Transcript Chapter 21: Electric Charge and Electric Field

Chapter 29: Maxwell’s Equations and
Electromagnetic Waves
Displacement Current & Maxwell’s Equations

Displacement current
Displacement Current & Maxwell’s Equations

Displacement current (cont’d)
Displacement Current & Maxwell’s Equations

Displacement current (cont’d)
Displacement Current & Maxwell’s Equations

Displacement current (cont’d)
Displacement Current & Maxwell’s Equations

Displacement current (cont’d)

ds
Displacement Current & Maxwell’s Equations

Displacement current : Example
Displacement Current & Maxwell’s Equations

Maxwell’s equations: Gauss’s law
Displacement Current & Maxwell’s Equations

Maxwell’s equations: Gauss’ law for magnetism
Displacement Current & Maxwell’s Equations

Maxwell’s equations: Farady’s law
Displacement Current & Maxwell’s Equations

Maxwell’s equations: Ampere’s law
Displacement Current & Maxwell’s Equations

Maxwell’s equations
Maxwell’s Equations and EM Waves
 Maxwell’s equations
Gauss’s law
  Qencl
 E  dA 
Gauss’s law for
magnetism
 
 B  dA  0
Farady’s law
Ampere’s law
0
 
d B
d  
 E  ds   dt   dt  B  dA
 
d E
d  
 B  ds  0 ( I   0 dt )encl  0 ( I   0 dt  E  dA)encl
Displacement Current & Maxwell’s Equations

Maxwell’s equations: Differential form
Maxwell’s Equations and EM Waves
 Oscillating electric dipole
First consider static electric field produced by
an electric dipole as shown in Figs.
(a) Positive (negative) charge at the top (bottom)
(b) Negative (positive) charge at the top (bottom)
Now then imagine these two charge are moving
up and down and exchange their position at every
half-period. Then between the two cases there is
a situation like as shown in Fig. below:
What is the electric field
in the blank area?
Maxwell’s Equations and EM Waves
 Oscillating electric dipole (cont’d)
Since we don’t assume that change propagate instantly once new position
is reached the blank represents what has to happen to the fields in meantime.
We learned that E field lines can’t cross and they need to be continuous except
at charges. Therefore a plausible guess is as shown in the right figure.
Maxwell’s Equations and EM Waves
 Oscillating electric dipole (cont’d)
What actually happens to the fields based on a precise calculate is shown in
Fig. Magnetic fields are also formed. When there is electric current, magnetic
field is produced. If the current is in a straight wire circular magnetic field is
generated. Its magnitude is inversely proportional to the distance from the
current.
Maxwell’s Equations and EM Waves
 Oscillating electric dipole (cont’d)
What actually happens to the fields based on a precise calculate is shown in
Fig.
Maxwell’s Equations and EM Waves
 Oscillating electric dipole (cont’d)
This is an animation of radiation of EM wave by an oscillating electric dipole
as a function of time.
Maxwell’s Equations and EM Waves
 Oscillating electric dipole (cont’d)
V(t)=Vocos(t)
• time t=0
+
+
-
B
-
• time t=/
one half cycle later
X
B
+
+
At a location far away from the source of the EM wave, the wave
becomes plane wave.
Maxwell’s Equations and EM Waves
 Oscillating electric dipole (cont’d)
+
+
-
x
z
y
Maxwell’s Equations and EM Waves
 Oscillating electric dipole (cont’d)
A qualitative summary of the observation of this example is:
1) The E and B fields are always at right angles to each other.
2) The propagation of the fields, i.e., their direction of travel away from the
oscillating dipole, is perpendicular to the direction in which the fields
point at any given position in space.
3) In a location far from the dipole, the electric field appears to form closed
loops which are not connected to either charge. This is, of course, always
true for any B field. Thus, far from the dipole, we find that the E and B
fields are traveling independent of the charges. They propagate away from
the dipole and spread out through space.
In general it can be proved that accelerating electric charges give rise to
electromagnetic waves.
Types of mechanical waves
 Periodic
waves
• When particles of the medium in a wave undergo periodic
motion as the wave propagates, the wave is called periodic.
wavelength
l
A = amplitude
t=0
x=0
t=T/4
t=T
period
x
Mathematical description of a wave
 Wave
function
• The wave function describes the displacement of particles
or change of E/B field in a wave as a function of time and
their position:
y  y( x, t ) ; y displacement at x, t
• A sinusoidal wave is described by the wave function:
sinusoidal wave moving in
y ( x, t )  A cos[ (t  x / v )]
+x direction
 A cos[ ( x / v  t )]
velocity of wave, NOT of
angular frequency
 A cos2 f ( x / v  t ) particles of the medium
  2 f
 A cos2 ( x / l  t / T )
fl  v wavelength
y ( x, t )  A cos[(t  v / x)]
period f  1 / T
sinusoidal wave moving in
-x direction v->-v
phase velocity
Mathematical description of a wave (cont’d)
 Wave
function (cont’d)
y ( x, t )  A cos2 ( x / l  t / T )
l
wavelength
 y( x  l , t )
 y ( x, t  T )
t=0
x=0
t=T/4
t=T period
x
Mathematical description of a wave (cont’d)
 Wave
number and phase velocity
wave number:
k  2 / l
y ( x, t )  A cos(kx  t )
phase
The speed of wave is the speed with which we have to
move along a point of a given phase. So for a fixed phase,
kx  t  const .
dx / dt   / k  v
phase velocity
Mathematical description of a wave (cont’d)
 Particle
velocity and acceleration in a sinusoidal wave
y ( x, t )  A cos(kx  t )
v y ( x, t )  y ( x, t ) / t  A sin(kx  t )
velocity
a y ( x, t )   2 y ( x, t ) / t 2   2 A cos(kx  t )
  2 y ( x, t )
Also
acceleration
2 y( x, t ) / x2  k 2 Acos(kx  t )  k 2 y( x, t )
 2 y ( x, t ) / x 2  (k 2 /  2 ) 2 y ( x, t ) / t 2
  y ( x, t ) / v t
2
2
2
wave eq.
Plane EM Waves and the Speed of Light
 Plane EM wave
y
x
z
Plane EM Waves and the Speed of Light
 Semi-qualitative description of plane EM wave
Consider a sheet perpendicular to the screen with current running toward
you. Visualize the sheet as many equal parallel fine wires uniformly spaced
close together.
The magnetic field from this current can be found using Ampere’s law
applied to a rectangle so that the rectangle’s top and bottom are equidistance
from the current sheet in opposite direction.
Plane EM Waves and the Speed of Light
 Semi-qualitative description of plane EM wave (cont’d)
B
d
B
L
Plane EM Waves and the Speed of Light
 Semi-qualitative description of plane EM wave (cont’d)
Applying Ampere’s law to the rectangular contour, there are contributions
only fromthe top and bottom because the contributions from the sides are

