Download The 4 Maxwell`s Equations Facts on EM waves Quick Quiz on EM

EM Waves
Friday Honor Lecture
This Lecture
!More on EM waves
!EM spectrum
!Polarization
Prof. D. Oertel
Dept. of Physiology
Interpreting Nerve Signals
From previous Lecture
Displacement currents
Maxwell’s equations
!EM Waves
!
!
1
Facts on EM waves
The 4 Maxwell’s Equations
charged particles create an
electric field
q
$ E" dA = #
S
(Gauss' Law)
0
There are no magnetic
monopoles
$ B" dA = 0
d&
$L E" ds = % dtB (Faraday - Henry)
S
d&
$ LB" ds = µ0I + µ0#0 dtE (Ampere - Maxwell law)
An electric field can be created by a
changing magnetic field
Consequence: induced current
!
Lorentz force
!
EM waves are solutions of Maxwell’s equations.
!
In empty space: sinusoidal wave propagating along x with
velocity
! E = E
max cos (kx – "t)
! B = B
max cos (kx – "t)
• E and B are perpendicular oscillating vectors
!
Currents create a magnetic field
!
A changing electric field can create a
magnetic field
F = q(E + v " B)
3
Transverse
! waves
E!B=0
•The direction of propagation is
perpendicular to E and B
E x B direction of c
!
Quick Quiz on EM waves
Quick Quiz pn EM waves
An electromagnetic wave propagates in the –y direction. The
electric field at a point in space is momentarily oriented in
the +x direction. The magnetic field at that point is
momentarily oriented in the
Which orientation will have the largest induced emf?
E
y
z
(a) –x direction
x
B
(b) +y direction
(c) +z direction
(d) –z direction
loop in xz
plane
loop in xy
plane
z
A
B
c
y
x
E
B
C
yz
in
p
o
lo ane
pl
Faraday’s law: Loops use B not E!
Only the loop in the xy plane will have a magnetic flux through
it as the wave passes. The flux will oscillate with time and
induce an emf.
EM Waves from an Antenna
EM waves from antennas
Sources of EM waves: oscillating charges, accelerated/decellerated charges,
electron transitions between energy levels in atoms, nuclei and molecules
- Electromagnetic radiation is greatest when
charges accelerate at right angles to the line of
sight.
- Zero radiation is observed when the charges
accelerate along the line of sight.
- These observations apply to electromagnetic
waves of all frequencies.
2 rods connected to AC generator; charges oscillate between the rods (a)
As oscillations continue, the rods become less charged, the field near the
charges decreases and the field produced at t = 0 moves away from the
rod (b)
The charges and field reverse (c)
The oscillations continue (d)
!
!
!
!
8
Relation between E and B
!
E = Emax cos (kx – "t)
!
B = Bmax cos (kx – "t)
!
First derivatives:
From:
!
Gamma rays: !~ 10-14- 10-10 m
Source: radioactive nuclei, cause
serious damage to living tissues
"E
= #kE max sin(kx # $t)
"x
"B
= $Bmax sin(kx # $t)
"t
"E
"B
=#
"x
"t
X-rays: ~10-12 -10-8 m
source: deceleration of high-energy
electrons striking a metal target
Diagnostic tool in medicine
Speed of em waves
(speed of light in vacuum)
UV !~ 6 x 10-10 - 4 x 10-7 m
Most UV light from the sun is absorbed
in the stratosphere by ozone
!
Sources: hot objects and molecules
Microwaves: ! ~10-4 -0.3 m
This relation comes from
Maxwell’s equations!
sources: electronic devices
radar systems, MW ovens
Energy density of E and B field
!
!
Rate at which energy flows through a unit area perpendicular
to direction of wave propagation
In a parallel plate capacitor: C =
1
1" A
U 1
U = CV 2 = 0 E 2 d 2 # uE =
= "0 E 2
2
2 d
Ad 2
This is the power per unit area (J/s.m2 = W/m2)
E/B=c
!
Radio:
! ~ 10 - 0.1 m
Sources:
charges
accelerating
through
conducting
wires
Radio and TV
Infrared: ! ~ 7 x 10-7-10-3 m
Poynting vector
!
Source:
atoms and
molecules
Human eye
Visible range
from red (700
nm) to violet
(400 nm)
The EM Spectrum
Its direction is the direction of propagation of the EM wave
Magnitude:
"0 A
d
True for any geometry
!
!
When a battery is connected to a circuit the current does not jump
instantaneously from 0 to the final value !/R because there is an induced emf
opposing to battery action. By calculating the work done in a solenoid against
induced emf arising from increasing B-flux as current goes from 0 to I we can
derive the energy density of the B-field associated to the current
EB E 2
=
µ0 cµ0
This is time dependent
! Its magnitude varies in time
! Its magnitude reaches a maximum at the same instant
as E and B do
S=
!
!
uB =
!
B2
2µ0
True for any geometry
Energy carried by EM waves
Intensity and Poynting vector
Total instantaneous energy density of EM waves
!
And the intensity (average P/area)
!
