{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Notes+9+fall+2004

# Notes+9+fall+2004 - Notes 9 fall 2004 Conductive materials...

This preview shows pages 1–3. Sign up to view the full content.

Notes 9 fall 2004.doc Conductive materials Ampere’s law in free space: E j H o G G G ωε = × In a conductive medium it becomes: ( ) E j E E j j H G G G G G ε′ ω + ε ′ ω = ε ′ ε′ ω = × Alternatively we’ve written it as E j E J J H d c G G G G G G ε′ ω + σ = + = × Comparing the first terms of the last two equations we can see that ε′ ω = σ , or ω σ = ε ′ We can then write the permittivity for conductive media as ω σ ε′ = ε′ ε′ = ε j j . Dividing the boxed equation by we find the loss tangent that we mentioned earlier: ε′ tangent loss ε′ ω σ = ε′ ε′ Note that the ratio of the magnitudes of the phasor conduction current density to the phasor displacement current density is ° ε′ ω σ = ε′ ω σ = ε′ ω σ = 90 J J j E j E J J d c d c which shows that the displacement current density leads the conduction current density by (think ICE) ° 90 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Notes 9 fall 2004.doc Slightly conducting media We can make a useful approximation if we recall that for 1 x << , . The approximation gets better as x gets smaller (try it yourself). ( ) nx 1 n x 1 ± ± A slightly conducting medium is defined by a loss tangent that is much less than one, i.e., 1 << ε′ ω σ or . In this case the complex propagation constant, , can be c d J J >> jk approximated by ( ) ( ) ε′ µ ω + ε′ µ σ = ε′ µ ω + ε′ ω ε′ µ ωσ = ε′ ω σ ε′ µ ω ε′ ω σ ε′ µ ω = ε′ µ ω = ε′ ε′ µ ω = µε ω = ω σ j 2 j 2 2 j 1 j j 1 j j j j j j jk 2 1 { } ε′ µ σ = = α 2 jk e { } ε′ µ ω = = β
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern