Physics 169 Luis anchordoqui Kitt Peak National Observatory Tuesday, April 24, 18 1 Maxwell’s Equations and Electromagnetic Waves
13.1 The Displacement Current
In Chapter 9, we learned that if a current-carrying wire possesses certain symmetry, the magnetic field could be obtained by using Ampere’s law:
! ! B ! d s = µ I . (13.1.1) "" 0 enc
The equationChapter states that the line integral of 11 a magnetic field around an arbitrary closed loop is equal to µ0 Ienc , where Ienc is the flux of the charge carries passing through the surface bound by the closed path. In addition, we also learned in Chapter 10 that, as a consequenceChapter of Faraday’s law of induction 11, an electric field is associated with a changing magnetic field, according to ! ! d ! ! E ! d s = # B ! dA . (13.1.2) Displacement"" dt "" Current and S Displacement Current and One might Maxwell’sthen wonder whether or not the converse Equations could be true, namely, a magnetic field is associated with a changing electric field. If so, then the right-hand side of Eq. (13.1.1) will11.1 have Displacement to be modified to reflect Current such “symmetry” between electric and ! ! magneticMaxwell’s fields, E and B . Equations We can use Ampere’ s law to calculate magnetic fields due to currents To see how magnetic fields are associated with a time-varying electric field, consider the process of 11.1chargingIntegral a capacitor. Displacement DuringB~ the d ~s charging around process, any Currentthe electric close field loop strength C is equal to µ0iincl increases with time as more chargeC is accumulated· on the plates. The conduction current that carries the charges alsoI produces a magnetic field. In order to apply Ampere’s law to calculate11.1 thisWewhere field, saw let us i incl Displacementchoose in Chap.7 curve= currentC1 as thatthe Amperian passing we loop, can Current shown an use inarea Figure 13.1.1.bounded by closed curve C Amp`ere’s law to calculate magnetic Wefields saw due in to Chap.7 currents. that we can use Amp`ere’s law to calculate magnetic We know that the integral C B d s fieldsaround due any to currents. close loop C is¸ equal· to Weµ i know, where thati the= integralcurrent passingB d an s 0 incl incl C · aroundarea bounded any close by the loop closedC curveis¸ equal C. to e.g. µ0iincl, where iincl = current passing an e.g. area bounded by the closed curve C. Figure 13.1.1 Surfaces S and S bound by curve C . = Flat surface1 2 bounded by1 loop C e.g. = Flat surface bounded by loop C
== Curved Flat surface surface bounded bounded by loopby loop C 13-C3 = Curved surface bounded by loop C If Ampere’ s law is true all time If Amp`ere’s= Curved law is true surface all the bounded time, then by loop the Ciincl determined should be inde- pendentthen of the i incl surface determined chosen. should be independent of surface chosen Tuesday, April 24, 18 2 If Amp`ere’s law is true all the time, then the iincl determined should be inde- pendent of the surface chosen. 11.1. DISPLACEMENT CURRENT 126
