A coaxial cable consists of a core cylindrical wire and a coaxial cylindrical shell, as illustrated in the Figure below. Consider a cable made of copper, which has resistivity of
ρ=1.70×10−8Ω·m. Thecorewirehasaradiusof r =1.0mm. Theinnerradiusofthe 1
outer shell is r = 5.0 mm, and the outer radius of the shell is r = 6.0 mm. The length of 23
the cable is L = 1.5 m. The electron number density in the material of the wire is n = 2.0×1022 1/m3. A battery with electromotive force of E = 1.50 V is connected
between the core and shell of the cable at one end, while at the other end a resistance-less wire connects the core and shell of the cable. Hint: Beware of unit conversion!
a. (6) What is the resistance Rc of the core wire? And what is the resistance Rs of the
b. (6) What is the current I through the core wire?
c. (4) The current density J is uniform in the conductors. What is its magnitude in the
core wire and in the shell?
This is just for practice, I have the answer key. I understand a but I’m not quite sure I totally understand the next two parts.
V = IR
J = I/A
The Attempt at a Solution
For a, Rc = 8.12E-3
Ra = .74E-3
For b.) V = IR
1.5 = I(Rc + Ra)
I don’t have a clear understanding of why the resistance is the sum of the resistances of both wires here? Are they somehow in series? If so, why?
And for part c I have a similar question. What is the area being used in the calculation and why?