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ELECTROMAGNETIC
FIELDS AND WAVES
w.1.
Model:
The net magnetic flux over a closed surface is zero.
Visualize:
Please refer to Ex34.1.
Solve:
Because we can't enclose a "net pole" within a surface,
Q,,,
=
fB.di
=
0.
Since the magnetic field is
uniform over each face of the box, the total magnetic flux around the box is
(1
cm
x
2 cm)(2 T
2
T
+
1 T
+
5 T)
t
(1 cm
x
1 cm)(2 T)+
Qunknown
=
0
j
j
E,,
.
arli
=
0.0002 T
m2
3
B~,,,,~
COS@
=
2.0 T
The angle 8 must be
180".
Because 8is the angle between
$
and the outward normal of
di,
the field
B
is directed
into the face.
34.2. Model:
Visualize:
magnetic field is in the direction of the negative zaxis.
Solve:
Electric and magnetic fields exert forces on charged particles. Assume the fields are uniform.
Please refer to Figure Ex34.2. The electric field
is
in the direction of the positive yaxis and the
Substituting into the Lorentz force law,
V
m
F,
=
q(
E
+
C
x
E)
=
(1.6
x
C)
1.0 x 106j

+
(2.0
x
1073
m
/
s)
x (0.10i
T))
=
(3.2;
+
1.6j)
x
N
3
=
dm
x
N
=
3.58
x
N
8
=
tad(
N
=
26.6"
1.6
x
3.2
x
101~
N
1
The angle
8
is
above the positive xaxis.
34.3. Model:
Solve:
Electric and magnetic fields exert forces on charged particles. Assume the fields are uniform.
Substituting into the Lorentz force law,
I$,,,
=
q(E+
G
x
=
(1.6
x
C)
5.0
x
IO6;
m
/
x
(0.lOi
T
=
(3.2
x 1014)(1
3)
N

(8.0
x
10l.S)
N=
(3.2
x

4.8
x
10l4j) N
34.4.
Model:
Visualize:
Please refer to Figure Ex34.4. The magnetic force on the negative electron by the righthand rule
directed
downward.
So
that the electron
is
undeflected, we must apply an electric field to cause an electric force
directed upward. That is, the electric field must point
downward.
Solve:
Assume that the electric and magnetic fields are uniform fields.
For the electron to not deflect,
F;,
=
4
a
clC
x
4
=
eE
=j
E
=
vBsin90"
=
(2.0
x
10' m
/
s)(O.OIO T)
=
2.0
x
10' V
/
m
34.5. Model:
Assume the electric and magnetic fields are uniform.
Visualize:
Solve:
The force on the proton. which
is
the sum of the electric and magnetic forces, is
Please refer to Figure Ex34.5.
F
=
+
=
Fcos30°f
+
Fsin30'3
=
(2.77;
+1.603)x
N
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Chapter 34
Since
v’
points out of the page, the magnetic force is
F,
=
e<
x
B
=
1.66
x
lOI33
N.Thus
.
=eE=FF,=2.77~10‘~~N~~=~,/e=1.73x10~~
V/m
That is, the electric field is
=
(1.73~10~
V/m, left).
34.6. Model:
Solve:
Assume that the electric and magnetic fields are uniform fields.
Substituting into the Lorentz force law,
=
q(E+
x
B)
3
(9.6
x
N)(;

i)
=
(1.60
x
.)[E+
(5.0
x
lo6; m
/
s)
x
(0.103 T)]
E
= (6.0 x
io5 N
/
c)(;


x
c)L
=
(6.0;
+
i.oL
) x
c
34.7. Model:
Apply the Galilean transformation of velocity.
Solve:
(a)
In the laboratory frame S, the speed of the proton is
v
=
j(1.41
x
lo6 m
/
s)’
+(1.41
x
/
s)’
=
2.0
x
lo6 m
/
s
The angle the velocity vector makes with the positive yaxis is
=
45”
1
1.41~10~
m/s
e
=
tan’
(b)
In the rocket frame
S’,
we need to fist determine the vector
V’.
Equation 34.5 yields:
V’
=
V

v
=
(1.41
x
lo6;
+
1.41
x
106y) m
/
s
(1.0
x
lo6;) m
/
s
=
(0.41
x
lo6;
+
1.41
x
106j) m
/
s
The speed of the proton is
v’=4(O.41x1O6 m/s)’ +(1.41x106 m/s)’ =1.47x106 m/s
The angle the velocity vector makes with the positive y’axis is
=
16.2”
1
0.41~10~
8’
=
tan’
34.8. Model:
Apply the Galilean transformation of fields.
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 Spring '08
 Medvedev

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