Let’s consider a simple case: charging a 11.1. DISPLACEMENT CURRENT 126 capacitorConsider. a simple case: charging a capacitor 11.1. DISPLACEMENT CURRENT 126 From Chap.5,Let’s consider we know a simple there case: is acharging current a t/RC We11.1. know DISPLACEMENT there is a current CURRENT flowing i(t)= E e 126 capacitor. t/RC R flowing i(t)=Let’s considerER e a simple, which case: charging leads a whichFrom leads Chap.5, to a magnetic we know therefield isobserved a currentB~ Let’s considercapacitor a simple. case: t/RCcharging⇤ a to a magneticflowing fieldi(t)= observedE e , whichB. With leads capacitor. From Chap.5,R we know there is a current From Chap.5, we know⇤ there is a~ currentt/RC ⇤ Amp`ere’sWith toAmpere’ law, aflowing magnetic sB lawi(t field)=d ⇤s = observedE µBe 0iincld~s,B.= which.µ With0iinc leads C t/RC R flowing i(t)= E e · ⇤, which leads· Amp`ere’sto a law, magneticR C B fieldCd⇤s = observedµ0iincl. B⇤ . With BUT WHAT IS¸ iincl? I· ⇤ to aBUT magnetic WHAT fieldi IS observed¸ iincl? B⇤. With BUT WHATAmp`ere’s IS ⇤ inc law, ? C B d⇤s = µ0iincl. Amp`ere’s law, C B d⇤s = µ0iincl.· BUT WHAT· IS¸ iincl? BUT WHAT IS¸ iincl? If we look at iinc = i(t) If we look at , iincl = i(t) If we look at , iincl = i(t) If we look atIf we look, i atincl = i(,t) iincl = i(t)
If we lookIf we at look at iinc, =0iincl =0 If we look at , i =0 If we look at If we look, atincli =0, iincl =0 (⇥ There is no chargeincl flow between the (⇥ There is( noThere charge is flow no between charge flow the between the capacitorcapacitor plates.)⇥ plates.) Amp`ere’scapacitor law plates.) is either WRONG or (⇥ There* ThereAmp`ere’s is nois no law charge charge is either flowflow WRONG between between or capacitor the plates INCOMPLETE.INCOMPLETE. Amp`ere’s law is either WRONG or capacitor plates.)INCOMPLETE. )TwoAmpere’Two observations: observations: s law is either WRONG or INCOMPLETE Amp`ere’s lawTwo observations: is either WRONG or 1. While1. While there is there no current is no between current the between capacitor’s the plates, capacitor’s there is plates, a time- there is a time- INCOMPLETE.Tuesday, Aprilvarying 24, 18 electric field between the plates of the capacitor. 3 varying1. While electric there field is no between current the between plates the of the capacitor’s capacitor plates,. there is a time- 2. We know Amp`ere’svarying electric law is mostly field between correct from the plates measurements of the capacitor of B-field. Two observations:around2. We circuits know. Amp`ere’s law is mostly correct from measurements of B-field 2. We know Amp`ere’s law is mostly correct from measurements of B-field around circuits. 1. While therearound is no circuits current. between the capacitor’s plates, there is a time- varying electricCan we field revise between Amp`ere’s the law plates to fix it? of the capacitor. Can we revise Amp`ere’s Q law to fix it? Electric field between capacitor’sCan plates: we reviseE = Amp`ere’s= , where lawQ = to charge fix it? on 2. We know Amp`ere’s law is mostly⇥0 ⇥0A correct from measurements of B-field capacitor’s plates, A = Area of capacitor’s plates. Q Q aroundElectricElectric circuits field fieldbetween. between capacitor’s capacitor’s plates: plates:E =E = = = , where, whereQ =Q charge = charge on on ⇥0 ⇥0A Q = ⇥0 E A = ⇥0 E ⇥0 ⇥0A capacitor’s plates, A = Area of capacitor’s· plates. capacitor’s plates, A = AreaElectric of flux capacitor’s plates. We can define ⇤ ⇥ Q⌅ = ⇥ E A = ⇥ Q0= ⇥0 E A 0= ⇥E0 E dQ d · · = ⇥ E = i ElectricElectric flux flux dt 0 dt disp ⇤ ⇥ ⌅ We canWe defineCan can define we revise Amp`ere’s⇤ ⇥ ⌅ law to fix it? where idisp is called Displacement Current (first proposed by Maxwell). dQ dQ d Ed E Maxwell first proposed that this is the missing= ⇥ term=0 ⇥ for= theidisp= Amp`ere’si law: dt dt dt0 dt disp Q Electric field between⇤ capacitor’sd E plates: E = = , where Q = charge on wherewhereidispB isiddisp⇤s called=isµ0 called(iDisplacementincl +Displacement⇥0 ) Amp`ere-Maxwell Current Current(first(first proposedlaw proposed by byMaxwell). Maxwell). ˛C · dt ⇥0 ⇥0A capacitor’sMaxwell plates,Maxwell first A proposed first = proposed Area that of that this capacitor’s this is the is missingthe missing plates term term for. forthe the Amp`ere’s Amp`ere’s law: law: ⇤ d Ed E B⇤ d⇤sB=dµ⇤s0=(iinclµ0(+iincl⇥0+ ⇥0 ) Amp`ere-Maxwell) Amp`ere-Maxwell law law ˛ ˛·C · Q =dt⇥ dtE A = ⇥ C 0 · 0 E Electric flux ⇤ ⇥ ⌅ We can define dQ d = ⇥ E = i dt 0 dt disp where idisp is called Displacement Current (first proposed by Maxwell). Maxwell first proposed that this is the missing term for the Amp`ere’s law:
d E B⇤ d⇤s = µ0(iincl + ⇥0 ) Amp`ere-Maxwell law ˛C · dt Two obser vations
① While there is no current between capacitor’s plates there is a time-varying electric field between plates of capacitor
② We know Ampere’s law is mostly correct from measurements of B-field around circuits ⇓ Can we revise Ampere’s law to fix it? Q Electric field between capacitor’s plates E = = ✏0 ✏0A Q = charge on capacitor’s plates A = area of capacitor’s plates Q = ✏ E A = ✏ ) 0 · 0 E Electric flux
Tuesday, April 24, 18 | {z } 4 ) We can define dQ d = ✏ E = i dt 0 dt dis
where i dis is called Displacement Current (first proposed by Maxwell) Maxwell first proposed that this is missing term for Ampere’s law
d E B~ d~s = µ (i + ✏ ) Ampere-Maxwell law · 0 inc 0 dt IC
i inc = current through any surface bounded by C
E = electric flux through that same surface bounded by curve C
= E~ d~a E · ZS
Tuesday, April 24, 18 5 11.2. INDUCED MAGNETIC FIELD 127
11.2 InducedWhere iincl Magnetic= current through Field any surface bounded by C, = electric flux through that same surface☛ boundedcharges by curve C, = E⇤ d⇤a. E E S · Electric field can be generated by ´ 11.2 Induced Magnetic☛ Fieldchanging magnetic flux Ampere-MaxwellWe learn law earlier that a that magnetic electric field field can can be begenerated generated by by charges . ☛ moving chargeschanging magnetic (current) flux We see from Amp`ere-Maxwell law that a magnetic field can be generated by ☛ changingmoving electric charges flux (current) . A change inchanging electric electric flux flux through a surface bounded by That is, a change in electric flux through a surface bounded by C canC lead to an induced magnetic field along the loop C. can lead to an induced magnetic field along loop C
Tuesday, April 24,Notes 18 The induced magnetic field is along the same direction as caused by the 6 changing electric flux.
Example What is the magnetic field strength inside a circular plate capacitor of radius R with a current I(t) charging it?
