Animated Solution for Physics - Current Electricity: In the given figure, a battery of emf $E$ is connected across a conductor $PQ$ of length $l$ and different area of cross-sections having radii $r_1$ and $r_2$ ($r_2<r_1$). Choose the correct option as one moves from $P$ to $Q$.

The physics of non-uniform conductors is a classic test of how well you understand the continuity of charge and the microscopic mechanisms of electric current. Let's dive into the elegant relationship between geometry, electric fields, and the speed of electrons!

A Funnel for Electrons

Imagine you are watering your garden with a hose. If you pinch the end of the hose, the water shoots out faster. Why? Because the total volume of water flowing through the hose every second must remain constant. If the opening is smaller, the water must speed up to get through in time.

This exact same principle applies to our conductor $PQ$. The battery establishes a steady state, meaning charge cannot pile up anywhere inside the material. The total electric current $I$—the number of coulombs passing through any cross-section per second—must be perfectly constant from end $P$ to end $Q$.

Slicing the Conductor

To understand what happens inside, let's take a microscopic slice of the conductor with a tiny length $dl$ and a local radius $r$.

The resistance of this tiny element is given by the standard formula:

According to Ohm's law, the potential drop $dV$ across this microscopic slice is simply the current multiplied by this resistance:

Electric Field and Radius

Now, we need to find the electric field $E$ driving the electrons. The electric field is the potential gradient—the rate at which voltage drops over distance:

Substituting our expression for $dV$, the $dl$ beautifully cancels out:

Look closely at this master equation. The current $I$ is constant, and the resistivity $ho$ is a fixed property of the material. Therefore, the electric field is inversely proportional to the square of the radius ($E\propto \frac{1}{r^2}$). As the conductor narrows, the electric field intensifies!

Drift Velocity in Action

Finally, how does this affect the electrons? The drift velocity $v_d$ of an electron is determined by the electric field accelerating it, balanced by collisions with the atomic lattice:

This tells us that $v_d\propto E$.

As we move from $P$ to $Q$, the radius $r$ decreases. A smaller radius creates a much stronger electric field. This intensified electric field exerts a greater force on the free electrons, causing their drift velocity $v_d$ to increase.

Pro-Tip: You can also solve this instantly using Current Density ($J$). Since $J=\frac{I}{A}$, a smaller area means a higher $J$. Because $E=\rho J$ and $v_d=\frac{J}{ne}$, both the electric field and drift velocity must increase as the area shrinks!