Question

What is meant by drift velocity? A current of 0.5 amperes flows in a wire of radius 0.5mm. If the number of free electrons in the wire be 4$ \times {10^{26}}$ per ${m^2}$, calculate their drift velocity.

Answer 1

Hint: The solution is in two parts. First the definition of drift velocity and second the numerical. Drifting means slow movement toward something that means it is all about movement and here it is said about the charged particle. The numerical is based on formula of drift velocity i.e. ${V_d}$ =$\dfrac{I}{{nAq}}$ 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 q is the charge.

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:
Now coming to the numerical part
We are given :
Current of 0.5 amperes = I
Radius is 0.5mm
The number of free electrons in the wire be 4$ \times {10^{26}}per$ ${m^2}$= n
We have to calculate drift velocity and formula for drift velocity is ${V_d}$ =$\dfrac{I}{{nAq}}$…….eqn(1)
 where ${V_d}$=drift velocity, I is equal to current, n equals the number of free electrons in the wire, A is the area of cross-section and q is the charge.
Here, it is important to note that a wire is in the form of a cylinder hence the formula for area will be used for the cylinder.
A=$\Pi {r^2}$ (r is the radius of wire given in question)
Calculating the area by putting the value radius is 0.5mm gives A=0.78×${10^{ - 6}}m$
Putting values in eqn(1).
Drift velocity is ${V_d}$ =$\dfrac{I}{{nAq}}$
Substituting the value we get, ${V_d}$ = $\dfrac{{0.5}}{{4 \times {{10}^{28}} \times 0.78 \times {{10}^{ - 6}} \times 1.6 \times {{10}^{ - 19}}}}$
Which is equals to 9.95$ \times {10^{ - 5}}$

The drift velocity ${V_d}$ =9.95$ \times {10^{ - 5}}$ $\dfrac{m}{{\sec }}$

Note: Depend on electric field: This shows that the drift velocity is either in the direction parallel or antiparallel to the direction of the electric field, increases with increase of charge and electric field but decreases with the increase in the mass of the particle.
Relation between current and drift velocity: Current is the flow of free charges, such as electrons and ions. Drift velocity ${V_d}$ is the average speed at which these charges move.