Answer Electric field of capacitor Q Q E = = 2 ⇥0A ⇥0 R Notes Induced magnetic field is along same direction as caused by ☛ changing electric flux Example What is magnetic field strength inside a circular plate capacitor of radius R with a currentI ( t ) charging it? Answer Electric field of capacitor Q Q 11.3. MAXWELL’S EQUATIONSE = = 2 128 ✏0A ✏0⇡R Electric flux inside capacitor through a loop C of radius r Ampere-MaxwellElectric Law fluxinside inside capacitor: capacitor through a loop C of radius r: d E B~ d~s = µ0(iincl + ✏0 ) 2 C · 2 Qr dt I E = E r = 2 ~ · ⇥0R Binduced d~s 2 * || d Qr 2⇡r Binduced = µ0✏0 2 | {zAmp`ere-Maxwell} Lawdt inside✏0R capacitor: Length of loop C 2 ✓ ◆ r dQ d E B⇤ d⇤s = µ ( i + ⇥ ) = µ0 2 0 incl 0 |{z} R ˛dtC · dt B d s *I(inducedt) µ0r ⇧ ⌅⇤ ⌃ ) Binduced = I(t) for r Outside the capacitor plate: Electric flux through loop C: E = E Q · R2 = ⇥0 d E B⇤ d⇤s = µ0(iincl + ⇥0 ) ˛C · dt 1 dQ 2 rB = µ ⇥ induced 0 0 ⇥ · dt 0 ⇥ µ I(t) B = 0 induced 2 r 11.3 Maxwell’s Equations The four equations that completely describe the behaviors of electric and magnetic fields. 11.3. MAXWELL’S EQUATIONS 128 Electric flux inside capacitor through a loop C of radius r: 2 2 Qr E = E r = 2 · ⇥0R Amp`ere-Maxwell Law inside capacitor: d E B⇤ d⇤s = µ0( iincl + ⇥0 ) ˛C · dt B d s * induced ⇧ ⌅⇤ ⌃ d Qr2 2 r Binduced = µ0⇥0 2 dt ⇥0R Length of loop C ⇥ ⇧⌅⇤⌃ r2 dQ = µ 0 R2 dt I(t) Outside capacitor plate ☛ µ r ⇧⌅⇤⌃ Electric flux throughB loop=C 0 I(t)forr 11.3 Maxwell’s Equations Tuesday, April 24, 18 8 The four equations that completely describe the behaviors of electric and magnetic fields. 11.3 Maxwell’s Equations Four equations that completely describe ☛ behaviors of electric and magnetic fields Q 1 E~ dA~ = inc = ⇢dV · ✏ ✏ IS 0 0 ZV B~ dA~ =0 · IS d E~ d~s = B~ dA~ · dt · IC ZS d B~ d~s = µ0iinc + µ0✏0 E~ Ad~A~ · dt · IC ZS One equation that describes ☛ how matter reacts to electric and magnetic fields F~ = q(E~ + ~v B~ ) ⇥ Tuesday, April 24, 18 9 Maxwell's equations in matter D~ dA~ = ⇢dV · IS ZV B~ dA~ =0 · I@⌦ d E~ d~s = B~ dA~ · dt · IC ZS d H~ d~s = J~ dA~ + D~ dA~ · · dt · IC ZS ZS Electric displacement field ☛ D~ = ✏0E~ + P~ vacuum ☛ D~ = ✏00.3E~ Summary ofisotropic boundary linear conditions dielectric ☛ D~ = ✏E~ Component General materials Linear materials Electric displacement Perpendicular D2,⊥ − D1,⊥ = σf D2,⊥ − D1,⊥ = σf D D 2,∥ 1,∥ D2 − D1 Boundary conditions ☛ Parallel ,∥ ,∥ =? ϵ2 = ϵ1 Electric field Perpendicular ϵ2E2,⊥ − ϵ1E1,⊥ =? ϵ2E2,⊥ − ϵ1E1,⊥ = σf Parallel E2,∥ = E1,∥ E2,∥ = E1,∥ tangential E perpendicular Tuesday, April 24, 18 In principle, then, we know the component of and the 10 component of D.Theothercomponentsdependonpolarisationwhichwedon’tapriori know (except for linear dielectrics). 4 Maxwell’s equations may also be written in differential form Important consequence of Maxwell’s equations prediction electromagnetic waves that travel @ speed of light Reason is due to the fact that changing electric field produces a magnetic field and vice versa Coupling between 2 fields leads to generation of electromagnetic waves Prediction was confirmed by Hertz in 1887 Tuesday, April 24, 18 11 ! ! """ E ! dA = 0, S ! ! d$ E ! d s = # B , #" dt ! ! (13.3.2) """ B ! dA = 0, S ! ! d$ B ! d s = µ % E . #" 0 0 dt An important consequence of Maxwell’s equations, as we shall see below, is the prediction of the existence of electromagnetic waves that travel with the speed of light c =1/ µ0!0 . The reason is due to the fact that a changing electric field is associated with a magnetic field and vice versa, and the coupling between the two fields leads to the generation of electromagnetic waves. In 1887, H. Hertz confirmed this prediction. 