This preview shows page 1. Sign up to view the full content.
Substituting the values given, we obtain
4
2
3
(30.0
m)(7.80 10 A)
0.70 m
5.22 10 A
2 (4.00 10
)
(35.0 m)(35.0 m
0.70 m)
i
π
−
Ω⋅
×
==
×
×Ω
+
.
34. We follow the procedure used in Sample Problem 265.
Since the current spreads uniformly over the hemisphere, the current density at any given
radius
r
from the striking point is
2
/2
JI r
=
. From Eq. 2610, the magnitude of the
electric field at a radial distance
r
is
2
2
w
w
I
EJ
r
ρ
,
where
30
m
w
=Ω
⋅
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Any
 Physics, Current

Click to edit the document details