zero. ( B  ds  0) The contribution from the top and bottom is 2BL.
Denoting the current density on the sheet is I A/m, the total current
enclosed by the rectangle is IL.
 
 B  ds  0 I encl  B  0 I / 2
Note that the B field strength is independent of the distance d from the sheet.
Now consider how the magnetic field develops if the current in the sheet is
suddenly switched on at time t=0. Here we assume that sufficiently close to the
sheet the magnetic field pattern found using Ampere’s law is rather rapidly
established. Further we assume that the magnetic field spreads out from the
sheet moving in both directions at some speed v so that after time
the field within distance vt of the sheet is the same as that found before for
the magnetostatic case, and beyond vt there is at that instant no magnetic present.
Plane EM Waves and the Speed of Light
 Semi-qualitative description of plane EM wave (cont’d)
For
 d <vt the previous result on the B field is still valid but for d > vt,
But there is definitely enclosed current!
B  ds  0.

We are forced to conclude that for Maxwell’s 4th equation to work, there must be
a changing electric field through the rectangular contour.
B
d
vt
B
L
Plane EM Waves and the Speed of Light
 Semi-qualitative description of plane EM wave (cont’d)
Maxwell’s 4th equation:
 
d
 B  ds  0 I encl  0 0 dt

 
 E  dA

source of changing
electric field
Now take a look at this electric field. It must have a component perpendicular
to the plane of the contour (rectangle), i.e., perpendicular to the magnetic
field. As other components do not contribute, let’s ignore them. We are ready
to apply Maxwell’s 4th equation:


 
 
