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KONGUNADU COLLEGE OF ENGINEERING AND TECHNOLOGY
DEPARTMENT OF ELECTRICALAND ELECTRONICS ENGINEERING
1Prepared by Mr.K.Karthik-AP/EEE
 Integral form  Differential form




V
ev
S
C
dvqsdD
ldE 0
evqD
E

 0
ED 
2Prepared by Mr.K.Karthik-AP/EEE
 Integral form  Differential form
0



S
SC
sdB
sdJldH
0

B
JH
HB 
3Prepared by Mr.K.Karthik-AP/EEE
 In the static case (no time variation), the
electric field (specified by E and D) and
the magnetic field (specified by B and H)
are described by separate and
independent sets of equations.
 In a conducting medium, both
electrostatic and magnetostatic fields
can exist, and are coupled through the
Ohm’s law (J = sE). Such a situation is
called electromagnetostatic.
4Prepared by Mr.K.Karthik-AP/EEE
 In an electromagnetostatic field, the
electric field is completely determined
by the stationary charges present in
the system, and the magnetic field is
completely determined by the current.
 The magnetic field does not enter into
the calculation of the electric field, nor
does the electric field enter into the
calculation of the magnetic field.
5Prepared by Mr.K.Karthik-AP/EEE
 Electric charges attract/repel each
other as described by Coulomb’s law.
 Current-carrying wires attract/repel
each other as described by Ampere’s law
of force.
 Magnetic fields that change with time
induce electromotive force as
described by Faraday’s law.
6Prepared by Mr.K.Karthik-AP/EEE
battery
switch
toroidal iron
core
compass
primary
coil
secondary
coil
7Prepared by Mr.K.Karthik-AP/EEE
 Upon closing the switch, current
begins to flow in the primary coil.
 A momentary deflection of the compass
needle indicates a brief surge of current
flowing in the secondary coil.
 The compass needle quickly settles back
to zero.
 Upon opening the switch, another
brief deflection of the compass needle is
observed.
8Prepared by Mr.K.Karthik-AP/EEE
 “The electromotive force induced
around a closed loop C is equal to the
time rate of decrease of the magnetic
flux linking the loop.”
C
S
dt
d
Vind


9Prepared by Mr.K.Karthik-AP/EEE
 
S
sdB
• S is any surface
bounded by C
 
C
ind ldEV
 
SC
sdB
dt
d
ldE
integral form
of Faraday’s
law
10Prepared by Mr.K.Karthik-AP/EEE







SS
SC
sd
t
B
sdB
dt
d
sdEldE
Stokes’s theorem
assuming a stationary surface S
11Prepared by Mr.K.Karthik-AP/EEE
 Since the above must hold for any S, we have
t
B
E



differential form
of Faraday’s law
(assuming a
stationary frame
of reference)
12Prepared by Mr.K.Karthik-AP/EEE
 Faraday’s law states that a changing magnetic
field induces an electric field.
 The induced electric field is non-conservative.
13Prepared by Mr.K.Karthik-AP/EEE
 “The sense of the emf induced by the
time-varying magnetic flux is such that
any current it produces tends to set up a
magnetic field that opposes the change
in the original magnetic field.”
 Lenz’s law is a consequence of
conservation of energy.
 Lenz’s law explains the minus sign in
Faraday’s law.
14Prepared by Mr.K.Karthik-AP/EEE
 “The electromotive force induced
around a closed loop C is equal to
the time rate of decrease of the
magnetic flux linking the loop.”
 For a coil of N tightly wound turns
dt
d
Vind


dt
d
NVind


15Prepared by Mr.K.Karthik-AP/EEE
 
S
sdB
• S is any surface
bounded by C
 
C
ind ldEV
C
S
16Prepared by Mr.K.Karthik-AP/EEE
 Faraday’s law applies to situations where
◦ (1) the B-field is a function of time
◦ (2) ds is a function of time
◦ (3) B and ds are functions of time
17Prepared by Mr.K.Karthik-AP/EEE
 The induced emf around a circuit can be
separated into two terms:
◦ (1) due to the time-rate of change of the B-field
(transformer emf)
◦ (2) due to the motion of the circuit (motional emf)
18Prepared by Mr.K.Karthik-AP/EEE
 






CS
S
ind
ldBvsd
t
B
sdB
dt
d
V
transformer emf
motional emf
19Prepared by Mr.K.Karthik-AP/EEE
 Consider a conducting bar moving
with velocity v in a magnetostatic field:
B
v
2
1
+
-
• The magnetic force on an
electron in the conducting
bar is given by
BveFm 
20Prepared by Mr.K.Karthik-AP/EEE
 Electrons are pulled
toward end 2. End 2
becomes negatively
charged and end 1
becomes + charged.
 An electrostatic
force of attraction is
established
between the two
ends of the bar.
B
v
2
1
+
-
21Prepared by Mr.K.Karthik-AP/EEE
 The electrostatic force on an electron due
to the induced electrostatic field is given
by
 The migration of electrons stops
(equilibrium is established) when
EeFe 
BvEFF me 
22Prepared by Mr.K.Karthik-AP/EEE
 A motional (or “flux cutting”) emf is produced
given by
  ldBvVind  
1
2
23Prepared by Mr.K.Karthik-AP/EEE
 Electrostatics:
 EE 0
scalar electric potential
24Prepared by Mr.K.Karthik-AP/EEE
 Electrodynamics:
 



















t
A
E
t
A
E
A
tt
B
E
0
AB 
25Prepared by Mr.K.Karthik-AP/EEE
 Electrodynamics:
t
A
E



scalar
electric
potential
vector
magnetic
potential
• both of these
potentials are now
functions of time.
26Prepared by Mr.K.Karthik-AP/EEE
 The differential form of Ampere’s law in the
static case is
 The continuity equation is
JH 
0



t
q
J ev
27Prepared by Mr.K.Karthik-AP/EEE
 In the time-varying case, Ampere’s law in the
above form is inconsistent with the continuity
equation
  0 HJ
28Prepared by Mr.K.Karthik-AP/EEE
 To resolve this inconsistency, Maxwell
modified Ampere’s law to read
t
D
JH c



conduction
current density
displacement
current density
29Prepared by Mr.K.Karthik-AP/EEE
 The new form of Ampere’s law is consistent
with the continuity equation as well as with
the differential form of Gauss’s law
    0


 HD
t
J c
qev
30Prepared by Mr.K.Karthik-AP/EEE
Derivation of Displacement Current
q EA I
dq
dt
d EA
dt
   0 0
( )
For a capacitor, and .
I
d
dt
E
 0
( )
Now, the electric flux is given by EA, so: ,
where this current , not being associated with charges, is
called the “Displacement current”, Id.
Hence:
and: B ds I I
B ds I d
dt
d
E


  
   

  
0
0 0 0
( )

I
d
dtd
E
  0 0

31Prepared by Mr.K.Karthik-AP/EEE
Derivation of Displacement Current
q EA I
dq
dt
d EA
dt
   0 0
( )
For a capacitor, and .
I
d
dt
E
 0
( )
Now, the electric flux is given by EA, so: ,
where this current, not being associated with charges, is
called the “Displacement Current”, Id.
Hence:
and: B dl I I
B dl I d
dt
d
E


  
   

  
0
0 0 0
( )

I
d
dtd
E
  0 0

32Prepared by Mr.K.Karthik-AP/EEE
 Ampere’s law can be written as
dc JJH 
where
)(A/mdensitycurrentntdisplaceme 2




t
D
J d
33Prepared by Mr.K.Karthik-AP/EEE
 Consider a parallel-plate capacitor with
plates of area A separated by a dielectric
of permittivity  and thickness d and
connected to an ac generator:
tVtv cos)( 0
+
-
z = 0
z = d

icA
id
z
34Prepared by Mr.K.Karthik-AP/EEE
 The displacement current is given by
c
d
S
dd
i
dt
dv
CtCV
tV
d
A
AJsdJi

 



sin
sin
0
0
conduction
current in
wire
35Prepared by Mr.K.Karthik-AP/EEE
 Consider a conducting medium
characterized by conductivity s and
permittivity .
 The conduction current density is
given by
 The displacement current density is
given by
EJ c s
t
E
J d


 
36Prepared by Mr.K.Karthik-AP/EEE
10
0
10
2
10
4
10
6
10
8
10
10
10
-4
10
-3
10
-2
10
-1
10
0
10
1
10
2
10
3
10
4
10
5
10
6
Frequency (Hz)
Humid Soil (r
= 30, s = 10-2
S/m)

s
good
conductor
good insulator
37Prepared by Mr.K.Karthik-AP/EEE
 We have been examining a variety of
electrical and magnetic phenomena
 James Clerk Maxwell summarized all of
electricity and magnetism in just four
equations
 Remarkably, the equations predict the
existence of electromagnetic waves
38Prepared by Mr.K.Karthik-AP/EEE
As you can see we need some vector calculus to use these equations. That isn’t
going to happen in this class, but I wanted you to see the equations anyway.
39Prepared by Mr.K.Karthik-AP/EEE
 Start with Ampere’s Law

B||l  0I
Earlier, we just went on a closed path
enclosing surface 1. But according to
Ampere’s Law, we could have
considered surface 2. The current
enclosed is the same as for surface 1.
We can say that the current flowing
into any volume must equal that
coming out.
40Prepared by Mr.K.Karthik-AP/EEE
 Suppose we have a charged capacitor and
it begins to discharge
Surface 1 works but
surface 2 has no current
passing through the
surface yet there is a
magnetic field inside the
surface.
41Prepared by Mr.K.Karthik-AP/EEE
Same problem here. Surface
1 works, but no current
passes through surface two
which encloses a magnetic
field. What is happening???
42Prepared by Mr.K.Karthik-AP/EEE
Maxwell solved this problem
by realizing that....
B EInside the capacitor there must
be an induced magnetic field...
How?.
43Prepared by Mr.K.Karthik-AP/EEE
Maxwell solved this problem
by realizing that....
B E
x
x x x x
x x x x x
x x
A changing
electric field
induces a
magnetic field
Inside the capacitor there must
be an induced magnetic field...
How?. Inside the capacitor there is a changing E 
E
B
44Prepared by Mr.K.Karthik-AP/EEE
Maxwell solved this problem
by realizing that....
B E
x
x x x x
x x x x x
x x
A changing
electric field
induces a
magnetic field
Inside the capacitor there must
be an induced magnetic field...
How?. Inside the capacitor there is a changing E 
where Id is called the
displacement current
B dl
d
dt
IE
d     0 0 0

E
B
45Prepared by Mr.K.Karthik-AP/EEE
B E
B dl I d
dt
E
     0 0 0

x
x x x x
x x x x x
x x
A changing
electric field
induces a
magnetic field
Inside the capacitor there must
be an induced magnetic field...
How?. Inside the capacitor there is a changing E 
where Id is called the
displacement current
Therefore, Maxwell’s revision
of Ampere’s Law becomes....
B dl
d
dt
IE
d     0 0 0

E
B
46Prepared by Mr.K.Karthik-AP/EEE
1
2
E dA
q
  
0
B dA   0
..monopole..
?
...there’s no
magnetic monopole....!!
Gauss’s Laws
47Prepared by Mr.K.Karthik-AP/EEE
4
E dl d
dt
B
    
B dl I   0
3
.. if you change a
magnetic field you
induce an electric
field.........
.......is the reverse
true..?
Faraday’s Law
Ampere’s Law
48Prepared by Mr.K.Karthik-AP/EEE
...lets take a look at charge flowing into a
capacitor...
E
...when we derived Ampere’s Law
we assumed constant current...
.. if the loop encloses one
plate of the capacitor..there is a
problem … I = 0
B
Side view:(Surface
is now like a bag:)
EB
B dl I   0
49Prepared by Mr.K.Karthik-AP/EEE
Consider these equations in a vacuum.....
......no mass, no charges. no currents.....
B dl
d
dt
E
    0 0

