Magnetic flux outside cylindrical conductor

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SUMMARY

The magnetic flux outside a cylindrical conductor, specifically in a coaxial cable configuration, is definitively zero. This conclusion is derived using Ampere's Law, which states that the magnetic field (B) outside the coaxial cable is zero due to equal and opposite currents flowing in the inner and outer conductors. The expressions for magnetic flux density B(r) in the regions defined by the radii a, b, and c are established, confirming that B = 0 for r > c. The calculations for the regions 0 < r < a, a < r < b, and b < r < c are provided, with the final region confirming zero magnetic flux density.

PREREQUISITES
  • Understanding of Ampere's Law
  • Familiarity with magnetic flux density concepts
  • Knowledge of coaxial cable structure and current distribution
  • Basic proficiency in mathematical expressions involving calculus
NEXT STEPS
  • Study the derivation of magnetic fields in different geometries, focusing on cylindrical conductors
  • Learn about the application of Ampere's Law in various electromagnetic scenarios
  • Explore the implications of magnetic field theory in real-world applications, such as coaxial cables
  • Investigate the relationship between current distribution and magnetic field strength in conductors
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Students and professionals in electrical engineering, physicists studying electromagnetism, and anyone involved in the design or analysis of coaxial cable systems.

mmh37
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Hey everyone,

I have failed to show that the magnetic flux outside a cylindrical conductor is zero.

The problem goes like this:

a coaxial cable consists of a solid inner cylindrical conductor of radius a and an outer cylindrical conductor of inner and outer radius b and c. Distributed currents of equal magnitude I flow in opposite directions in the two conductors. Derive expressions for the magnetic flux density B(r) for each of these regions

1) 0<r <a

by amperes law I showed: [tex]B = \frac {K *l*r} {2*pi*a^2}[/tex]

where K is the permeability of free space, don't know how to write that with latex

2) a<r<b

again by amperes law: [tex]B = \frac {K *l} {2*pi*r}[/tex]

3) b<r<c

again by amperes law: [tex]B = \frac {K *l*(r^2-b^2)} {2*pi*r*(c^2-b^2)}[/tex]

4) c<r

I got stuck here, surely it must be zero...but how can that be shown?

thanks for your help - it's very much appreciated
 
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With the coaxial cable coming out of the paper at you, draw a circular closed path around the outside of it.

INTEG[B dot dL] around that closed path = mu * I(enclosed)

The total I enclosed by the circle is zero, since the current up the inner conductor is the same as the current down the outer conductor, so B (or H) has to be identically zero everywhere outside the coax.


Edit -- BTW, check your answer for (3). When r=c, you should get zero for H.
 
Last edited:
Thanks Berkeman - that's explained it and it's very much appreciated! :-) As for the third one it surely has to be (c^2-r^2)/(c^2-b^2)
 

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