Displacement Current & Maxwell’s Equations Displacement current Chapter 29: Maxwell’s Equations and Electromagnetic Waves

Displacement Current & Maxwell’s Equations Displacement current

Chapter 29: Maxwell’s Equations and Electromagnetic Waves

Displacement Current & Maxwell’s Equations

Displacement current (cont’d)







sd


Displacement current : Example


Maxwell’s equations: Gauss’s law


Maxwell’s equations: Gauss’ law for magnetism


Maxwell’s equations: Farady’s law


Maxwell’s equations: Ampere’s law


Maxwell’s equations

Maxwell’s Equations and EM Waves Maxwell’s equations

enclenclE

B

encl

AdEdt

dI

dt

dIsdB

AdBdt

d

dt

dsdE

AdB

QAdE

)()(

0

0000

0

�

Gauss’s law

Gauss’s law for magnetism

Farady’s law

Ampere’s law


Maxwell’s equations: Differential form

Oscillating electric dipole

First consider static electric field produced byan 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 movingup and down and exchange their position at everyhalf-period. Then between the two cases there isa situation like as shown in Fig. below:

What is the electric fieldin 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 positionis 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 exceptat charges. Therefore a plausible guess is as shown in the right figure.



What actually happens to the fields based on a precise calculate is shown inFig. Magnetic fields are also formed. When there is electric current, magneticfield is produced. If the current is in a straight wire circular magnetic field isgenerated. Its magnitude is inversely proportional to the distance from thecurrent.


Oscillating electric dipole (cont’d)What actually happens to the fields based on a precise calculate is shown inFig.



This is an animation of radiation of EM wave by an oscillating electric dipoleas a function of time.




At a location far away from the source of the EM wave, the wave becomes plane wave.

++

--

--

++

V(t)=Vocos(t)

• time t=0 • time t=/one half cycle later

XBB

++

--

x

z

y




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 toelectromagnetic 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.

x=0 x

t=0

A

t=T/4

t=T

period

= amplitudewavelength

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:

txatntdisplacemeytxyy ,;),(• A sinusoidal wave is described by the wave function:

)//(2cos

)/(2cos

)]/(cos[

)]/(cos[),(

TtxA

tvxfA

tvxA

vxtAtxy

sinusoidal wave moving in

+x direction

angular frequencyf 2

velocity of wave, NOT ofparticles of the medium

wavelengthperiod

vf Tf /1

)]/(cos[),( xvtAtxy sinusoidal wave moving in-x direction v->-v

phase velocity

Mathematical description of a wave (cont’d)

Wave function (cont’d)

x=0 x

t=0

t=T/4

t=T period

wavelength

)//(2cos),( TtxAtxy ),(

),(

Ttxy

txy


Wave number and phase velocity

)cos(),( tkxAtxy

vkdtdx //

wave number: /2k

The speed of wave is the speed with which we have tomove along a point of a given phase. So for a fixed phase,

phase

.consttkx

phase velocity


Particle velocity and acceleration in a sinusoidal wave

)cos(),( tkxAtxy

),(

)cos(/),(),(

)sin(/),(),(

2

222

txy

tkxAttxytxa

tkxAttxytxv

y

y

velocity

acceleration

Also ),()cos(/),( 2222 txyktkxAkxtxy

222

222222

/),(

/),()/(/),(

tvtxy

ttxykxtxy

wave eq.

Plane EM wave

Plane EM Waves and the Speed of Light

y

x

z

Semi-qualitative description of plane EM wave


Consider a sheet perpendicular to the screen with current running towardyou. Visualize the sheet as many equal parallel fine wires uniformly spacedclose together.

The magnetic field from this current can be found using Ampere’s lawapplied to a rectangle so that the rectangle’s top and bottom are equidistancefrom the current sheet in opposite direction.

Semi-qualitative description of plane EM wave (cont’d)


B

B

L

d



Applying Ampere’s law to the rectangular contour, there are contributionsonly from the top and bottom because the contributions from the sides arezero. The contribution from the top and bottom is 2BL.Denoting the current density on the sheet is I A/m, the total currentenclosed by the rectangle is IL.

)0( sdB

2/00 IBIsdB encl

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 issuddenly 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 rapidlyestablished. Further we assume that the magnetic field spreads out from thesheet moving in both directions at some speed v so that after timethe field within distance vt of the sheet is the same as that found before forthe magnetostatic case, and beyond vt there is at that instant no magnetic present.



B

B

L

dvt

For d < vt the previous result on the B field is still valid but for d > vt, But there is definitely enclosed current!

We are forced to conclude that for Maxwell’s 4th equation to work, there must bea changing electric field through the rectangular contour.

.0 sdB



Maxwell’s 4th equation:

AdEdt

dIsdB encl

000

source of changingelectric field

Now take a look at this electric field. It must have a component perpendicularto the plane of the contour (rectangle), i.e., perpendicular to the magneticfield. As other components do not contribute, let’s ignore them. We are readyto apply Maxwell’s 4th equation:

LIAdEdtdLIIsdB encl

/;0 00

As long as the outward moving front of the B field, traveling at v, has not reachedthe top and bottom, the E field through contour increases linearly with time, butincrease drops to zero the moment the front reaches the top and bottom.



The simplest way to achieve the behavior of the E field just described is tohave an electric field of strength E, perpendicular to the magnetic fieldeverywhere there is a magnetic field so that the electric field also spreadsoutwards at speed v!

