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:

Use the force on an electron between collisions to get the drift velocity, then use I = nAev_d. For the last part the current is the same, so v_d is proportional to 1/r^2.

Answer 1

Step 1: Understanding the Concept:
Free electrons in a metal move randomly with high thermal speeds. Their average velocity is zero, so there is no net current without a field. They collide with the positive ions again and again. Let \(\tau\) be the average time between two collisions (the relaxation time).

Step 2: Motion of an electron in the field (part i):
The field \(\vec{E}\) gives each electron a force \(-e\vec{E}\). So each electron has an acceleration:
\[ \vec{a} = -\frac{e\vec{E}}{m} \]
Take the i-th electron. Its velocity just after its last collision was \(\vec{u}_i\). If it has moved for a time \(t_i\) since then, its velocity is:
\[ \vec{v}_i = \vec{u}_i - \frac{e\vec{E}}{m}\,t_i \]

Step 3: Average velocity does not depend on time:
Take the average over all N electrons. The thermal velocities \(\vec{u}_i\) point in all directions, so their average is zero. The average of \(t_i\) is the relaxation time \(\tau\). So:
\[ \vec{v}_d = \langle \vec{v}_i \rangle = -\frac{e\vec{E}}{m}\,\tau \]
After every collision the electron loses the extra velocity it gained and starts again from a random direction. So the gain is the same in every interval. This is why the average velocity settles to a constant value, called the drift velocity, and does not grow with time.

Step 4: 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 \(v_d \Delta t\) of a cross-section cross it.
Number of electrons crossing = \(n A v_d \Delta t\).
Charge crossing = \(n A v_d \Delta t \cdot e\).
\[ I = \frac{\Delta Q}{\Delta t} = n e A v_d \]

Step 5: Why current starts almost instantly (part ii):
The drift speed is only a few mm per second, so one electron takes a very long time to travel along the wire. But the electric field is set up in the whole circuit at nearly the speed of light, about \(3 \times 10^{8}\ \text{m s}^{-1}\).
Free electrons are already present everywhere in the wire. As soon as the circuit is closed, the field acts on all of them at once, and all start drifting together. So current flows in every part of the circuit almost at the same instant.

Step 6: Ratio of drift velocities (part iii):
The wires are in series, so the same current I flows in both. Both are copper, so n and e are the same. Using \(I = n e A v_d\) with \(A = \pi r^2\):
\[ v_d \propto \frac{1}{r^2} \]
\[ \frac{v_{d1}}{v_{d2}} = \frac{r_2^2}{r_1^2} = \frac{2^2}{3^2} = \frac{4}{9} \]

Final Answer:
The average velocity is \(\vec{v}_d = -e\vec{E}\tau/m\), the current is \(I = n e A v_d\), and the drift velocities in the two wires are in the ratio 4 : 9. \[ \boxed{v_{d1} : v_{d2} = 4 : 9} \]