13.4 Plane Traveling Electromagnetic Waves To examine the properties of the electromagnetic waves, let’s consider an ! electromagnetic wave propagating in the +x- direction, with a uniform electric field E ! pointing in the + y- direction and a uniform magnetic field B in the +z- direction. At any ! ! instant both E and B are uniform over any yz -plane perpendicular to the direction of 11.4 Plane Travelingpropagation. Electromagnetic This means that for anyWaves value of x , the electric and magnetic fields are the Consider electromagneticsame at all wavepoints propagating yz -plane perpendicular in +x direction to that value of x . The electric and magnetic with uniform E ~ pointingfield are independentin + y -direction of the ( yand,z) coordinates. B ~ in + z -directionTherefore the electric and magnetic fields ! ~ ˆ ! ˆ At any instant bothare onlyE functionsand ~ of☛ theindependent (x,t) coordinates, of ( y, E ( z x ), tcoordinates) = Ey (x,t)j and B(x,t) = Bz (x,t)k . B E~ (x, t)=Ey(x, t)ˆ| yz B~ (x, t)=Bz(x, t)kˆ This (non-physical) Figureelectric 13. and4.1 Electricmagnetic and field magnetic yield fields a plane at a fe wavew selected points along the x -axis because at any instant both E~ andassociated B ~ with a plane electromagnetic wave . are uniform over any plane perpendicular to direction of propagation Tuesday, April 24, 18 12 13-7 ! In the representation shown in Figure 13.4.1, at time t = 0 , consider the vectors E(0,0) ! and B(0,0) , corresponding to the electric and magnetic fields on the plane x = 0 . We ! show two additional pairs of electric and magnetic vectors representing E(x ,ct ) and ! ! ! 1 1 B(x ,ct ) on the plane x = ct , and E(x ,ct ) and B(x ,ct ) on the plane x = ct . 1 1 1 1 2 2 2 2 2 2 This (non-physical) electric and magnetic field is called a plane wave because at any ! ! instant both E and B are uniform over any plane perpendicular to the direction of propagation. In addition, the wave is transverse because both fields are perpendicular to ! ! the direction of propagation, which points in the direction of the cross product E!B . Using Maxwell’s equations, we may obtain the relationship between the magnitudes of the fields and their derivatives. To see this, consider a rectangular loop that lies in the xy- plane, with the left side of the loop at x and the right at x + !x . The bottom side of the loop is located at y and the top of the loop is located at y + !y as shown in Figure 13.4.2. Let the unit vector normal to the loop be in the positive z-direction, nˆ = kˆ . Consider a rectangular loop that lies in xy -plane left side of loop at x and right at x + x bottom at y and top at y + y as shown in ☛ Unit vector normal to loop positive z-direction nˆ = kˆ Recall Faraday’s law d E~ d~s = B~ dA~ (11.4.1) · dt · ! I ZZ Figure 13.4.2 Spatial variation of the electric field E To evaluate LHS of (11.4.1) ☛ integrate around closed path Recall Faraday’s law ! ! d ! ! E~ d~s = E (x + x) y E (x) y "# E!ds = " ## B!(11.4.2)dA . (13.4.1) y y dt · useI Taylor expansion to approximateIn order to evaluate the left-hand-side of Eq. (13.4.1), we integrate counterclockwise @aroundE the closed path shown in Figure 13.4.2, y (11.4.3) Ey(x + x)=Ey(x)+ x + ! ! @x E ! ds = Ey (x + #x)#y $ Ey (x)#y (13.4.2) ··· "" