velocity in the copper wire (conductor) carrying a current of 1.34 Amperes having a cross-section area of $1{\text{ m}}{{\text{m}}^2}$ with the number of the free electrons per $c{m^2}$ is $8.4 \times {10^{22}}$ is 0.1 mm/sec.

So, the correct answer is “Option C”.

Note:
Students should be careful while using the units of the measuring terms, here as we can see that the area is given in square millimetres and the number of conductors is given in per centimetres square, so, we need to convert either of them before substituting in the formula.