After time t, the E field flux through the rectangular contour will be just fieldtimes area, E(2vtL), and the rate of change will be 2EvL:

From the previous analysis, we know that:

LILE )2(0

EIB 000 2/



L

dvt

I

EB

EB

Now we use Maxwell’s 3rd equation: 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 theonly contribution to the integral is EL from the first side. The area of the rectanglethe B flux is passing through will be increasing at a rate Lv as the B field spreadsoutwards. Then,

AdBdtdsdE

/

.BELBEL !/1 and 0000 cBEEB

Qualitative description of plane EM wave in vacuum


E

B

y

dx

Maxwell’s equations when Q=0,I=0 (in vacuum) :

dt

dsdB

dt

dsdE

AdBAdE

EB00;

0;0

Apply Farady’s law (3rd equation) to the rectangularpath shown in Fig. No contributions from the topand bottom as the E field is perpendicular to the path.

ydxdt

dB

dt

dydEyEydEEsdE B

; )(

t

B

x

E

dt

dB

dx

dEydx

dt

dBydE

Qualitative description of plane EM wave in vacuum (cont’d)


E

B

z

dx

Maxwell’s equations when Q=0,I=0 (in vacuum) :

dt

dsdB

dt

dsdE

AdBAdE

EB00;

0;0

Apply Ampere’s law (4th equation) to the rectangularpath shown in Fig. No contributions from the shortsides as the B field is perpendicular to the contour.

zdxdt

dB

dt

dzdBzdBBzBsdB E

0000 ; )(

t

E

x

B

dt

dE

dx

dBzdx

dt

dEzdB

000000



2

2

00

2

00 t

E

xt

B

t

E

x

B

Take the derivative of the 2nd differential equation with respect to t:

Then take the derivative of the 1st differential equation with respect to x:

xt

B

x

E

t

B

x

E

2

2

2

2

2

2

2

00

2

x

E

t

E

xt

B



xt

E

x

B

t

E

x

B

2

002

2

00

Take the derivative of the 2nd differential equation with respect to x:

Then take the derivative of the 1st differential equation with respect to t:

2

22

t

B

xt

E

t

B

x

E

2

2

2

2

00

2 1

t

B

x

B

xt

E

In both cases, if we replace with , two differential equations

become equations that describe a wave traveling with speed

2

00

1



2

2

002

2 1

x

B

t

B

2

2

002

2 1

x

E

t

E

Solve these equations assuming that the solutions are sine waves:

)sin( and )sin(0 tkxBBtkxEEE zy Insert these solutions to the differential equations :

0000

0

00000

0

000

)cos()cos(

)cos()cos(

kE

BtkxEtkxkB

kB

EtkxBtkxkE

c00

1

Speed of light in vacuum!

00 cBE

EM wave in matter


Maxwell’s equations for inside matter change from those in vacuumby change 0 and 0 to = m0 and 0:

mm

c

00

11

For most of dielectrics the relative permeability m is close to 1 except forinsulating ferromagnetic materials :

mm

c

00

11

mnc

Index of refraction

Total energy density in vacuum

Energy and Momentum in Electromagnetic Waves

2

0

20 2

1

2

1BEu

energy density storedin electric field

energy density storedin magnetic field

EcEB 00/

20Eu

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

))(( 20 AcdtEudVdU

• This energy passes through the area A in time dt. So the energy flow per unit time per unit area in vacuum is:

20

1cE

dt

dU

AS

Electromagnetic energy flow and Poynting vector (cont’d)


• We can also rewrite this quantity in terms of B and E as:

0

20

1

EBcE

dt

dU

AS 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:

BES

0

1

Poynting vector

• The total energy flow per unit time (power P) out of any closed surface is:

AdSP

Electromagnetic energy flow and Poynting vector (cont’d)


• Intensity of the sinusoidal wave = time averaged value of S :

itkxBE

kjtkxBE

txBtxEtxS

ktkxBtxBjtkxEtxE

ˆ)(sinˆˆ)(sin),(),(1

),(

,ˆ)sin(),(,ˆ)sin(),(For

2

0

002

0

00

0

00

• Time averaged value of S :

200

20

0

0

0

20

0

00

0

002

0

00

2

0

00

0

2

1

2

1

22

)](2cos1[2

)(sin),(

ˆ)(sin),(),(1

),(

cEEc

EBESI

tkxBE

tkxBE

txS

itkxBE

txBtxEtxS

av

x

z

y

x

Electromagnetic momentum flow and radiation pressure


• It also can be shown that electromagnetic waves carry momentum p with corresponding momentum density of magnitude :

220 c

S

c

EB

dV

dp

• Similarly a corresponding momentum flow rate can be obtained:

c

EB

c

S

dt

dp

AAcdtdV

c

S

c

EB

dV

dp

022

0

1,

• The average rate of momentum transfer per unit area is obtained by replacing S by Sav=I.

momentum carried per unit volume

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.

c

I

c

Sp avav radiation pressure, wave totally absorbed

• If the wave is totally reflected, the momentum change is:

c

I

c

Sp avav

22 radiation pressure, wave totally reflected

The value of I for direct sunlight, before it passes through the Earth’satmosphere, is approximately 1.4 kW/m2:

Pa 107.4m/s 100.3

W/m104.1 68

23

c

I

c

Sp avav

Electromagnetic spectrum


400-700 nm

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Displacement Current & Maxwell’s Equations Displacement current Chapter 29: Maxwell’s Equations and Electromagnetic Waves