Download Electricity and Magnetism Maxwell`s Laws Electromagnetic Radiation

Electricity and Magnetism
Maxwell’s Laws
Electromagnetic Radiation
Lana Sheridan
De Anza College
Dec 1, 2015
Last time
• alternating current (AC)
• transformers
Overview
• Maxwell’s equations
• Ampère-Maxwell law
• electromagnetic radiation
Transformers
which is alm
Transformers change ∆Vrms and Irms simultaneously, while keeping
rule: Transm
the average power Pavg = Irms ∆Vrms constant (conservation of
energy).
The Ideal
ΦB
The transm
for efficient
Np
and consum
Vp
Vs
R
lower (for u
Ns
sentially con
by Faraday’
Primary
Secondary
The ide
Fig. 31-18 An ideal transformer (two
bers of turn
This workscoils
via mutual
If theincurrent
wound inductance.
on an iron core)
a basicin the first coil
In use, the
did not constantly
change
(AC)An
thisacwould
not work.
transformer
circuit.
generator
progenerator w
duces current in the coil at the left (the priS
Ns
∆vp (the secondary)
mary). The coil ∆v
at the
s = right
Np
is connected to the resistive load R when
round a common iron core that is found inside all
Transformers
The circuit diagram for a transformer looks like:
I1
Circuit diaormer.
I2
RL
!v1
N1
!v2
N2
RL is the load resistance.
The equivalent resistance, as seen by the generator on the primary
side (Req = ∆Vp /Ip ) is:
2
Np
Req =
RL
Ns
Transformers
Np 2
RL
Req =
Ns
Since the “effective resistance” is different from the actual load
resistance RL , transformers are also used for load matching.
33.8 The Transformer and Power Transmission
101
Maximum power is delivered when the emf source’s internal
Figure 33.20 Electronic devic
primary winding in this transformer is
are often powered by AC adaptor
resistance, r = RL . The
attached to the prongs of the plug, whereas
containing transformers such as
this one. These adaptors alter the
AC voltage. In many applications
the adaptors also convert alterna
ing current to direct current.
the secondary winding is connected to the
power cord on the right.
nsformer is smaller than the one in
ning photograph of this chapter. In
, it is a step-down transformer. It
. Cengage Learning/George Semple
. Cengage Learning/George Semple
Sometimes, this is not possible, but using a transformer we can
make Req = r .
Transformer Question
An alternating-current emf device in a certain circuit has a smaller
resistance than that of the resistive load in the circuit; to
increase the transfer of energy from the device to the load, a
transformer will be connected between the two.
(i) Should Ns be greater than or less than Np?
(ii) Will that make it a step-up or step-down transformer?
A (i) greater; (ii) step-up
B (i) greater; (ii) step-down
C (i) less; (ii) step-up
D (i) less; (ii) step-down
Transformer Question
An alternating-current emf device in a certain circuit has a smaller
resistance than that of the resistive load in the circuit; to
increase the transfer of energy from the device to the load, a
transformer will be connected between the two.
(i) Should Ns be greater than or less than Np?
(ii) Will that make it a step-up or step-down transformer?
A (i) greater; (ii) step-up
←
B (i) greater; (ii) step-down
C (i) less; (ii) step-up
D (i) less; (ii) step-down
Maxwell’s Laws
Amazingly, we can summarize the majority of the relations that we
have talked about in this course in a set of just 4 equations.
These are together called Maxwell’s equations.
I
E · dA =
I
qenc
B · dA = 0
I
E · ds = −
I
B · ds = µ0 0
dΦB
dt
dΦE
+µ0 Ienc
dt
Gauss’s Law for Magnetic Fields
The first of Maxwell’s equations is Gauss’s Law for E-fields:
I
qenc
E · dA =
The second is for Gauss’s Law for B-fields:
I
B · dA = 0
Faraday’s Law of Induction
Faraday’s Law of Induction is the third of Maxwell’s laws.
I
dΦB
E · ds = −
dt
This tells us that a changing magnetic field will induce an electric
field.
Faraday’s Law of Induction
Faraday’s Law of Induction is the third of Maxwell’s laws.
I
dΦB
E · ds = −
dt
This tells us that a changing magnetic field will induce an electric
field.
But what about the reverse? A changing electric field inducing a
magnetic field?
Faraday’s Law of Induction
Faraday’s Law of Induction is the third of Maxwell’s laws.
I
dΦB
E · ds = −
dt
This tells us that a changing magnetic field will induce an electric
field.
