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 .