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CHAPTER
29
Sources of the Magnetic Field
1*
∙
Compare the directions of the electric and magnetic forces between two positive charges, which move along
parallel paths (
a
) in the same direction, and (
b
) in opposite directions.
(
a
) The electric forces are repulsive; the magnetic forces are attractive (the two charges moving in the same
direction act like two currents in the same direction).
(
b
) The electric forces are again repulsive; the magnetic forces are also repulsive.
2
∙
At time
t
= 0, a particle with charge
q
= 12
µ
C is located at
x
= 0,
y
= 2 m; its velocity at that time is
v
=
30 m/s
i
. Find the magnetic field at (
a
) the origin; (
b
)
x
= 0,
y
= 1 m; (
c
)
x
= 0,
y
= 3 m; and (
d
)
x
= 0,
y
= 4 m.
(
a
) Use Equ. 291;
r
= –2 m
j
,
v
= 30 m/s
i
(
b
)
r
= –1 m
j
(
c
)
r
= 1 m
j
(
d
)
r
= 2 m
j
B
= 10
–7
q
(
v
×
r
)/
r
3
;
B
= –9
×
10
–12
T
k
B
= –3.6
×
10
–11
T
k
B
= 3.6
×
10
–11
T
k
B
= 9
×
10
–12
T
k
3
∙
For the particle in Problem 2, find the magnetic field at (
a
)
x
= 1 m,
y
= 3 m; (
b
)
x
= 2 m,
y
= 2 m; and (
c
)
x
=
2 m,
y
= 3 m.
(
a
) Use Equ. 291;
r
= 1 m
i
+ 1 m
j
;
v
= 30 m/s
i
(
b
)
r
= 2 m
i
(
c
)
r
= 2 m
i
+ 1 m
j
B
= 10
–7
q
(
v
r
)/
r
3
;
B
= 1.27
×
10
–11
T
k
B
= 0
B
= 3.22
×
10
–12
T
k
4
∙
A proton (charge +
e
) traveling with a velocity of
v
= 1
×
10
4
m/s
i
+ 2
×
10
4
m/s
j
is located at
x
= 3 m,
y
= 4 m at
some time
t
. Find the magnetic field at the following positions: (
a
)
x
= 2 m,
y
= 2 m; (
b
)
x
= 6 m,
y
= 4 m; and (
c
)
x
= 3 m,
y
= 6 m.
(
a
) Use Equ. 291;
q
= 1.6
×
10
–19
C;
r
= (–
i
–2
j
) m;
v
= (
i
+ 2
j
)
×
10
4
m/s;
B
= 10
–7
q
(
v
r
)/
r
3
;
(
b
)
r
= 3 m
i
(
c
)
r
= 2 m
j
B
= 0 (
v
and
r
are colinear)
B
= –3.56
×
10
–23
T
k
B
= 4
×
10
–23
T
k
5*
∙
An electron orbits a proton at a radius of 5.29
×
10
–11
m. What is the magnetic field at the proton due to the
orbital motion of the electron?
1. Determine the speed of the electron;
mv
2
/
r
=
ke
2
/
r
2
2. Use Equ. 291;
B
= (
0
e
2
/4
π
r
2
)
r
/m
k
; evaluate
v
=
r
/m
k
e
B
= 12.5 T
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View Full DocumentChapter 29
Sources of the Magnetic Field
B
6
∙∙
Two equal charges
q
located at (0, 0, 0) and (0,
b
, 0) at time zero are moving with speed
v
in the positive
x
direction (
v
<<
c
). Find the ratio of the magnitudes of the magnetic and electrostatic force on each.
Note that
v
and
r
, where
r
is the vector from one charge to the other, are at right angles. The field
B
due to the
charge at the origin at the location (0,
b
, 0) is perpendicular to
v
and
r
, i.e., in the
z
direction, and its magnitude is
B
= (
µ
0
/4
π
)
qv
/
b
2
.
The magnitude of the force on the moving charge at (0,
b
, 0) is
F
B
=
qvB
= (
0
/4
)
q
2
v
2
/
b
2
.
Applying the right hand rule the direction of the force is toward the charge at the origin; i.e., the magnetic
interaction between the two moving charges is attractive. The electrostatic force is repulsive and of magnitude
F
E
= (1/4
πε
0
)
q
2
/
b
2
. Thus
F
B
/
F
E
=
ε
0
0
v
2
=
v
2
/
c
2
, where
c
is the speed of light in vacuum.
7
∙
The BiotSavart law is similar to Coulomb's law in that both
(
a
) are inverse square laws.
(
b
) deal with forces on charged particles.
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 Spring '08
 drake
 Charge, Force

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