Magnetic field intensity, flux density and magnetization of coax cable

  • Thread starter AndrewC
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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.
 

Answers and Replies

  • #2
mitochan
294
133
Hi. Area integration would be
[tex]\int_0^a e^{-(r/a)^2} dA

= 2\pi \int_0^a e^{-(r/a)^2} r dr

= 2\pi a^2 \int_0^1 e^{-r^2} r dr

= \pi a^2 \int_0^1 e^{-r^2} d(r^2)=...[/tex]
 

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