 B  ds  0 ; I encl  LI  0 0 d / dt  E  dA   LI
As long as the outward moving front of the B field, traveling at v, has not reached
the top and bottom, the E field through contour increases linearly with time, but
increase drops to zero the moment the front reaches the top and bottom.
Plane EM Waves and the Speed of Light
 Semi-qualitative description of plane EM wave (cont’d)
The simplest way to achieve the behavior of the E field just described is to
have an electric field of strength E, perpendicular to the magnetic field
everywhere there is a magnetic field so that the electric field also spreads
outwards at speed v!
After time t, the E field flux through the rectangular contour will be just field
times area, E(2vtL), and the rate of change will be 2EvL:
 0 E(2L)  LI
From the previous analysis, we know that:
B  0 I / 2  0 0E
Plane EM Waves and the Speed of Light
 Semi-qualitative description of plane EM wave (cont’d)
Now we use Maxwell’s 3rd equation:

 
 
 E  ds  d / dt  B  dA

We apply this equation to a rectangular contour with sides parallel to the E field,
one side being within vt of the current sheet, the other more distant so that the
only contribution to the integral is EL from the first side. The area of the rectangle
the B flux is passing through will be increasing at a rate Lv as the B field spreads
outwards. Then, EL  LB  E  B.
B  0 0E and E  B    1 / 0 0  c!
L
d
vt
B
E
I
E
B
Plane EM Waves and the Speed of Light
 Qualitative description of plane EM wave in vacuum
Maxwell’s equations when Q=0,I=0 (in vacuum) :
E
Dy
 
 E  dA  0
 
;  B  dA  0
 
 
d B
d E
 E  ds   dt ;  B  ds  0 0 dt
dx
B
Apply Farady’s law (3rd equation) to the rectangular
path shown in Fig. No contributions from the top
and bottom as the E field is perpendicular to the path.
 
d B dB
E

d
s

(
E

dE
)
D
y

E
D
y

dE
D
y
;

dxDy

dt
dt
dEDy  
dB
dE
dB
E
B
dxDy 



dt
dx
dt
x
t
Plane EM Waves and the Speed of Light
 Qualitative description of plane EM wave in vacuum (cont’d)
Maxwell’s equations when Q=0,I=0 (in vacuum) :
E
 
 E  dA  0
dx
Dz
B
 
;  B  dA  0
 
 
d B
d E
 E  ds   dt ;  B  ds  0 0 dt
Apply Ampere’s law (4th equation) to the rectangular
path shown in Fig. No contributions from the short
sides as the B field is perpendicular to the contour.
 
d E
dB
B

d
s

B
D
z

(
B

dB
)
D
z


dB
D
z
;





dxDz
0 0
0 0

dt
dt
 dBDz  0 0
dE
dB
dE
B
E
dxDz 
   0 0

   0 0
dt
dx
dt
x
t
Plane EM Waves and the Speed of Light
 Qualitative description of plane EM wave in vacuum (cont’d)
Take the derivative of the 2nd differential equation with respect to t:
B
E
2B
2E
  0 0

  0 0 2
x
t
tx
t
Then take the derivative of the 1st differential equation with respect to x:
E
B
2E
2B

 2 
x
t
x
tx
2B
2E 2E

 0 0 2  2
tx
t
x
Plane EM Waves and the Speed of Light
 Qualitative description of plane EM wave in vacuum (cont’d)
Take the derivative of the 2nd differential equation with respect to x:
B
E
2B
2E
  0 0
 2   0 0
x
t
x
tx
Then take the derivative of the 1st differential equation with respect to t:
E
B
2E
2B


 2
x
t
tx
t
2E
1 2B 2B


 2
2
tx 0 0 x
t
In both cases, if we replace
1
0 0
with
 2 , two differential equations
become equations that describe a wave traveling with speed .
Plane EM Waves and the Speed of Light
 Qualitative description of plane EM wave in vacuum (cont’d)
2B
1 2B

2
t
0 0 x 2
2E
1 2E

2
t
0 0 x 2
Solve these equations assuming that the solutions are sine waves:
E  Ey  E0 sin(kx  t ) and B  Bz sin(kx  t )
Insert these solutions to the differential equations :
kE0 cos(kx  t )  B0 cos(kx  t ) 
E0 
 
B0 k
B0 0 0
kB0 cos(kx  t )  0 0E0 cos(kx  t ) 

