Magnetic Forces in and around a coaxial cable

AI Thread Summary
The discussion revolves around calculating the magnetic field in a coaxial cable consisting of an inner solid wire and an outer thin tube, both carrying equal but opposite currents. The participant is uncertain about modeling the outer tube, questioning whether it should be treated as a solenoid and how to apply Ampere's Law correctly. It is clarified that the outer tube does not have turns and can be modeled as many thin wires. The key takeaway is that when applying Ampere's Law, the magnetic field calculations should consider only the inner wire's current inside the wire, and both currents cancel each other outside the tube. This leads to the conclusion that the magnetic field outside the coaxial cable is zero.
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Homework Statement


So I'm doing a problem set and the question refers to a coaxial cable, I am told that the cable consists of an inner solid cable radius (R1), which is surrounded by a outer thin tube with radius (R2) , both have the same current though them but opposite direction. I have to create equations for the magnetic field inside the inner cable, between the two, and outside the tube. The picture I am given is a filled in circle inside of a larger circle.
My questions about doing the problem are:
1. Should the thin tube be modeled as a solenoid? But then I don't know the number of turns
2. If I use Amperes Law, then wouldn't I get a I_through of zero?

Homework Equations


magnetic field inside a current carrying wire is =[(µIr)/2piR2]

The Attempt at a Solution


Honestly I don't even know where to start because I don't know how to model the outer circle
 
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welcome to pf!

hi 032050! welcome to pf! :smile:

(have a mu: µ and try using the X2 and X2 icons just above the Reply box :wink:)
032050 said:
1. Should the thin tube be modeled as a solenoid? But then I don't know the number of turns

no …

there's no turns …

if you like, you can model it as a million very thin wires running along the tube, each with one-millionth of the current :wink:
 
Thanks, so if I model the outer tube using Ampere's Law to calculate the Ithrough wouldn't that give zero current and then zero magnetic field?
 
maybe :rolleyes:

where? :smile:
 
ohhh..
is it inside the wire I would only use the current component of the inner wire, between the two I would still only use the inner wire, and then outside i would add the two to get Ithrough=0?
 
yup! :biggrin:
 
Thankyouuuu! :approve::smile::biggrin:
 
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