Question

A 2-ampere current flows in a conductor which has 1×${10^{24}}$ free electrons per meter. What is their average drift velocity?
A) 1.25$\dfrac{m}{s}$
B) 125000$\dfrac{m}{s}$
C) 3×${10^8}$ $\dfrac{m}{s}$
D) 1.25×${10^{ - 5}}$ $\dfrac{m}{s}$

Answer 1

Hint: This question is based on the formula of the drift velocity and it is given as drift velocity is ${V_d}$ =$\dfrac{I}{{\eta eA}}$ where, ${V_d}$=drift velocity, I is equal to current, n equals to number of free electrons in the wire, $A$ is the area of cross-section and e=$1.6 \times {10^{ - 24}}$.

Complete step by step answer:
Step 1:
The definition of drifting will solve the whole of the question itself. Before we start it is important to know about the drifting. Drifting is to become driven or carried along (as by a current of water, wind, or air) a balloon drifting in the wind. b: to move or float smoothly and effortlessly.
Definition of drifting velocity:
The average velocity attained by random moving electrons when the external electric field is applied, which causes the electrons to move towards one direction is called the drift velocity.
It is given by=$ neJ$,
Where $J$ is the Current density, $e$ is the charge on an electron and $n$ is the electron density.

Step 2:
Drift velocity is ${V_d}$ =$\dfrac{I}{{\eta eA}}$
We have to calculate drift velocity and the formula for drift velocity is ${V_d}$ =$\dfrac{I}{{\eta eA}}$ ………………..(1)
 Where,
${V_d}$ is the drift velocity,
$I$ is equal to current,
$n$ equal to the number of free electrons in the wire,
$A$ is the area of cross-section and $e=1.6 \times {10^{ - 24}}$
Now putting the values in equation (1)
$ \Rightarrow $ ${V_d}$ =$\dfrac{2}{{1 \times {{10}^{24}} \times 1.6 \times {{10}^{ - 24}} \times 1}}$ which give a value of drift velocity

Now on solving the upper part we will get ${V_d}$ =1.25$\dfrac{m}{s}$. Hence option A is correct.

Note:
Remember the formula for drift velocity because it has many questions based on it and any value can be asked to find depending on the formula. The area must be taken care of every time because with the change in object the area will change. For example, for wire, the area will be of the cylinder and different for other shapes.