
330 Forensic Engineering Fundamentals
For a solid conductor, D = 0. So i
z
= CI
o
(ν) and external radii = a. Applying boundary con-
ditions at
ρ = 0, B = 0,
(),|∇× =
i
z
δρ
0
(10.51)
at
ρ
=
za
0
iCIv C
Iv
oo
o
=⇒=
=
()
()|
ρ
(10.52)
then
i
Iv
Iv
Ijp
Ijpa
i
z
oi
oa
o
o
o
o
==
()|
()|
[(
[(
1
2
1
2
ρ
(10.53)
For numerical computation, since I
o
is complex, it must be broken down into its real and
imaginary parts, so for argument x,
Ijxber xjbe
ooo
[( )] ()
1
2
=+
(10.54)
Using ber and bei functions and multiplying by e
jωt
and taking the real part, the current
density in a solid wire of radius “a” and i
Β
= 1 is
i
berp beip
berpabei p
z
oo
oo
=
+
+
22
22
1
2
1
2
1
2
1
2
1
2
() ()
()