Magnetic field intensity, flux density and magnetization of coax cable

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SUMMARY

The discussion focuses on calculating the total current enclosed by the inner conductor of a coaxial cable with specific parameters: an inner conductor radius of 1 cm and an outer conductor radius of 10 cm. The conductivity in the region between the conductors is zero, with a relative permeability (๐œ‡r) of 100. The enclosed current is derived using cylindrical coordinates and involves integrating the current density function ๐‘ฑ(๐‘Ÿ) = (10^4)(๐‘’^-(r/a)^2)(az). The integration process utilizes substitution to simplify the calculation, confirming the application of Ampรจre's Law in this context.

PREREQUISITES
  • Cylindrical coordinate system
  • Understanding of Ampรจre's Law
  • Basic calculus for integration
  • Electromagnetic theory, specifically concepts of magnetic field intensity and flux density
NEXT STEPS
  • Study the application of Ampรจre's Law in cylindrical coordinates
  • Learn about magnetic field intensity and flux density in coaxial cables
  • Explore advanced integration techniques in calculus
  • Investigate the effects of varying conductivity and permeability in electromagnetic fields
USEFUL FOR

Electrical engineers, physics students, and professionals working with electromagnetic theory and coaxial cable design will benefit from this discussion.

AndrewC
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Homework Statement
A cylindrical conducting rod of radius a = 1 cm has a non-uniform current density ๐‘ฑ(๐‘Ÿ) = ๐’‚z J0 ๐‘’^-(r/a)^2 (A/m2) and is surrounded by a cylindrical conducting surface of radius b = 10 cm carrying a current I0 in the opposite (-az) direction. The region between the two conductors is filled with a material having conductivity sigma = 0 and ๐œ‡r = 100, whereas ๐œ‡r = 1 for the conductors. Assuming J0 = 1 x 10^4 A/m2 and I0 = 1 A, find:

a) The magnetic field intensity H, flux density B and magnetization M for r < a
b) The magnetic field intensity H, flux density B and magnetization M for a < r < b
c) The magnetic field intensity H, flux density B and magnetization M for r > b
Relevant Equations
Amperes circuital law:
โˆฎ๐โˆ™d๐ฅ= ๐œ‡0 ๐ผ๐‘’๐‘›๐‘
โˆฎ๐‡โˆ™d๐ฅ= ๐ผ๐‘’๐‘›๐‘
Magnetization:
๐‡= ๐๐œ‡0โˆ’๐Œ
๐Œ=๐œ’๐‘š ๐‡
Inner conductor radius = 1cm
outer conductor radius = 10cm
region between conductors has conductivity = 0 & ๐œ‡r = 100
๐œ‡r = 1 for inner and outer conductor
Io = 1A(-az)
๐‘ฑ(๐‘Ÿ) = (10^4)(๐‘’^-(r/a)^2)(az)

Problem has cylindrical symmetry, use cylindrical coordinate system.

Find the total current enclosed by inner conductor for r<a:

Ienc = (0,r)โˆซ (10^4)(๐‘’^-(r/a)^2)(2ฯ€r)dr

= 2ฯ€*10^4(โˆซ(r๐‘’^-(r/a)^2)dr

let t = (r/a)^2, dt = (2rdr)/a^2

Ienc = a^2(ฯ€*10^4)โˆซ(e^-t)dt from 0 to โˆšt(a^2)

Ienc = a^2(ฯ€*10^4)[-e^-t] from 0 to โˆšt(a^2)

At this point I started questioning whether I was doing this right. Would appreciate any pointers on proper setup of amperes law.
 
Physics news on Phys.org
Hi. Area integration would be
[tex]\int_0^a e^{-(r/a)^2} dA<br /> <br /> = 2\pi \int_0^a e^{-(r/a)^2} r dr<br /> <br /> = 2\pi a^2 \int_0^1 e^{-r^2} r dr<br /> <br /> = \pi a^2 \int_0^1 e^{-r^2} d(r^2)=...[/tex]
 

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