But what about the reverse? A changing electric field inducing a
magnetic field?
It does happen!
Maxwell’s Law of Induction
I
B · ds = µ0 0
dΦE
dt
Maxwell’s Law of Induction
864
+
–
+
–
The changing of the
electric field betwee
dΦE (a)
CHAPTER 32 MAXWELL’S
MAGNETISM OF
B · ds = µ0 EQUATIONS;
0
the plates creates a
dt
magnetic field.
2
2
a changing electric flux will alwa
assume that the charge on our c
+E–
rate by a Bconstant current i in th
E
+ –
1
tude between
the plates must als
1
+ –
Figure 32-5b is a view of th
plates. Ther electric field is direc
B
+ – i
i
through point 1 in Figs. 32-5a and
+ –
and has a radius smaller
than tha
B
+ –
loop is changing,
the
electric
flux t
R
Eq.
32-3,
this
changing
electric
flu
+ –
B Experiment proves that a
+ –
The changing of the
a loop, directed as shown. This
electric field between point (b)
around the loop and thu
(a)
the plates creates a
the capacitor plates (the axis ex
I
Ampere-Maxwell Law
However, a changing electric field is not the only cause of a
magnetic field.
We know from Ampere’s Law:
I
B · ds = µ0 ienc
that a moving charge (current) causes a magnetic field also.
distribution
of currents
Reminder:
Ampère’s Law
due to a current-length
l the elements.Again we
I
However, if the distribre’s law to find the magB · ds = µ0 Ienc
an be derived from the
André-Marie Ampère
owever, The
the law
actually
line
integral of the magnetic field around
a closed loop is
proportional to the current that flows through the loop.1
Only the currents
encircled by the
loop are used in
Ampere's law.
(29-14)
.
:
product B ! ds: is to be
p. The current ienc is the
Amperian
loop
i1
its integral, let us first
θ
i3
B
The figure shows cross
ds
i2, and i3 either directly
i2
op lying in the plane of
Direction of
. The counterclockwise
integration
sen direction of integraFig. 29-11 Ampere’s law applied to an
1
That is, the current
thatAmperian
flows through
anyencircles
surfacetwo
bounded by the loop.
arbitrary
loop that
I d ; P0
Maxwell’s Law of Induction
Path P
!q
dt
I
q
S2
I
S1
conduction current I in the
Surfaces S1 and SThe
2 have different currents flowing through them!
wire passes only through S1, which
leads to a contradiction in
Maxwell’s Law of Induction
Maxwell realized that there should be another term in Ampère’s
law.
He introduced the notion of a displacement current:
Id = 0
dΦE
dt
Maxwell’s Law of Induction
Maxwell realized that there should be another term in Ampère’s
law.
He introduced the notion of a displacement current:
Id = 0
dΦE
dt
Note: The displacement “current” is not a current and has
nothing to do with displacement. However, it does have units of
Amps.
This completes Ampere’s law as:
I
B · ds = µ0 (Ienc + Id )
The Ampère-Maxwell’s Law
The fourth and last of Maxwell’s equations:
The Ampère-Maxwell’s Law
I
B · ds = µ0 0
dΦE
+µ0 Ienc
dt
Differential form:
∇ × B = µ0 0
∂E
+ µ0 J
∂t
Maxwell’s Law of Induction
864
+
–
+
–
The changing of the
electric field betwee
dΦE (a)
CHAPTER 32 MAXWELL’S
MAGNETISM OF
B · ds = µ0 EQUATIONS;
0
the plates creates a
dt
magnetic field.
2
2
a changing electric flux will alwa
assume that the charge on our c
+E–
rate by a Bconstant current i in th
E
+ –
1
tude between
the plates must als
1
+ –
Figure 32-5b is a view of th
plates. Ther electric field is direc
B
+ – i
i
through point 1 in Figs. 32-5a and
+ –
and has a radius smaller
than tha
B
+ –
loop is changing,
the
electric
flux t
R
Eq.
32-3,
this
changing
electric
flu
+ –
B Experiment proves that a
+ –
The changing of the
a loop, directed as shown. This
electric field between point (b)
around the loop and thu
(a)
the plates creates a
the capacitor plates (the axis ex
I
butMaxwell’s
no change inLaw
electric
(such as with
a wire
offlux
Induction
question
), the first
term
on
the
right
side
of
Eq.
