Ampere's Law and Magnetic Fields

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SUMMARY

The discussion centers on determining the outside current density (Jout) in a coaxial cable configuration to achieve a net magnetic field of zero for radii greater than c. The participants confirm that the magnetic field (H) can be zero if the net current is zero, which occurs when the currents in the inner and outer conductors are equal in magnitude but opposite in direction. The relationship between the current densities is established as (area of inner conductor) * Jo = (area of outer conductor) * Jout, emphasizing the need to know the cross-sectional areas of both conductors to solve the problem accurately.

PREREQUISITES
  • Understanding of Ampere's Law and its application in magnetic fields
  • Knowledge of current density concepts in electromagnetism
  • Familiarity with coaxial cable configurations and their properties
  • Ability to calculate areas of circular cross-sections
NEXT STEPS
  • Study the application of Ampere's Law in different geometries
  • Learn about magnetic field calculations in coaxial cables
  • Explore the concept of current density and its implications in electromagnetic theory
  • Investigate the effects of varying current densities on magnetic fields
USEFUL FOR

Students of electromagnetism, electrical engineers, and anyone involved in the design or analysis of coaxial cables and their magnetic field interactions.

datran
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Homework Statement


I have a coaxial cable with current density Jo in the center, with radius a, going in -z_hat direction. This generates a magnetic field. The outside of the cable, radius c, also carries a current density Jout going in the +z_hat direction. This generates its own magnetic field.

Find the value of the outside current density to make the magnetic field 0 for r > c

Homework Equations


I used Ampere's equation.


The Attempt at a Solution



I do not use Ampere's equation explicitly (starting from dot product and such), but conclude that H = 0 if the net current = 0. and then find out Jout in terms of Jo.

Is this correct thinking? I mean, if I drew a loop that was outside of the coaxial cable, the Inet would be 0, but there is still an H-field being contributed by both the current densities.

How else would I go solving this problem then?

Thanks!
 
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We need to know the cross-sectional area of the outide conductor.

You are right in assuming the net current = 0 for H to be 0. If the inside conductor cross-sectional area = a then you know the inside current is Joπa2 but we also need to know the outside area & it can't be πc2 obviously.
 
Ok thank you! This was more of a conceptual question than anything. So if the two current densities are in opposite direction but equal to each other in magnitude, this will cancel out the magnetic field entirely?

I'm lost in the understanding because if we take an amperian loop outside of the coaxial cable and assume that the currents were different (meaning they each generate their own magnetic field, which does not cancel each other out), the enclosed current will be 0, but that does not necessarily guarantee the magnetic field is 0 right?
 
datran said:
Ok thank you! This was more of a conceptual question than anything. So if the two current densities are in opposite direction but equal to each other in magnitude, this will cancel out the magnetic field entirely?

I'm lost in the understanding because if we take an amperian loop outside of the coaxial cable and assume that the currents were different (meaning they each generate their own magnetic field, which does not cancel each other out), the enclosed current will be 0, but that does not necessarily guarantee the magnetic field is 0 right?

It's not the current densities that cancel each other to attain H = 0, it's the currents. That's why you have to know the cross-sectional area of the outer conductor. Then (area of inner conductor) * Jo = (area of outer conductor) * Jout to give ∫Hds = I = net current = 0 so by symmetry H = 0 everywhere along any closed loop outside the outer conductor.
 

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