E dl
d
dt
B
   

E dA
q
  
0
B dA   0
B dl I d
dt
E
     0 0 0

E dA   0
E dl
d
dt
B
   

B dA   0
50Prepared by Mr.K.Karthik-AP/EEE
 So, a magnetic field will be produced in
space if there is a changing electric field
 But, this magnetic field is changing since
the electric field is changing
 A changing magnetic field produces an
electric field that is also changing
 We have a self-perpetuating system
51Prepared by Mr.K.Karthik-AP/EEE
Close switch and current flows briefly.
Sets up electric field. Current flow
sets up magnetic field as little circles
around the wires. Fields not
instantaneous, but form in time.
Energy is stored in fields and cannot
move infinitely fast.
52Prepared by Mr.K.Karthik-AP/EEE
Picture a shows first half cycle. When current reverses in picture b, the fields reverse.
See the first disturbance moving outward. These are the electromagnetic waves.
53Prepared by Mr.K.Karthik-AP/EEE
Notice that the electric and magnetic
fields are at right angles to one another!
They are also perpendicular to the
direction of motion of the wave.
54Prepared by Mr.K.Karthik-AP/EEE
 Now that we have shown how the waves
are formed from oscillating charges, we
need to see if we can predict how fast the
move
 We move far away from the source so that
the wave fronts are essentially flat
 Just like dropping a rock in a pond and
looking at the waves a few hundred feet
away from the impact point
55Prepared by Mr.K.Karthik-AP/EEE
This picture defines the coordinate system we will use in our discussion. Wave
propagates along the x-axis. The electric field varies in the y-direction and the
magnetic field in the z-direction.
56Prepared by Mr.K.Karthik-AP/EEE
We are going to apply Faraday’s Law to the imaginary moving rectangle abcd. Compute
the magnetic flux change
emf 
B
t

BA
t

By0vt
t
 By0v
57Prepared by Mr.K.Karthik-AP/EEE
 We can say the emf around the loop is the
sum of the individual emfs going along
each straight line segment in the loop
 We look at the work done in moving a test
charge around the loop
 emf = W/q = Fd/q = Ed
 emf = Ey0 = By0v
 E = Bv
58Prepared by Mr.K.Karthik-AP/EEE
Now we are going to look at the change in electric flux. Set a new imaginary rectangle
and play the same game as before.
B||l  00
E
t
  00
Ez0vt
t
 00Ez0v
59Prepared by Mr.K.Karthik-AP/EEE
Bz0  00Ez0v
B  00Ev
E  Bv
B  00(Bv)v
1 00v2
v 
1
00
v 
1
8.851012
 4 107
v  3108
60Prepared by Mr.K.Karthik-AP/EEE
The Electromagnetic Spectrum
61Prepared by Mr.K.Karthik-AP/EEE
...lets take a look at charge flowing into a
capacitor...
...when we derived Ampere’s Law
we assumed constant current...
EB
B dl I   0
62Prepared by Mr.K.Karthik-AP/EEE
B dl
d
dt
E
    0 0
 E dl
d
dt
B
   

Faraday’s law: dB/dt electric field
Maxwell’s modification of Ampere’s law
dE/dt magnetic field
These two equations can be solved simultaneously.
The result is:
E(x, t) = EP sin (kx-t)
B(x, t) = BP sin (kx-t) ˆz
ˆj
63Prepared by Mr.K.Karthik-AP/EEE
B dl
d
dt
E
    0 0

E dl
d
dt
B
   

B
E
dE
dt

dB
dt

64Prepared by Mr.K.Karthik-AP/EEE
B dl
d
dt
E
    0 0

E dl
d
dt
B
   

B
E
Special case..PLANE WAVES...
satisfy the wave equation
   A tsin( )Maxwell’s Solution
v
 
 
 

2
2 2
2
2
1
x t

 
E E x t j B B x t ky z ( , ) ( , )
dE
dt

dB
dt

65Prepared by Mr.K.Karthik-AP/EEE
x
Ey
Bz
E(x, t) = EP sin (kx-t)
B(x, t) = BP sin (kx-t) ˆz
ˆj
c
66Prepared by Mr.K.Karthik-AP/EEE
Static wave
F(x) = FP sin (kx + )
k = 2  
k = wavenumber
 = wavelength
F(x)
x

