Transcript Lecture 5: Time-varying EM Fields

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Lecture 5: Time-varying EM Fields
Instructor:
Dr. Gleb V. Tcheslavski
Contact:
[email protected]
Office Hours: Room
2030
Class web site:
www.ee.lamar.edu/gleb/e
m/Index.htm
ELEN 3371 Electromagnetics
Fall 2008
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Introduction
Although “Time-varying EM fields” is the title of the present
lecture only, the rest of our course will be devoted to timevarying fields! These fields are generalizations of
electrostatic and magnetostatic fields studied earlier.
ELEN 3371 Electromagnetics
Fall 2008
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Faraday’s law of induction
We consider an electric
potential across an
inductor inserted in a
circuit.
The voltage across the inductor
can be expressed as
dI (t )
V (t ) L
dt
where L is the inductance [H], I(t) – time varying current passing through the
inductor.
ELEN 3371 Electromagnetics
Fall 2008
(5.3.1)
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Faraday’s law of induction
We have learned that a constant current induces magnetic
field and a constant charge (or a voltage) makes an electric
field. Similar phenomena happen when currents/voltages
are changing in time.
The relation between the electric and magnetic field
components is governed by the Faraday’s law…
d  m (t )
V (t )
dt
where
m (t )
(5.4.1)
Lentz’s law
is the total time-varying magnetic flux passing through a surface.
Voltage is induced in a closed loop that completely surrounds
the surface, through which the magnetic flux passes.
ELEN 3371 Electromagnetics
Fall 2008
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Faraday’s law of induction
Therefore, we consider an
individual turn of a coil (inductor)
as a closed loop.
At a glance: an electric current (an emf) is generated in a closed loop
when any change in magnetic environment around the loop occurs
(Faraday’s law). When an emf is generated by a change in magnetic
flux according to Faraday's Law, the polarity of the induced emf is such
that it produces a current whose magnetic field opposes the change
which produces it. The induced magnetic field inside any loop of wire
always acts to keep the magnetic flux in the loop constant.
ELEN 3371 Electromagnetics
Fall 2008
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Faraday’s law of induction
The total magnetic flux passing through the loop is
 m (t ) 
 B ds
(5.6.1)
s
The terminals of the loop are separated by an infinitely small distance,
therefore, we assume the loop as closed.
Recall from (3.20.2) that the electric potential can be expressed as
V (t ) 
 E dl
(5.6.2)
Combining two last equations with (5.4.1) yields the integral form of Faraday’s law:
d
 E dl   dt
ELEN 3371 Electromagnetics
Fall 2008
 B ds
s
(5.6.3)
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Faraday’s law of induction
Application of the Stokes’s theorem to the integral form of the Faraday’s
law of induction leads to
B
ds
t
s
 E dl    E ds   
s
(5.7.1)
Equality of integrals implies an equality of integrands, therefore:
B
 E  
t
Is a differential form of the Faraday’s law of induction.
ELEN 3371 Electromagnetics
Fall 2008
(5.7.2)
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Faraday’s law of induction (Ex)
A good illustration of the
Faraday’s law of induction is an
ideal transformer.
Find the voltage V2 at the loop 2
(secondary coil) if the voltage V1 is
applied to the loop 1 (primary coil).
The “ideal” assumption implies no
loss in the core and that the core
has infinite permeability.
V1   N1
d m
d m
;V2   N 2
dt
dt
V1 N 2
V2 
N1
ELEN 3371 Electromagnetics
Fall 2008
(5.8.1)
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Faraday’s law of induction
We assumed previously that the area of the loop s is not changing. It
does not always hold.
Let us consider a conductive bar
moving at a speed v through a
constant uniform magnetic field B.
The Lorentz force will cause a charge
separation within the conductor.
Charge separation will create an
electric field. Since the net force on the
bar is zero, the electric and magnetic
contributions must cancel each other.
Therefore:
Electric field is an induced field acting
along conductor and producing:
F
E   v B
q
b
V    v  B  dl
a
ELEN 3371 Electromagnetics
Fall 2008
(5.9.1)
(5.9.2)
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Faraday’s law of induction
Example: a Faraday’s disc generator
consists of a metal disc rotating with a
constant angular velocity  = 600 1/s in a
uniform time-independent magnetic field
with a magnetic flux density B = B0uz,
where B0 = 4 T. Determine the induced
voltage generated between the brush
contacts at the axis and the edge of the
disc, whose radius a = 0.5 m.
An electron at a radius  from the center has a velocity .
Therefore, the potential generated is:
B
a
a
0
V   E dl    v  B  dl   u  B0uz  d u   B0  d  
A
0
ELEN 3371 Electromagnetics
0
a
Fall 2008
 B0 a2
2
 300V (5.10.1)
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Faraday’s law of induction
We can evaluate a divergence for both sides of the Faraday’s law:
B

   E  
   B 
t
t
(5.11.1)
Since divergence of the curl is zero:
 B0
Or in the integral form:

B ds  0
(5.11.2)
(5.11.3)
Same as for MS fields, there is no magnetic monopole and magnetic
lines are continuous.
ELEN 3371 Electromagnetics
Fall 2008
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Equation of Continuity
 v
 J 
t
(5.12.1)
Current density is the movement of charge density. The continuity equation
says that if charge is moving out of a differential volume (i.e. divergence of
current density is positive) then the amount of charge within that volume is
going to decrease, so the rate of change of charge density is negative.
Therefore the continuity equation amounts to a conservation of charge.
ELEN 3371 Electromagnetics
Fall 2008
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Equation of Continuity
Example: charges are introduced into the interior of a conductor during the
time t < 0. Calculate the time needed for charges to move to the surface of the
conductor so the interior charge density v = 0 and interior electric field E = 0.
Let us combine the Ohm’s law and the continuity equation:
J   E   E   
The Poisson’s equation:
Therefore:
The solution will be:
ELEN 3371 Electromagnetics
v
 E

d v v

0
dt

v  v 0 e
Fall 2008


 v
t
(5.13.1)
(5.13.2)
(5.13.3)
 t
(5.13.4)
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Equation of Continuity
The initial charge density v decays to 1/e  37% of its initial value in a
relaxation time  = /.
(5.14.1)
For copper, the relaxation time is:
 0 8.85411012
19
 

1.5

10
s
7

5.8 10
(5.14.2)
For dielectric materials, the relaxation time can be quite big: hours or even days.
ELEN 3371 Electromagnetics
Fall 2008
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Displacement current
Maxwell’s experiment:
Connect together an AC voltage
source, an AC ammeter reading a
constant current I and an ideal
parallel plate capacitor…
1. How can the ammeter read any values if the capacitor is an open circuit?
2. What happens to the time-varying magnetic field that is created by the
current and surrounds the wire as we pass through the vacuum between the
capacitor’s plates?
The Ampere’s law:
ELEN 3371 Electromagnetics
 B  0 J
Fall 2008
(5.15.1)
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Displacement current
Applying the divergence operation to both sides of the Ampere’s law, we get:
and
  B  0
(5.16.1)
0 J  0
(5.16.2)
However, the last expression disagrees with the equation of continuity…
To overcome this contradiction, we introduce an additional current called a
displacement current, whose current density will be Jd. The continuity equation:
  D
v
D 

 J
0 J 
 J 

t
t
t 

ELEN 3371 Electromagnetics
Fall 2008
(5.16.2)
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Displacement current
Therefore, the displacement current density is:
D
Jd 
t
(5.17.1)
This is the current passing between the plates of the capacitor.
As a consequence, the postulate of Magnetostatic we discussed previously
must be modified as follows:
D

J
0
t
B
Differential form:
Integral form:
ELEN 3371 Electromagnetics

D 

dl    J 
 ds
0
t 
s
B
Fall 2008
(5.17.2)
(5.17.3)
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Displacement current
Example 1: Verify that the conduction current in the
wire equals to the displacement current between the
plates of the parallel plate capacitor in Figure. The
(5.18.1)
voltage source supplies: Vc = V0 sin t.
The conduction current in the wire is
dVc
Ic  C
 CV0cos t
dt
(5.18.2)
The capacitance of a parallel plate capacitor is:
Vc
E
d
The electric field between the plates:
The displacement flux density is:
C
A
d
(5.18.4)
V0
D   E   sin t
d
 V

   0 sin t 
D
d
 ds   A V  cos t  CV  cos t  I
ds   
Finally: I d   J d ds  
0
0
c

t

t
d
A
A
A
ELEN 3371 Electromagnetics
Fall 2008
(5.18.3)
(5.18.5)
(5.18.6)
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Displacement current
Example 2: Find the displacement current, the displacement charge density, and
the volume charge density associated with the magnetic flux density in a vacuum:
B  Bxux  B0 cos(2x)cos(t   y)ux
(5.19.1)
There is no conductors, therefore, only the displacement flux exists.
ux
uy
uz
D 1
1 

 B 
t 0
0 x
Bx
The displacement flux density:

y
0
 B0


cos(2 x) sin(t   y )u z
z
0
0
Jd 
 B

 B0
D   J d dt     0 cos(2 x)sin(t   y )uz  dt 
cos(2 x) cos(t   y)uz
0
 0

From the Gauss’s law:
ELEN 3371 Electromagnetics
v   D  0
Fall 2008
(5.19.2)
(5.19.3)
(5.19.4)
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Displacement current
Example 3: In a lossy dielectric with a conductivity  and a relative permittivity
r, there is a time-harmonic electric field E = E0sin t. Compare the magnitudes
of the conduction current density and the displacement current density.
Conduction current density:
Displacement current density:
The ratio of the magnitudes is:
J c   E   E0 sin t
Jd 
D
  E0 cos t
t
Jc
Jd


 0 r
HW 4 is ready.
ELEN 3371 Electromagnetics
Fall 2008