Question

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 : 

 

• A. Drift velocity of electron increases.
• B. Electron current decreases.
• C. Electric field decreases.
• D. All of these

Hint:

In a non-uniform conductor with a steady current, remember these key proportionalities: Current (I) is constant. Current Density (J) and Electric Field (E) are inversely proportional to the cross-sectional area ($J, E \propto 1/A$). Drift Velocity ($v_d$) is also inversely proportional to the area ($v_d \propto 1/A$).

Answer 1

The Correct Option is A

As we move from point P to point Q, the radius of the conductor decreases from $r_1$ to $r_2$. Consequently, the cross-sectional area $A = \pi r^2$ also decreases.
In a steady state, the electric current (I) is constant throughout the conductor due to the conservation of charge. Electron current is therefore also constant. So, option (B) is incorrect.
The relationship between current (I), drift velocity ($v_d$), and cross-sectional area (A) is given by the equation:
$I = n e A v_d$
where n is the number density of free electrons and e is the charge of an electron, both of which are constants for the material.
Since I, n, and e are constant, we can write $v_d \propto \frac{1}{A}$.
As one moves from P to Q, the area A decreases. Therefore, the drift velocity $v_d$ must increase. Option (A) is correct.
The relationship between electric field (E), current density (J), and conductivity ($\sigma$) is $J = \sigma E$. Current density is $J = I/A$.
So, $E = \frac{J}{\sigma} = \frac{I}{\sigma A}$.
Since I and $\sigma$ are constant, $E \propto \frac{1}{A}$.
As area A decreases from P to Q, the electric field E must increase. So, option (C) is incorrect.
Since only option (A) is correct, option (D) is incorrect.