32-5
is
zero,
and
The figure shows graphs of the electric field
magnitude E versus
32-4, Ampere’s law. When there is a change in electric
time t for four uniform electric fields, all contained within identical
inside or outside the gap of a charging capacitor), the
circular regions as in the circular-plate capacitor. Rank the E-fields
side of Eq. 32-5 is zero, and so Eq. 32-5 reduces to
according to the magnitudes of the magnetic fields they induce at
nduction.
the edge of the region, greatest first.
d
E
he electric field magnitude E
m electric fields, all contained
ns as in Fig. 32-5b. Rank the
tudes of the magnetic fields
region, greatest first.
a
c
b
t
A a, b, c, d
B a, c, b, d
Sample Problem
C d, b, c, a
Magnetic field induced by changing electric field
D d, c, a, b
1
Halliday,
Walker,
865. side of Eq. 32-6: We assume that th
with circular
platesResnick,
of radius
R is pageRight
butMaxwell’s
no change inLaw
electric
(such as with
a wire
offlux
Induction
question
), the first
term
on
the
right
side
of
Eq.
32-5
is
zero,
and
The figure shows graphs of the electric field
magnitude E versus
32-4, Ampere’s law. When there is a change in electric
time t for four uniform electric fields, all contained within identical
inside or outside the gap of a charging capacitor), the
circular regions as in the circular-plate capacitor. Rank the E-fields
side of Eq. 32-5 is zero, and so Eq. 32-5 reduces to
according to the magnitudes of the magnetic fields they induce at
nduction.
the edge of the region, greatest first.
d
E
he electric field magnitude E
m electric fields, all contained
ns as in Fig. 32-5b. Rank the
tudes of the magnetic fields
region, greatest first.
a
c
b
t
A a, b, c, d
B a, c, b, d ← Sample Problem
C d, b, c, a
Magnetic field induced by changing electric field
D d, c, a, b
1
Halliday,
Walker,
865. side of Eq. 32-6: We assume that th
with circular
platesResnick,
of radius
R is pageRight
Ampere-Maxwell Law and Displacement “Current”
displacement “current”
Id = 0
dΦE
dt
This lets us rewrite the Ampere-Maxwell law as:
I
B · ds = µ0 Id + µ0 Ienc
Looking at it this way can give us some insights.
B-field around a charging capacitor
32-4 DISPLACEME
Suppose a capacitor is being charged with a constant current, i.
During charging
field is created
the real and fict
Before charging, there
is no magnetic field.
i
i
(a)
(b)
+
B
the
How does i During
relate tocharging,
id the “current”
from the E-field between the
right-hand rule works for both
plates?
the real and fictional currents.
B
32-4 DISPLACEMENT CURRENT
867
B-field around a charging capacitor
Suppose a capacitor is being charged with a constant current, i.
A
During charging, magnetic
field is created by both
the real and fictional currents.
ng, there
c field.
id
i
(b)
i
–
+
B
B
B
How does i relate to id the “current” from the E-field between the
for both
plates?
urrents.
B-field around a charging capacitor
Suppose a capacitor
with a constant
current, i.
B is being charged
B
B
or both
urrents.
After charging, there
is no magnetic field.
i
(d )
–
+
–
B
How does i relate to id the “current” from the E-field between the
plates?
B-field around a charging capacitor
id = 0
dΦE
dE
= 0 A
dt
dt
Gauss’s law allows us to relate q, the charge on one plate of the
capacitor to the flux:
I
q
= E · dA = EA
0
The current is the rate of flow of charge:
i=
dq
dE
= 0 A
dt
dt
So, id = i !
The B-field outside a circular capacitor looks the same as the
B-field around the wire leading up to the capacitor.
Magnetic Field around a circular capacitor
Can be calculated just like the field around a wire!
Magnetic Field around a circular capacitor
Can be calculated just like the field around a wire!
Outside the capacitor at radius r from the center:
B=
µ0 id
2πr
Inside the capacitor (plates have radius R) at radius r from the
center:
µ0 id
r
B=
2πR 2
But remember: id is not a current. No current flows across the
gap between the plates.
Maxwell’s Equations
I
E · dA =
I
qenc
B · dA = 0
I
E · ds = −
I
B · ds = µ0 0
1
dΦB
dt
dΦE
+µ0 ienc
dt
Strictly, these are Maxwell’s equations in a vacuum.
Maxwell’s Equations Differential Form
∇·E =
ρ
0
∇·B = 0
∂B
∂t
∂E
∇ × B = µ0 0
+ µ0 J
∂t
∇×E = −
Another Implication of Maxwell’s Equations
Using all 4 equations (in their differential form) it is possible to
reach a pair of wave equations for the electric and magnetic fields:
∇2 E =
∇2 B =
1 ∂2 E
c 2 ∂t2
1 ∂2 B
c 2 ∂t2
with wave solutions:
E = E0 sin(k · r − ωt)
B = B0 sin(k · r − ωt)
where c =
ω
k .
