30: Problem 30.55

INTRO:
The two wires in the figure (Figure 1) are made of the same material.

Figure 1

-----------------------------------------------------------------------------------------------------
PART A:
What is the current in the 2.0-mm-diameter segment of the wire?

SOLUTION:
Since both segments are connected in series, current is the same 
I₁ = I₂ = 2 A

-----------------------------------------------------------------------------------------------------
PART B:
What is the electron drift speed in the 2.0-mm-diameter segment of the wire?

SOLUTION:
Using eq. (30.13)... J = I/A = n_e⋅e⋅v_d
rearranging... I = A⋅n_e⋅e⋅v_d
I₁ = I₂ 

→ A₁⋅n_e1⋅e⋅v_d1 = A₂⋅n_e2⋅e⋅v_d2​

The electron charge cancels out, and also the intro says that the wires are the same material, so n_e1= n_e2 ... 

→ A₁⋅v_d1 = A₂⋅v_d2
→v_d2 = A₁⋅v_d1 / A₂

= (π/4⋅d₁²)(v_d1) / (π/4⋅d₂²)
= (d₁²)(v_d1) / (d₂²)
= (1×10^-3 m)²(2×10^-4 m/s) / (2×10^-3 m)²
⇒ v_d2 = 5×10^-5 m/s

= 50 μm/s​