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 $$Q$$.

We need to determine how physical properties like drift velocity, electric field, and electron current change as we move from point $$P$$ to point $$Q$$ along a non-uniform conductor.

1. Analyze the Geometry of the Conductor

From the problem details :

• The conductor has a tapering cross-section.
• The radius at $$P$$ is $$r_1$$, and the radius at $$Q$$ is $$r_2$$, given that $$r_2 < r_1$$.
• As you move from $$P$$ to $$Q$$, the radius decreases, which means the cross-sectional area ($$A = \pi r^2$$) decreases.

2. Analyze individual Physical Quantities

• Electron Current ($$I$$):
  The conductor is connected in a single series loop with a battery. Due to the conservation of charge, the steady-state electric current must remain constant at every cross-section along the conductor.

  $$\text{Current } I = \text{constant}$$

  Therefore, the statement "Electron current decreases" is incorrect.


• Drift Velocity ($$v_d$$):
  The relationship between electric current and drift velocity is given by the formula:

  $$I = n e A v_d \implies v_d = \frac{I}{n e A}$$

  Where $$n$$ is the free electron density and $$e$$ is the elementary charge (both constant for a given material). Since current $$I$$ is constant and the area $$A$$ decreases as we approach $$Q$$, the drift velocity must increase:

  $$v_d \propto \frac{1}{A}$$



• Electric Field ($$E$$):
  According to the microscopic form of Ohm's Law, current density ($$J$$) is related to the electric field by $$J = \sigma E$$, where $$\sigma$$ is the electrical conductivity. Since $$J = \frac{I}{A}$$:

  $$E = \frac{I}{\sigma A} \implies E \propto \frac{1}{A}$$

  As area $$A$$ decreases from $$P$$ to $$Q$$, the electric field strength must increase. Therefore, the statement "Electric field decreases" is incorrect.

Conclusion

As one moves from $$P$$ to $$Q$$, the narrowing area causes the drift velocity of the electrons to increase.