Moving wave
F(x, t) = FP sin (kx - t )
 = 2  f
 = angular frequency
f = frequency
v =  / k
F(x)
x

v
67Prepared by Mr.K.Karthik-AP/EEE
x
v
Moving wave
F(x, t) = FP sin (kx - t )
What happens at x = 0 as a function of time?
F(0, t) = FP sin (-t)
F
For x = 0 and t = 0  F(0, 0) = FP sin (0)
For x = 0 and t = t  F (0, t) = FP sin (0 – t) = FP sin (– t)
This is equivalent to: kx = - t  x = - (/k) t
F(x=0) at time t is the same as F[x=-(/k)t] at time 0
The wave moves to the right with speed /k
68Prepared by Mr.K.Karthik-AP/EEE
x
Ey
Bz
Notes: Waves are in Phase,
but fields oriented at 900.
k=2.
Speed of wave is c=/k (= f
At all times E=cB.
c m s  1 3 100 0
8/ / 
E(x, t) = EP sin (kx-t)
B(x, t) = BP sin (kx-t) ˆz
ˆj
c
69Prepared by Mr.K.Karthik-AP/EEE

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ELECTRODYNAMIC FIELDS

  • 1. KONGUNADU COLLEGE OF ENGINEERING AND TECHNOLOGY DEPARTMENT OF ELECTRICALAND ELECTRONICS ENGINEERING 1Prepared by Mr.K.Karthik-AP/EEE
  • 2.  Integral form  Differential form     V ev S C dvqsdD ldE 0 evqD E   0 ED  2Prepared by Mr.K.Karthik-AP/EEE
  • 3.  Integral form  Differential form 0    S SC sdB sdJldH 0  B JH HB  3Prepared by Mr.K.Karthik-AP/EEE
  • 4.  In the static case (no time variation), the electric field (specified by E and D) and the magnetic field (specified by B and H) are described by separate and independent sets of equations.  In a conducting medium, both electrostatic and magnetostatic fields can exist, and are coupled through the Ohm’s law (J = sE). Such a situation is called electromagnetostatic. 4Prepared by Mr.K.Karthik-AP/EEE
  • 5.  In an electromagnetostatic field, the electric field is completely determined by the stationary charges present in the system, and the magnetic field is completely determined by the current.  The magnetic field does not enter into the calculation of the electric field, nor does the electric field enter into the calculation of the magnetic field. 5Prepared by Mr.K.Karthik-AP/EEE
  • 6.  Electric charges attract/repel each other as described by Coulomb’s law.  Current-carrying wires attract/repel each other as described by Ampere’s law of force.  Magnetic fields that change with time induce electromotive force as described by Faraday’s law. 6Prepared by Mr.K.Karthik-AP/EEE
  • 8.  Upon closing the switch, current begins to flow in the primary coil.  A momentary deflection of the compass needle indicates a brief surge of current flowing in the secondary coil.  The compass needle quickly settles back to zero.  Upon opening the switch, another brief deflection of the compass needle is observed. 8Prepared by Mr.K.Karthik-AP/EEE
  • 9.  “The electromotive force induced around a closed loop C is equal to the time rate of decrease of the magnetic flux linking the loop.” C S dt d Vind   9Prepared by Mr.K.Karthik-AP/EEE
  • 10.   S sdB • S is any surface bounded by C   C ind ldEV   SC sdB dt d ldE integral form of Faraday’s law 10Prepared by Mr.K.Karthik-AP/EEE
  • 12.  Since the above must hold for any S, we have t B E    differential form of Faraday’s law (assuming a stationary frame of reference) 12Prepared by Mr.K.Karthik-AP/EEE
  • 13.  Faraday’s law states that a changing magnetic field induces an electric field.  The induced electric field is non-conservative. 13Prepared by Mr.K.Karthik-AP/EEE
  • 14.  “The sense of the emf induced by the time-varying magnetic flux is such that any current it produces tends to set up a magnetic field that opposes the change in the original magnetic field.”  Lenz’s law is a consequence of conservation of energy.  Lenz’s law explains the minus sign in Faraday’s law. 14Prepared by Mr.K.Karthik-AP/EEE
  • 15.  “The electromotive force induced around a closed loop C is equal to the time rate of decrease of the magnetic flux linking the loop.”  For a coil of N tightly wound turns dt d Vind   dt d NVind   15Prepared by Mr.K.Karthik-AP/EEE
