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6/15/2011
1
Chapter
31
Electromagnetic Oscillations and Alternating
Current
In this chapter we will cover the following topics:
Electromagnetic oscillations in an LC circuit
Alternating current (AC) circuits with capacitors
Resonance in RCL circuits
Power in ACcircuits
Transformers, AC power transmission
(31  1)
A horizontal power line carries a current of 5000 A
from south to north. Earth's magnetic eld (60
μ
T) is
directed towards north and inclined downward 50
degrees to the horizontal. Find the magnitude and
direction of the magnetic force on 100 m of the line
due to the Earth's field.
(1)23 N, west
(2)
23 N, east
(3) 30N, west
(4) 30N, east
(5) none of these
50
B
N
I
Suppose this page is perpendicular to a uniform magnetic field
and the magnetic flux through it is 5Wb. If the page is turned
by 30
◦
around an edge the flux through it will be:
A. 2.5Wb
B. 4.3Wb
C. 5Wb
D. 5.8Wb
E. 10Wb
A car travels northward at 75 km/h along a straight road in a
region where Earth’s magnetic field has a vertical component of
0.50 × 10
4
T. The emf induced between the left and right
side, separated by 1.7m, is:
A. 0
B. 1.8mV
C. 3.6mV
D. 6.4mV
E. 13mV
L
C
The circuit shown in the figure consists of a capacitor
and an inductor .
We give the capacitor an initial
chanrge
and then abserve what happens.
The capacitor
will discharge th
C
L
Q
LC
Oscillations
rough the inductor resulting in a time
dependent current .
i
We will show that the charge
on the capacitor plates as well as the current
1
in the inductor oscillate with constant amplitude at an angular frequency
The total energy
in the circuit is t
q
i
LC
U
ϖ
=
2
2
he sum of the energy stored in the electric field
of the capacitor and the magnetic field of the inductor.
.
2
2
The total energy of the circuit does not change with time.
Thus
E
B
q
Li
U
U
U
C
dU
=
+
=
+
2
2
2
2
0
0.
1
0
dt
dU
q dq
di
dq
di
d q
Li
i
dt
C d
d q
L
t
dt
dt
q
dt
dt
dt
C
+
=
=
+
=
=
→
→
=
=
(31  2)
L
C
2
2
2
2
1
0
(
)
This is a homogeneous, second order, linear differential equation
which we have encountered previously. We used it to d
1
0
escribe
the simple harmonic oscillat
o
d q
L
q
dt
C
d q
q
dt
LC
+
=
=
→
+
eqs.1
2
2
2
0
with sol
r (SHO)
(
)
ution: ( )
cos(
)
d x
x
dt
x t
X
t
φ
+
=
=
+
eqs.2
(
29
If we compare eqs.1 with eqs.2 we find that the solution to the differential
equation that describes the LCcircuit (eqs.1)
is:
1
( )
cos
where
, and
is the phase angle.
The current
q t
Q
t
LC
=
+
=
(
29
sin
dq
i
Q
t
dt
=
= 
+
(
29
( )
co
s
q t
Q
t
=
+
1
LC
=
(31  3)
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2
L
C
(
29
(
29
(
29
2
2
2
2
2
2
2
2
2
2
The energy stored in the electric field of the capacitor
cos
2
2
The energy stored in the magnetic field of the inductor
sin
sin
2
2
2
The total energy
2
E
B
E
B
q
Q
U
t
C
C
Li
L
Q
Q
U
t
t
C
U
U
U
Q
U
ϖ
φ
=
=
+
=
=
+
=
+
=
+
=
(
29
(
29
2
2
2
cos
sin
2
The total energy is constant;
Q
t
t
C
C
+
+
+
=
energy is conserved
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This note was uploaded on 12/05/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Current, Force, Power

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