Magnetic field intensity, flux density and magnetization of coax cable

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 2K views
AndrewC
Messages
6
Reaction score
0
Homework Statement
A cylindrical conducting rod of radius a = 1 cm has a non-uniform current density 𝑱(𝑟) = 𝒂z J0 𝑒^-(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 𝜇r = 100, whereas 𝜇r = 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:
∮𝐁∙d𝐥= 𝜇0 𝐼𝑒𝑛𝑐
∮𝐇∙d𝐥= 𝐼𝑒𝑛𝑐
Magnetization:
𝐇= 𝐁𝜇0−𝐌
𝐌=𝜒𝑚 𝐇
Inner conductor radius = 1cm
outer conductor radius = 10cm
region between conductors has conductivity = 0 & 𝜇r = 100
𝜇r = 1 for inner and outer conductor
Io = 1A(-az)
𝑱(𝑟) = (10^4)(𝑒^-(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)∫ (10^4)(𝑒^-(r/a)^2)(2πr)dr

= 2π*10^4(∫(r𝑒^-(r/a)^2)dr

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

Ienc = a^2(π*10^4)∫(e^-t)dt from 0 to √t(a^2)

Ienc = a^2(π*10^4)[-e^-t] from 0 to √t(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]