  • 16.   S sdB • S is any surface bounded by C   C ind ldEV C S 16Prepared by Mr.K.Karthik-AP/EEE
  • 17.  Faraday’s law applies to situations where ◦ (1) the B-field is a function of time ◦ (2) ds is a function of time ◦ (3) B and ds are functions of time 17Prepared by Mr.K.Karthik-AP/EEE
  • 18.  The induced emf around a circuit can be separated into two terms: ◦ (1) due to the time-rate of change of the B-field (transformer emf) ◦ (2) due to the motion of the circuit (motional emf) 18Prepared by Mr.K.Karthik-AP/EEE
  • 20.  Consider a conducting bar moving with velocity v in a magnetostatic field: B v 2 1 + - • The magnetic force on an electron in the conducting bar is given by BveFm  20Prepared by Mr.K.Karthik-AP/EEE
  • 21.  Electrons are pulled toward end 2. End 2 becomes negatively charged and end 1 becomes + charged.  An electrostatic force of attraction is established between the two ends of the bar. B v 2 1 + - 21Prepared by Mr.K.Karthik-AP/EEE
  • 22.  The electrostatic force on an electron due to the induced electrostatic field is given by  The migration of electrons stops (equilibrium is established) when EeFe  BvEFF me  22Prepared by Mr.K.Karthik-AP/EEE
  • 23.  A motional (or “flux cutting”) emf is produced given by   ldBvVind   1 2 23Prepared by Mr.K.Karthik-AP/EEE
  • 24.  Electrostatics:  EE 0 scalar electric potential 24Prepared by Mr.K.Karthik-AP/EEE
  • 26.  Electrodynamics: t A E    scalar electric potential vector magnetic potential • both of these potentials are now functions of time. 26Prepared by Mr.K.Karthik-AP/EEE
  • 27.  The differential form of Ampere’s law in the static case is  The continuity equation is JH  0    t q J ev 27Prepared by Mr.K.Karthik-AP/EEE
  • 28.  In the time-varying case, Ampere’s law in the above form is inconsistent with the continuity equation   0 HJ 28Prepared by Mr.K.Karthik-AP/EEE
  • 29.  To resolve this inconsistency, Maxwell modified Ampere’s law to read t D JH c    conduction current density displacement current density 29Prepared by Mr.K.Karthik-AP/EEE
  • 30.  The new form of Ampere’s law is consistent with the continuity equation as well as with the differential form of Gauss’s law     0    HD t J c qev 30Prepared by Mr.K.Karthik-AP/EEE
  • 31. Derivation of Displacement Current q EA I dq dt d EA dt    0 0 ( ) For a capacitor, and . I d dt E  0 ( ) Now, the electric flux is given by EA, so: , where this current , not being associated with charges, is called the “Displacement current”, Id. Hence: and: B ds I I B ds I d dt d E              0 0 0 0 ( )  I d dtd E   0 0  31Prepared by Mr.K.Karthik-AP/EEE
  • 32. Derivation of Displacement Current q EA I dq dt d EA dt    0 0 ( ) For a capacitor, and . I d dt E  0 ( ) Now, the electric flux is given by EA, so: , where this current, not being associated with charges, is called the “Displacement Current”, Id. Hence: and: B dl I I B dl I d dt d E              0 0 0 0 ( )  I d dtd E   0 0  32Prepared by Mr.K.Karthik-AP/EEE
  • 33.  Ampere’s law can be written as dc JJH  where )(A/mdensitycurrentntdisplaceme 2     t D J d 33Prepared by Mr.K.Karthik-AP/EEE
  • 34.  Consider a parallel-plate capacitor with plates of area A separated by a dielectric of permittivity  and thickness d and connected to an ac generator: tVtv cos)( 0 + - z = 0 z = d  icA id z 34Prepared by Mr.K.Karthik-AP/EEE
  • 35.  The displacement current is given by c d S dd i dt dv CtCV tV d A AJsdJi       sin sin 0 0 conduction current in wire 35Prepared by Mr.K.Karthik-AP/EEE
  • 36.  Consider a conducting medium characterized by conductivity s and permittivity .  The conduction current density is given by  The displacement current density is given by EJ c s t E J d     36Prepared by Mr.K.Karthik-AP/EEE
