50.
(a) We denote the copper wire with subscript c and the aluminum wire with subscript a.
R = ρ
a
L
A
=
(2.75
× 10
−8
Ω
·m)(1.3 m)
(5.2
× 10
−3
m)
2
= 1.3
× 10
−3
Ω .
(b) Let R = ρ
c
L/(πd
2
/4) and solve for the diameter d of the copper wire:
d =
4ρ
c
L
πR
=
4(1.69
× 10
−8
Ω
·m)(1.3 m)
π(1.3
× 10
−3
Ω)
= 4.6
× 10
−3
m .