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Chapter 34
The Laws of Electromagnetism
Maxwell’s Equations
Displacement Current
Electromagnetic Radiation
The Electromagnetic Spectrum
The Equations of Electromagnetism
(at this point …)
E dA
q
  
0
B dA
   0
E dl d
dt
B
    
B dl I
   0
Gauss’ Law for Electrostatics
Gauss’ Law for Magnetism
Faraday’s Law of Induction
Ampere’s Law
1
2
The Equations of Electromagnetism
E dA
q
  
0
B dA
   0
..monopole..
?
...there’s no
magnetic monopole....!!
Gauss’s Laws
4
The Equations of Electromagnetism
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
...lets take a look at charge flowing into a capacitor...
...when we derived Ampere’s Law
we assumed constant current...
E
B
B dl I
   0
...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:)
E
B
B dl I
   0
Maxwell solved this problem
by realizing that....
B E
Inside the capacitor there must
be an induced magnetic field...
How?.
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
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
I
E
d
   
  
0 0 0

E
B
Maxwell solved this problem
by realizing that....
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
I
E
d
   
  
0 0 0

E
B
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
dt
d
E
  
0 0

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
dt
d
E
  
0 0

Maxwell’s Equations of Electromagnetism
E dA
q
  
0
B dA
   0
E dl d
dt
B
    
Gauss’ Law for Electrostatics
Gauss’ Law for Magnetism
Faraday’s Law of Induction
Ampere’s Law B dl I
d
dt
E
   
  
0 0 0

Maxwell’s Equations of Electromagnetism
in Vacuum (no charges, no masses)
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
Maxwell’s Equations of Electromagnetism
in Vacuum (no charges, no masses)
E dA
   0
B dA
   0
E dl
d
dt
B
   

B dl
d
dt
E
    
0 0

B dl
d
dt
E
    
0 0
 E dl
d
dt
B
   

Electromagnetic Waves
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) ẑ
ĵ
B dl
d
dt
E
    
0 0

E dl
d
dt
B
   

B
E
Electromagnetic Waves
dE
dt

dB
dt

B dl
d
dt
E
    
0 0

E dl
d
dt
B
   

B
E
Special case..PLANE WAVES...
satisfy the wave equation
  
 
A t
sin( )
Maxwell’s Solution
Electromagnetic Waves
v
 
 
 

2
2 2
2
2
1
x t

 
E E x t j B B x t k
y z
 
( , ) ( , )
dE
dt

dB
dt

Plane Electromagnetic Waves
x
Ey
Bz
E(x, t) = EP sin (kx-t)
B(x, t) = BP sin (kx-t) ẑ
ĵ
c
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
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
Plane Electromagnetic Waves
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 10
0 0
8
/ /
 
E(x, t) = EP sin (kx-t)
B(x, t) = BP sin (kx-t) ẑ
ĵ
c

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Maxwell's Equations Reveal Electromagnetic Waves

  • 1. Chapter 34 The Laws of Electromagnetism Maxwell’s Equations Displacement Current Electromagnetic Radiation
  • 3. The Equations of Electromagnetism (at this point …) E dA q    0 B dA    0 E dl d dt B      B dl I    0 Gauss’ Law for Electrostatics Gauss’ Law for Magnetism Faraday’s Law of Induction Ampere’s Law
  • 4. 1 2 The Equations of Electromagnetism E dA q    0 B dA    0 ..monopole.. ? ...there’s no magnetic monopole....!! Gauss’s Laws
  • 5. 4 The Equations of Electromagnetism 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
  • 6. ...lets take a look at charge flowing into a capacitor... ...when we derived Ampere’s Law we assumed constant current... E B B dl I    0
  • 7. ...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:) E B B dl I    0
  • 8. Maxwell solved this problem by realizing that.... B E Inside the capacitor there must be an induced magnetic field... How?.
  • 9. 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
  • 10. 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 I E d        0 0 0  E B
  • 11. Maxwell solved this problem by realizing that.... 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 I E d        0 0 0  E B
  • 12. 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 dt d E    0 0 
  • 13. 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 dt d E    0 0 
  • 14. Maxwell’s Equations of Electromagnetism E dA q    0 B dA    0 E dl d dt B      Gauss’ Law for Electrostatics Gauss’ Law for Magnetism Faraday’s Law of Induction Ampere’s Law B dl I d dt E        0 0 0 
  • 15. Maxwell’s Equations of Electromagnetism in Vacuum (no charges, no masses) 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
  • 16. Maxwell’s Equations of Electromagnetism in Vacuum (no charges, no masses) E dA    0 B dA    0 E dl d dt B      B dl d dt E      0 0 
  • 17. B dl d dt E      0 0  E dl d dt B      Electromagnetic Waves 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) ẑ ĵ
  • 18. B dl d dt E      0 0  E dl d dt B      B E Electromagnetic Waves dE dt  dB dt 
  • 19. B dl d dt E      0 0  E dl d dt B      B E Special case..PLANE WAVES... satisfy the wave equation      A t sin( ) Maxwell’s Solution Electromagnetic Waves v        2 2 2 2 2 1 x t    E E x t j B B x t k y z   ( , ) ( , ) dE dt  dB dt 
  • 20. Plane Electromagnetic Waves x Ey Bz E(x, t) = EP sin (kx-t) B(x, t) = BP sin (kx-t) ẑ ĵ c
  • 21. 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
  • 22. 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
  • 23. Plane Electromagnetic Waves 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 10 0 0 8 / /   E(x, t) = EP sin (kx-t) B(x, t) = BP sin (kx-t) ẑ ĵ c

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