  • 37. 10 0 10 2 10 4 10 6 10 8 10 10 10 -4 10 -3 10 -2 10 -1 10 0 10 1 10 2 10 3 10 4 10 5 10 6 Frequency (Hz) Humid Soil (r = 30, s = 10-2 S/m)  s good conductor good insulator 37Prepared by Mr.K.Karthik-AP/EEE
  • 38.  We have been examining a variety of electrical and magnetic phenomena  James Clerk Maxwell summarized all of electricity and magnetism in just four equations  Remarkably, the equations predict the existence of electromagnetic waves 38Prepared by Mr.K.Karthik-AP/EEE
  • 39. As you can see we need some vector calculus to use these equations. That isn’t going to happen in this class, but I wanted you to see the equations anyway. 39Prepared by Mr.K.Karthik-AP/EEE
  • 40.  Start with Ampere’s Law  B||l  0I Earlier, we just went on a closed path enclosing surface 1. But according to Ampere’s Law, we could have considered surface 2. The current enclosed is the same as for surface 1. We can say that the current flowing into any volume must equal that coming out. 40Prepared by Mr.K.Karthik-AP/EEE
  • 41.  Suppose we have a charged capacitor and it begins to discharge Surface 1 works but surface 2 has no current passing through the surface yet there is a magnetic field inside the surface. 41Prepared by Mr.K.Karthik-AP/EEE
  • 42. Same problem here. Surface 1 works, but no current passes through surface two which encloses a magnetic field. What is happening??? 42Prepared by Mr.K.Karthik-AP/EEE
  • 43. Maxwell solved this problem by realizing that.... B EInside the capacitor there must be an induced magnetic field... How?. 43Prepared by Mr.K.Karthik-AP/EEE
  • 44. Maxwell solved this problem by realizing that.... B E x x x x x x x x x x x x A changing electric field induces a magnetic field Inside the capacitor there must be an induced magnetic field... How?. Inside the capacitor there is a changing E  E B 44Prepared by Mr.K.Karthik-AP/EEE
  • 45. Maxwell solved this problem by realizing that.... B E x x x x x x x x x x x x A changing electric field induces a magnetic field Inside the capacitor there must be an induced magnetic field... How?. Inside the capacitor there is a changing E  where Id is called the displacement current B dl d dt IE d     0 0 0  E B 45Prepared by Mr.K.Karthik-AP/EEE
  • 46. B E B dl I d dt E      0 0 0  x x x x x x x x x x x x A changing electric field induces a magnetic field Inside the capacitor there must be an induced magnetic field... How?. Inside the capacitor there is a changing E  where Id is called the displacement current Therefore, Maxwell’s revision of Ampere’s Law becomes.... B dl d dt IE d     0 0 0  E B 46Prepared by Mr.K.Karthik-AP/EEE
  • 47. 1 2 E dA q    0 B dA   0 ..monopole.. ? ...there’s no magnetic monopole....!! Gauss’s Laws 47Prepared by Mr.K.Karthik-AP/EEE
  • 48. 4 E dl d dt B      B dl I   0 3 .. if you change a magnetic field you induce an electric field......... .......is the reverse true..? Faraday’s Law Ampere’s Law 48Prepared by Mr.K.Karthik-AP/EEE
  • 49. ...lets take a look at charge flowing into a capacitor... E ...when we derived Ampere’s Law we assumed constant current... .. if the loop encloses one plate of the capacitor..there is a problem … I = 0 B Side view:(Surface is now like a bag:) EB B dl I   0 49Prepared by Mr.K.Karthik-AP/EEE
  • 50. Consider these equations in a vacuum..... ......no mass, no charges. no currents..... B dl d dt E     0 0  E dl d dt B      E dA q    0 B dA   0 B dl I d dt E      0 0 0  E dA   0 E dl d dt B      B dA   0 50Prepared by Mr.K.Karthik-AP/EEE
  • 51.  So, a magnetic field will be produced in space if there is a changing electric field  But, this magnetic field is changing since the electric field is changing  A changing magnetic field produces an electric field that is also changing  We have a self-perpetuating system 51Prepared by Mr.K.Karthik-AP/EEE
