the nearest one should contribute an upward component to the field at
P
. The current
elements are all equivalent, as is reflected in the horizontaltranslational symmetry built
into this problem; therefore, all vertical components should cancel in pairs. The field at
P
must be purely horizontal, as drawn.
(b) The path used in evaluating
G
G
Bd
s
z
⋅
is rectangular, of horizontal length
∆
x
(the
horizontal sides passing through points
P
and
P'
respectively) and vertical size
δ
y >
∆
y
.
The vertical sides have no contribution to the integral since
G
B
is purely horizontal (so the
scalar dot product produces zero for those sides), and the horizontal sides contribute two
equal terms, as shown next. Ampere’s law yields
00
1
2.
2
Bx
x
B
µ
λµλ
∆=
∆ ⇒
=
79. The “current per unit
x
length” may be viewed as current density multiplied by the
thickness
∆
y
of the sheet; thus,
λ
=
J
∆
y
. Ampere’s law may be (and often is) expressed in
terms of the current density vector as follows
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Current

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