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291
28.2.
IDENTIFY:
A moving charge creates a magnetic field as well as an electric field.
SET UP:
The magnetic field caused by a moving charge is
0
2
sin
4
qv
B
r
µφ
π
=
, and its electric field is
2
0
1
4
e
E
r
=
P
since
q = e
.
EXECUTE:
Substitute the appropriate numbers into the above equations.
71
9
6
0
21
1
2
sin
4
10 T m/A (1.60 10
C)(2.2 10 m/s)sin90
44(
5
.
3
1
0
m
)
qv
B
r
ππ
−−
−
×⋅
×
×
°
==
×
= 13 T, out of the page.
92
2
1
9
11
1
2
0
1
(9.00 10
燦
m/C)
(1
.60 10
燙
)
5.1 10
燦
/C,
4(
5
.
3
1
0
爉
)
e
E
r
−
−
×
=
×
×
P
toward the electron.
EVALUATE:
There are enormous fields within the atom!
28.8.
IDENTIFY:
Both moving charges create magnetic fields, and the net field is the vector sum of the two. The
magnetic force on a moving charge is
mag
sin
Fq
v
B
φ
=
and the electrical force obeys Coulomb’s law.
SET UP:
The magnetic field due to a moving charge is
0
2
sin
4
qv
B
r
µ
=
.
EXECUTE:
(a)
Both fields are into the page, so their magnitudes add, giving
B
=
B
e
+
B
p
=
0
22
ep
sin90
4
ev
ev
rr
+
°
B
=
()
19
0
11
1.60 10
燙
845,000 m/s
4
(5.00 10
爉
)
(4.00 10
爉
)
−
×+
××
B
= 1.39
×
10
–3
T = 1.39 mT, into the page.
(b)
Using
0
2
sin
4
qv
B
r
=
, where
r
=
41 nm and
= 180°
−
arctan(5/4) = 128.7°, we get
9
4
41
0
T
m
/
A
(
1
.
6
1
0
燙
)(845,000 m/s)sin128.7
?
.58 10 T,
4
(4
0
爉
)
B
−
−
×
°
×
×
into the page.
(c)
19
4
17
mag
sin90
(1.60 10
C)(845,000 m/s)(2.58 10 T)
3.48 10
N,
v
B
−
=°
=
×
×
=
×
in the +
x
direction.
2
1
9
2
12
elec
0
(9.00 10
燦
燙
)
(1/ 4
) /
5.62 10
N,
0
爉
)
Fe
r
−
−
−
×
=
×
×
P
at 51.3° below the +
x
axis measured
clockwise.
EVALUATE:
The electric force is much stronger than the magnetic force.
28.23.
IDENTIFY:
The net magnetic field at the center of the square is the vector sum of the fields due to each wire.
SET UP:
For each wire,
0
2
I
B
r
=
and the direction of
B
r
is given by the righthand rule that is illustrated in
Figure 28.6 in the textbook.
EXECUTE:
(a)
and
(b)
B
= 0 since the magnetic fields due to currents at opposite corners of the square cancel.
(c)
The fields due to each wire are sketched in Figure 28.23.
0
cos45
cos45
cos45
cos 45
4
cos45
4
cos45
2
abcd
a
μI
BB
B
B
B
B
π
r
+°
=
°
=
°
.
(10 cm)
(10 cm)
10 2 cm
0.10 2 m
r
=+
=
=
, so
7
4
(4
10
T m A) (100 A)
4
cos 45
4.0 10
T, to the left.
2 (0.10 2 m)
π
B
−
−
=
×
EVALUATE:
In part (c), if all four currents are reversed in direction, the net field at the center of the square would
be to the right.
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Chapter 29
Figure 28.23
28.30.
IDENTIFY:
The magnetic field at the center of a circular loop is
0
2
I
B
R
µ
=
. By symmetry each segment of the loop
that has length
l
∆
contributes equally to the field, so the field at the center of a semicircle is
1
2
that of a full loop.
SET UP:
Since the straight sections produce no field at
P
, the field at
P
is
0
4
I
B
R
=
.
EXECUTE:
0
4
I
B
R
=
. The direction of
B
r
is given by the righthand rule:
B
r
is directed into the page.
EVALUATE:
For a quartercircle section of wire the magnetic field at its center of curvature is
0
8
I
B
R
=
.
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This note was uploaded on 09/22/2011 for the course PHY 09C 9c taught by Professor Zhu during the Spring '11 term at UC Davis.
 Spring '11
 Zhu

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