 0 0
E0
k

1
0 0
c
Speed of light in vacuum!
E0  cB0
Plane EM Waves and the Speed of Light
 EM wave in matter
Maxwell’s equations for inside matter change from those in vacuum
by change 0 and 0 to  = km0 and   k0:

1


1
0 0 k mk
c

k mk
For most of dielectrics the relative permeability km is close to 1 except for
insulating ferromagnetic materials :

c

1


1
0 0 k mk
 n  k mk  k

c
k mk
Index of refraction
Energy and Momentum in Electromagnetic
Waves
 Total energy density in vacuum
1
1 2
2
u  0E 
B
2
20
energy density stored
in electric field
energy density stored
in magnetic field
B  E / c  0 0 E
u  0E2
Energy and Momentum in Electromagnetic
Waves
 Electromagnetic energy flow and Poynting vector
• E and B fields advance with time into regions where
originally no fields were present and carry the energy
density u with them as they advance.
• The energy transfer is described in terms of energy
transferred per unit time per unit area.
• The wave front moves in a time dt by dx=vdt=cdt.
And the volume the wave front sweeps is Adx. So
the energy in this volume in vacuum is:
area A
dU  udV  ( 0 E 2 )( Acdt)
• This energy passes through the area A in time dt. So the energy flow per
unit time per unit area in vacuum is:
S
1 dU
  0 cE 2
A dt
Energy and Momentum in Electromagnetic
Waves
 Electromagnetic energy flow and Poynting vector (cont’d)
• We can also rewrite this quantity in terms of B and E as:
S
1 dU
EB
  0cE 2 
A dt
0
units J/(s m2) or W/m2
• We can also define a vector that describes both the magnitude and direction
of the energy flow as:
 1  
S
EB
0
Poynting vector
• The total energy flow per unit time (power P) out of any closed surface is:
 
P   S  dA
Energy and Momentum in Electromagnetic
Waves
 Electromagnetic energy flow and Poynting vector (cont’d)
• Intensity of the sinusoidal wave = time averaged value of S :


ˆ
For E ( x, t )  E0 sin(kx  t ) j , B( x, t )  B0 sin(kx  t )kˆ,


EB
EB
1 
S ( x, t ) 
E ( x, t )  B( x, t )  0 0 sin 2 (kx  t ) ˆj  kˆ  0 0 sin 2 (kx  t )iˆ
0
0
0
y
• Time averaged value of S :


EB
1 
S ( x, t ) 
E ( x, t )  B( x, t )  0 0 sin 2 (kx  t )iˆ
0
0
z
EB
EB
 S x ( x, t )  0 0 sin 2 (kx  t )  0 0 [1  cos 2(kx  t )]
0
20
E0 B0
E02
1 0 2 1
 I  S av 


E0   0 cE02
20
20c 2 0
2
x
Energy and Momentum in Electromagnetic
Waves
 Electromagnetic momentum flow and radiation pressure
• It also can be shown that electromagnetic waves carry momentum p with
corresponding momentum density of magnitude :
dp
EB
S
momentum carried per unit volume


2
2
dV 0c
c
• Similarly a corresponding momentum flow rate can be obtained:
dp
EB
S
1 dp S EB


,
dV

Acdt

 
dV 0c 2 c 2
A dt c 0c
• The average rate of momentum transfer per unit area is obtained by replacing
S by Sav=I.
Energy and Momentum in Electromagnetic
Waves
 Electromagnetic momentum flow and radiation pressure
• When an electromagnetic wave is completely absorbed by a surface,
the wave’s momentum is also transferred to the surface. dp/dt, the rate
at which momentum is transferred to the surface is equal to the force on
the surface. The average force per unit area due to the wave (radiation)
is the average value of dp/dt divided by the absorbing area A.
S av I
pav 

c
c
radiation pressure, wave totally absorbed
• If the wave is totally reflected, the momentum change is:
2S av 2 I
pav 

c
c
radiation pressure, wave totally reflected
The value of I for direct sunlight, before it passes through the Earth’s
atmosphere, is approximately 1.4 kW/m2:
S av I 1.4 103 W/m2
pav 
 
 4.7 106 Pa
8
c
c
3.0 10 m/s
Energy and Momentum in Electromagnetic
Waves
 Electromagnetic spectrum
400-700 nm