  • 52. Close switch and current flows briefly. Sets up electric field. Current flow sets up magnetic field as little circles around the wires. Fields not instantaneous, but form in time. Energy is stored in fields and cannot move infinitely fast. 52Prepared by Mr.K.Karthik-AP/EEE
  • 53. Picture a shows first half cycle. When current reverses in picture b, the fields reverse. See the first disturbance moving outward. These are the electromagnetic waves. 53Prepared by Mr.K.Karthik-AP/EEE
  • 54. Notice that the electric and magnetic fields are at right angles to one another! They are also perpendicular to the direction of motion of the wave. 54Prepared by Mr.K.Karthik-AP/EEE
  • 55.  Now that we have shown how the waves are formed from oscillating charges, we need to see if we can predict how fast the move  We move far away from the source so that the wave fronts are essentially flat  Just like dropping a rock in a pond and looking at the waves a few hundred feet away from the impact point 55Prepared by Mr.K.Karthik-AP/EEE
  • 56. This picture defines the coordinate system we will use in our discussion. Wave propagates along the x-axis. The electric field varies in the y-direction and the magnetic field in the z-direction. 56Prepared by Mr.K.Karthik-AP/EEE
  • 57. We are going to apply Faraday’s Law to the imaginary moving rectangle abcd. Compute the magnetic flux change emf  B t  BA t  By0vt t  By0v 57Prepared by Mr.K.Karthik-AP/EEE
  • 58.  We can say the emf around the loop is the sum of the individual emfs going along each straight line segment in the loop  We look at the work done in moving a test charge around the loop  emf = W/q = Fd/q = Ed  emf = Ey0 = By0v  E = Bv 58Prepared by Mr.K.Karthik-AP/EEE
  • 59. Now we are going to look at the change in electric flux. Set a new imaginary rectangle and play the same game as before. B||l  00 E t   00 Ez0vt t  00Ez0v 59Prepared by Mr.K.Karthik-AP/EEE
  • 60. Bz0  00Ez0v B  00Ev E  Bv B  00(Bv)v 1 00v2 v  1 00 v  1 8.851012  4 107 v  3108 60Prepared by Mr.K.Karthik-AP/EEE
  • 61. The Electromagnetic Spectrum 61Prepared by Mr.K.Karthik-AP/EEE
  • 62. ...lets take a look at charge flowing into a capacitor... ...when we derived Ampere’s Law we assumed constant current... EB B dl I   0 62Prepared by Mr.K.Karthik-AP/EEE
  • 63. B dl d dt E     0 0  E dl d dt B      Faraday’s law: dB/dt electric field Maxwell’s modification of Ampere’s law dE/dt magnetic field These two equations can be solved simultaneously. The result is: E(x, t) = EP sin (kx-t) B(x, t) = BP sin (kx-t) ˆz ˆj 63Prepared by Mr.K.Karthik-AP/EEE
  • 64. B dl d dt E     0 0  E dl d dt B      B E dE dt  dB dt  64Prepared by Mr.K.Karthik-AP/EEE
  • 65. B dl d dt E     0 0  E dl d dt B      B E Special case..PLANE WAVES... satisfy the wave equation    A tsin( )Maxwell’s Solution v        2 2 2 2 2 1 x t    E E x t j B B x t ky z ( , ) ( , ) dE dt  dB dt  65Prepared by Mr.K.Karthik-AP/EEE
  • 66. x Ey Bz E(x, t) = EP sin (kx-t) B(x, t) = BP sin (kx-t) ˆz ˆj c 66Prepared by Mr.K.Karthik-AP/EEE
  • 67. Static wave F(x) = FP sin (kx + ) k = 2   k = wavenumber  = wavelength F(x) x  Moving wave F(x, t) = FP sin (kx - t )  = 2  f  = angular frequency f = frequency v =  / k F(x) x  v 67Prepared by Mr.K.Karthik-AP/EEE
  • 68. x v Moving wave F(x, t) = FP sin (kx - t ) What happens at x = 0 as a function of time? F(0, t) = FP sin (-t) F For x = 0 and t = 0  F(0, 0) = FP sin (0) For x = 0 and t = t  F (0, t) = FP sin (0 – t) = FP sin (– t) This is equivalent to: kx = - t  x = - (/k) t F(x=0) at time t is the same as F[x=-(/k)t] at time 0 The wave moves to the right with speed /k 68Prepared by Mr.K.Karthik-AP/EEE
  • 69. x Ey Bz Notes: Waves are in Phase, but fields oriented at 900. k=2. Speed of wave is c=/k (= f At all times E=cB. c m s  1 3 100 0 8/ /  E(x, t) = EP sin (kx-t) B(x, t) = BP sin (kx-t) ˆz ˆj c 69Prepared by Mr.K.Karthik-AP/EEE

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