Question

Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section $5\; mm^2$, is $v$. If the electron density in copper is $9\times 10^{28}/m^3$ the value of $v$ in mm/s is close to (Take charge of electron to be $=1.6\times 10^{-19}C$)
A.)0.2
B.)3
C.)2
D.)0.02

Answer 1

Hint: This question can be solved, if we use the formula for drift velocity as the value of other parameters have already been given. But we need to keep in mind to convert the given units into SI systems.

Formula used:
Drift velocity, $v_d=\dfrac{I}{nqA}$.

Complete step by step solution:
When a particle is placed in an electric field, the average velocity attained by it due to the presence of the field is known as the drift velocity. Since, in a conductor, there are so many free charges, the drift velocity is comparably very small.

So, when a current is passed through a wire of cross-sectional area A of a conductor having a charge density $n$, and if each carrier will have a charge on it as q, then the drift velocity, $v_d$ of the carriers is given by.

$v_d=\dfrac{I}{nqA}$.

Now, according to the question current, I = 1.5 A, area of cross-section of the wire, A $=5\; mm^2 =5\times 10^{-6}\; m^2$, electron density $n=9\times 10^{28}/m^3$ and charge of each electron, $q=1.6\times 10^{-19}V$.

Thus, drift velocity, $v=\dfrac{1.5}{9\times 10^{28}\times 5\times 10^{-6}\times 1.6\times 10^{-19}}=0.02\times 10^{-3}\; m/s = 0.02\; mm/s$

Hence, option d is the correct answer.

Note: The unit conversion should be done to bring them into the SI system. The electric field generated is due to the flow of current through the wire. Without the presence of an electric field, the electrons move randomly inside a conductor and at fermi velocity and these random movements result in a zero-average velocity. The application of electric field provides a common direction to the electrons and thus average velocity turns to drift velocity.