Induced Voltages and Inductance
183
20.18
From
B v
ε
=
, the required speed is
A
(
)(
)
(
)(
)
0.500 A
6.00
1.00 m s
2.50 T
1.20 m
v
Ω
=
A
A
IR
B
B
ε
=
=
=
20.19
B
v
ε
⊥
=
A
v
G
, where
is the component of the magnetic field perpendicular to the velocity
. Thus,
B
⊥
(
)
(
)
(
)
6
50.0
0
T
sin58.0
60.0 m
300 m s
0.763 V
1
ε
−
°
=
=
×
20.20
The speed of the beam after falling freely for 9.00 m, starting from rest
(
)
, is
0
0
y
v
=
(
)
(
)
(
)
2
2
0
2
0
2
9.80 m s
9.00 m
13.3 m s
y
y
y
v
v
a
y
=
+
∆
=
+
−
−
=
B
v
Since the induced emf is
ε
⊥
=
A
B
, where
⊥
is the component of the magnetic field
perpendicular to the velocity
v
G
, we find
(
)
(
)
(
)
6
3
18.0
10
T
13.3 m
2.87
10
V
2.87 mV
12.0 m
s
ε
−
−
=
×
=
×
=
20.21
(a)
Observe that only the horizontal component,
B
h
, of Earth’s magnetic field is
effective in exerting a vertical force on charged particles in the antenna. For the
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 Spring '09
 Physics, Inductance, Work, Magnetic Field, Electric charge, Bh, ε

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