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Chapter 30
197
P30.26
(a)
B
NI
r
inner
TmA
A
m
T
==
×⋅
×
=
−
µ
π
0
73
2
41
0
9
0
0
1
4
0
1
0
20
7
0
0
360
ej
af
.
.
.
(b)
B
NI
r
outer
A
m
T
×
=
−
0
2
21
0
9
0
01
4
0
130
194
.
.
.
*P30.27
We assume the current is vertically upward.
(a)
Consider a circle of radius
r
slightly less than
R
. It encloses no current so from
Bs
⋅=
z
dI
0i
n
s
i
d
e
Br
bg
=
we conclude that the magnetic field is
zero .
(b)
Now let the
r
be barely larger than
R
. Ampere’s law becomes
BR
I
2
0
πµ
=
,
so
B
I
R
=
0
2
.
FIG. P30.27(a)
The field’s direction is
tangent to the wall of the cylinder in a counterclockwise sense .
(c)
Consider a strip of the wall of width
dx
and length
A
. Its width is so small
compared to 2
R
that the field at its location would be essentially
unchanged if the current in the strip were turned off.
The current it carries is
I
Idx
R
s
=
2
up.
The force on it is
FB
u
p
=×
=
F
H
G
I
K
J
×=
I
Idx
R
I
R
Id
x
R
s
AA
A
22
4
00
2
intopage
radiallyinward
.
FIG. P30.27(c)
The pressure on the strip and everywhere on the cylinder is
P
F
A
x
Rd
x
I
R
=
0
2
0
2
2
4
2
A
A
inward .
The pinch effect makes an effective demonstration when an aluminum can crushes itself as
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics, Current

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