Another Implication of Maxwell’s Equations
∇2 E =
1 ∂2 E
c 2 ∂t2
The constant c appears as the wave speed and
1
c=√
µ0 0
c = 3.00 × 108 m/s, is the speed of light.
The values of 0 and µ0 together predict the speed of light!
Another Implication of Maxwell’s Equations
Wave solutions:
E = E0 sin(k · r − ωt)
B = B0 sin(k · r − ωt)
where c =
ω
k .
These two solutions are in phase. There is no offset in the angles
inside the sine functions.
The two fields peak at the same point in space and time.
At all times:
E
=c
B
Relation between Electric and Magnetic Fields
These oscillating electric and magnetic fields make up light.
Faraday’s Law of Induction
A changing magnetic field gives rise to an electric field.
Ampere-Maxwell Law of Induction
A changing electric field gives rise to an magnetic field.
Light
Faraday’s Law ⇒ a changing magnetic field causes an electric field.
Maxwell’s Law ⇒ a changing electric field causes a magnetic field.
Light (Electromagnetic Radiation)
All light waves in a vacuum travel at the same speed, the speed of
light, c = 3.00 × 108 m s−1 .
Maxwell’s equations possess the ‘wrong’ symmetry for Gallilean
transformations between observers; they are Lorentz-invariant.
This gave Einstein an important idea.
All observers, no matter how they move relative to one another all
agree that any light wave travels at that same speed.
Light (Electromagnetic Radiation)
Light travels at this fixed speed, c = 3.00 × 108 m s−1 .
For any wave, if v is the wave propogation speed:
v =fλ
For light:
c =fλ
So, if the frequency of the light is given, you also know the
wavelength, and vice versa.
λ=
c
;
f
f =
c
λ
Electromagnetic spectrum
Electromagnetic spectrum
Electromagnetic Radiation
EM radiation carries both energy and momentum!
1045
.7 The Spectrum of Electromagnetic Waves
The Poynting vector expresses the power per unit area of em
radiation, together with its direction of propogation.
na. The source of this
The distance from the origin to
the time-varying maga point on the edge of the tan
-varying electric field,
shape is proportional
to the
1
S = of theE Poynting
×B
magnitude
etic fields produced in
µ0
vector and the intensity of
he result is an outward
units are W/m2
ed by a dipole antenna
er radiated are a maxig through its midpoint.
s axis. A mathematical
ws that the intensity of
he axis of the antenna.
ceiving antenna. The
a maximum when the
d zero when the axis is
radiation in that direction.
y
u
S
S
x
Electromagnetic Radiation
Momentum, p, is related to the Poynting vector also:
p
S
= 2
unit-Vol
c
Because the fields carry momentum, they exert a pressure (force
per unit area) on an illuminated surface:
Prad =
S
c
(complete absorption on ⊥ surface)
Prad =
2S
c
(complete reflection by ⊥ surface)
or
Radiation Pressure Question
Radiation pressure can be used as a means of propulsion in solar
sails.
To maximize the radiation pressure on the sails of a spacecraft
using solar sailing, the sheets should be
(A) very black to absorb as much sunlight as possible
(B) very shiny to reflect as much sunlight as possible
Radiation Pressure Question
Radiation pressure can be used as a means of propulsion in solar
sails.
To maximize the radiation pressure on the sails of a spacecraft
using solar sailing, the sheets should be
(A) very black to absorb as much sunlight as possible
(B) very shiny to reflect as much sunlight as possible
←
Radiation Pressure Question
IKAROS (Interplanetary Kite-craft Accelerated by Radiation Of the
Sun), a Japan Aerospace Exploration Agency (JAXA) spacecraft,
launched on 21 May 2010, passed by Venus and still operating.
1
Depiction of IKAROS by Andrzej Mirecki, Wikipedia.
Summary
• Maxwell’s equations
• Ampère-Maxwell law
• electromagnetic radiation
Homework
Serway & Jewett:
• PREV:Ch 33, onward from page 1021. Obj. Qs: 12, 13; Conc.
Qs.: 8; Probs: 1, 3, 5, 49, 51, 57
• NEW: Ch 34, onward from page 1048. Obj. Qs: 3; Probs: 1,
3, 5, 9, 15, 21, 57