Home / Math and Physics Files / Physics / Transmission Lines / Notes By Chapter and Appendix / Appendix B MagJ
App B_7 reader exercise REVIEWED
DOCX · 20.9 KB
Open DOCX file
Handwritten-style calculation notes in Phil's own words, a reader exercise for Appendix B section B.7. It computes the vector potential A_z from a surface magnetization current and a volume current using the 3D integral of 1/R with a log cutoff Λ. It compares the result to the earlier 2D method of B.6, finding the non-Λ terms agree and working out the leftover Λ terms. Some equations are lost in text extraction.
AI-written summary; may contain errors. This description is approximate.
Extracted text (machine-read; may contain errors)
These are notes for my Appendix B exercise in B.7. Keep !!
2A(x) = -μ1Jm(x) -μ2Jc(x) // compare to (B.6.1) which is 2D version
A(x) = (1/4π) ∫dV' [μ1Jm(x') +μ2Jc(x')] (1/R) R = |x-x'|
Az(x) = (1/4π) (1/2πa) I { (μ1-μ2) ∫dS' (1/R) + (2/a) μ2∫dV' (1/R) } // compare (B.6.4)
∫surfdS' (1/R) = ∫(adθ') !Syntax Error, Idz' (1/R) = a !Syntax Error, I dθ' !Syntax Error, Idz' (1/ )
∫dV' (1/R) = ∫(r'dθ') ∫dr'!Syntax Error, Idz' (1/R) = !Syntax Error, I r'dr' !Syntax Error, I dθ' !Syntax Error, Idz' (1/ )
!Syntax Error, Idz' (1/ ) = -2ln(s/Λ)
∫dS' (1/R) = - 2a !Syntax Error, I dθ' ln(s/Λ)
∫dV' (1/R) = - 2!Syntax Error, I r'dr' !Syntax Error, I dθ' ln(s/Λ)
s2 = (x-x')2+ (y-y')2 = R2
∫dS' (1/R) = - a !Syntax Error, I dθ' ln(R2) + a!Syntax Error, I dθ' ln(Λ2) = - a !Syntax Error, I dθ' ln(R2) + 2πa ln(Λ2)
∫dV' (1/R) = - !Syntax Error, I r'dr' !Syntax Error, I dθ' ln(R2) + !Syntax Error, I r'dr' !Syntax Error, I dθ' ln(Λ2)
= - !Syntax Error, I r'dr' !Syntax Error, I dθ' ln(R2) + ln(Λ2) 2π (1/2)a2
Now
!Syntax Error, I dθ' ln(R2) = 2 !Syntax Error, Idx ln [r'2 +r2-2rr' cosx] = 2 Q(a,r)
!Syntax Error, I r'dr' !Syntax Error, I dθ' ln(R2) = !Syntax Error, Ir' dr' 2 Q(r',r)
So we then have
Az(x) = (1/4π) (1/2πa) I { (μ1-μ2) ∫dS' (1/R) + (2/a) μ2∫dV' (1/R) }
= (1/4π) (1/2πa) I { (μ1-μ2) [ - a !Syntax Error, I dθ' ln(R2) + 2πa ln(Λ2)]
+ (2/a) μ2 [- !Syntax Error, I r'dr' !Syntax Error, I dθ' ln(R2) + ln(Λ2) 2π (1/2)a2] }
= (1/4π) (1/2πa) I { (μ1-μ2) [ - a 2 Q(a,r) + 2πa ln(Λ2)]
+ (2/a) μ2 [-!Syntax Error, Ir' dr' 2 Q(r',r) + ln(Λ2) 2π (1/2)a2] }
= - (I/4π2a) { (μ1-μ2) [ a Q(a,r) - πa ln(Λ2)]
+ (1/a) μ2 [!Syntax Error, Ir' dr' 2 Q(r',r) - ln(Λ2) π a2] }
= - (I/4π2a2) { (μ1-μ2) [ a2 Q(a,r) - πa2 ln(Λ2)]
+ 2 μ2 [!Syntax Error, Ir' dr' Q(r',r) - ln(Λ2) π a2] }
Using the 2D method I got instead
Az = - [I/(4π2a2)] { (μ1-μ2) (a/2) ∫dS' ln(R2) + μ2∫dV' ln(R2) }
= - [I/(4π2a2)] { (μ1-μ2) (a/2) 2a Q(a,r) + μ22 !Syntax Error, Ir' dr' Q(r',r) }
= - [I/(4π2a2)] { (μ1-μ2) (a/2) 2a Q(a,r) + μ22 !Syntax Error, Ir' dr' Q(r',r) }
= - [I/(4π2a2)] { (μ1-μ2) a2 Q(a,r) + μ2 2 !Syntax Error, Ir' dr' Q(r',r) }
The non-Λ terms agree!!!! The Λ terms are these
= - (I/4π2a2) { (μ1-μ2) [- πa2 ln(Λ2)] + 2μ2 [- ln(Λ2) π a2] }
= + (I/4π2a2) πa2 ln(Λ2) { (μ1-μ2) + 2μ2 }
= + (I/4π) ln(Λ2) {μ1+ μ2 }