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 r₁ and r₂ (r₂ < r₁).
Choose the correct option as one moves from P to Q :

Solution

Physics sir · 2 steps

The current stays the same all along PQ, so as the wire gets thinner, the electrons must move faster.

1. Current is fixed. PQ is in series with the battery, so the same current I flows through every cross-section from P to Q. Charge cannot pile up anywhere in between.

   I = neAv_d ⇒ v_d = I/(ne π r²)

   Quantity	Depends on	From P to Q (r falls from r₁ to r₂)
   Drift velocity v_d	I/neA	increases
   Electric field E	ρ J = (ρ I)/A	increases
   Current I	same at every section	stays constant

2. So. The area A = π r² shrinks from P to Q while I, n and e stay fixed, so the drift velocity increases. The answer is (a).

The idea to keep. In a wire of varying thickness, the current is the same everywhere. Current density, electric field and drift velocity all vary as 1/A, so they are largest where the wire is thinnest.

Answer from the NTA final answer key. Solution by PrepWiser.