E/B=c
u =uE + uB = 1/2 $oE2 + B2 /(2µo)
uE = uB
Since B = E/c and
!
Iav =
1
B2
E2
uE = "0 E 2 = uB =
= 2
2
2µ0 2c µ0
!
!
!
In a given volume, the energy is shared equally by the two fields
!
!
Pav =
U av uav AL
=
= uav Ac
"t
"t
!
power per unit area
!
!
"p (energy absorbed)/"t Power
=
=
"t
c
c
F Power / A I
=
=
A
c
c
from
EM waves transport momentum# pressure on a surface
Complete absorption on a surface:
total transported energy U
!
in time interval %t # total momentum p = U / c and prad=Sav/c
(b) very shiny, to reflect as much sunlight as possible
Which is the value of the radiation pressure in the above
case?
(a) P = 2S/c
Perfectly reflecting surface: momentum of incoming and
reflected light p = U/c # total transferred momentum p = 2U/c
and prad = 2Sav/c
Direct sunlight pressure ~5 x 10-6 N/m2
(b) P = S/c
Circular and elliptical polarization
Polarization of Light Waves (34.8)
!
ExB
(a) very black to absorb as much sunlight as possible
Power
I=
= Save
c
!
!
EB E 2
=
µ0 cµ0
To maximize the radiation pressure on the sails of a spacecraft
using solar sailing, the sheets must be
momentum p = U / c
!
!
S=
Quick Quiz on radiation pressure
Radiation Pressure = force per unit area
prad =
I & E2
!
Radiation pressure and momentum
F=
B 2 EB
=
µ0 µ0c
E max Bmax
2µ0
!
!
u = uE + uB = "0 E 2 =
Wave intensity I = time average over one or more cycle
<sin2(kx - "t)> = 1/2 then <E2> = Emax2/2 and <B2> = Bmax2/2
!
Iav = uav c =
Let’s consider a cylinder with axis along x of area A and length L and the
time for the wave to travel L is !t=L/c
The average power is:
!
Pav
= uav c
A
Linearly polarized waves: E-field oscillates at all times in the
plane of polarization
Unpolarized light: E-field in random
directions. Superposition
of waves with E vibrating
in many different
directions
Linearly polarized
light: E-field has one
spatial orientation
!
Circularly polarized light: superposition of 2 waves of equal
amplitude with orthogonal linear polarizations, and 90˚ out of phase.
The tip of E describes a circle (counterclockwise = RH and
clockwise=LH depending on y component ahead or behind)
!
The electric field rotates in time with constant magnitude.
If amplitudes differ # elliptical polarization
!
DEMO with MW generator and metal grid
Producing polarized light
!
Polarization by selective absorption: material that transmits
waves whose E-field vibrates in a plain parallel to a certain
direction and absorbs all others
pick up antenna connected to Ammeter
Metal grid
MW generator
This polarization
absorbed
This polarization
transmitted
transmission axis
Long-chain hydrocarbon
molecules
Polaroid sheet
(E. Land 1928)
If the wires of the grid are parallel to the plane of polarization the grid absorbs
the E-component (electrons oscillate in the wire).
The same thing happens to a polaroid: the component parallel
to the direction of the chains of hydrocarbons is absorbed.
If the grid is horizontal the Ammeter will measure a
This
not null current since the wave reaches the antenna
polarization
absorbed
pick-up
This polarization
transmitted
transmission axis
Detecting polarized light: Malus’ law
Polarization by selective absorption
If linearly polarized light of intensity I0 passes through a polarizing
filter with transmission axis at an angle ' along y
y
Polaroid sheet
'
Einc = E0sin' i + E0 cos' j
x transmission axis
After the polarizer
Etransm = E0cos' j
E0cos'
So the intensity transmitted is
Itransm = E02 cos2' = (0cos2'
Long-chain
hydrocarbon
molecules
A polarizer is used to produce
!
Ideal polarizer transmits waves with E parallel to transmission axis and
absorbs those with E ) axis
!
Relative orientation of axis of polarizer and analyzer determines intensity of
transmitted light.
Transmitted intensity: I = I0cos2' I0 = intensity of polarized beam on analyzer
(Malus’ law)
!
Allowed component
parallel to analyzer axis
polarized light of intensity I0
and an analyzer rotated at an
angle ': transmission 100%
when ' = 0 and zero when ' =
90°
Polaroid sheets
Quick Quiz on Polarization
Law of Malus Example
I1= 0.5 I0
I2= I1cos2(45)
1) Light transmitted by first polarizer is vertically polarized. I1 = I0
cos245=I0/2
A
B
2) Angle between it and second polarizer is '=45º. I2= I1 cos2 (45º)
=0.5I1=0.25 I0
3) Light transmitted through second polarizer is polarized 45º
from vertical. Angle between it and third polarizer is '=45º. I3= I2
cos2(45º) = 0.125I0
Relative orientation of polarizers
!
!
!
Transmitted amplitude is Eocos'
(component of polarization along polarizer axis)
Transmitted intensity is Iocos2'
( square of amplitude)
Perpendicular polarizers give zero intensity.