Question

(a) An electric field $\vec{E}$ is established across the ends of a cylindrical conductor of length $L$ and area of cross-section $A$. Discuss how electrons attain an average velocity, independent of time. Hence, obtain a relation between current in the conductor and this ‘average velocity’ of electrons.

(b) (i) This ‘average velocity’ is found to be a few $\text{mm}\,\text{s}^{-1}$ for currents in range of a few amperes. How then is current established almost the instant a circuit is closed?

(ii) Two copper wires having their radii in the ratio of $3:2$ are connected in series across a battery. Find the ratio of the drift velocities of the electrons in the wires.

Hint:

Drift velocity is the acceleration times the mean collision time, and I = nAe v_d. For wires in series I and n are equal, so v_d goes as 1/r^2.

Answer 1

Step 1: Understanding the Question:
The question has three tasks: explain why electrons reach a steady drift velocity, link current to that velocity, and answer two follow-up points. The second follow-up is a numerical on two wires in series.

Step 2: Part (i), why the average velocity does not change with time:
The field \( \vec{E} \) pushes each free electron with a force \( -e\vec{E} \), so its acceleration is \( a = -eE/m \). Between two collisions, its velocity grows with time.
The electron then collides with the lattice ions and loses its extra velocity. The collisions occur at random times, with an average gap \( \tau \) called the relaxation time.
Right after a collision, the velocity is random, and its average is zero. So the average velocity gained from the field is the acceleration times the average time between collisions:
\[ v_d = a\tau = -\dfrac{eE}{m}\tau \]
\( v_d \) is called the drift velocity. It stays constant, because the gain from the field is balanced on average by the loss in collisions.

Step 3: Relation between current and drift velocity:
Let \( n \) be the number of free electrons per unit volume. In a time \( \Delta t \), all electrons within a length \( v_d\Delta t \) cross a section of the conductor.
The charge crossing is \( \Delta Q = n e A v_d \Delta t \). So the current is:
\[ I = \dfrac{\Delta Q}{\Delta t} = n e A v_d \]

Step 4: Part (ii)(first), why current starts instantly:
The drift speed is small, but free electrons are present everywhere in the wire. When the circuit is closed, the electric field is set up in the whole wire at nearly the speed of light.
Every electron in the wire begins to drift at the same instant, so the current starts everywhere at once. No electron has to travel from the battery to the bulb.

Step 5: Part (ii)(second), two wires in series:
In series, the same current \( I \) flows in both wires. Both are copper, so \( n \) is the same. From \( I = neAv_d \), we get \( v_d \propto 1/A \propto 1/r^2 \).
\[ \dfrac{v_{d1}}{v_{d2}} = \dfrac{A_2}{A_1} = \dfrac{r_2^2}{r_1^2} = \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9} \]

Final Answer:
The drift velocity is \( v_d = -eE\tau/m \) and the current is \( I = neAv_d \). Current starts at once because the field is set up everywhere at nearly the speed of light. The ratio of drift velocities is \[ \boxed{v_{d1} : v_{d2} = 4 : 9} \]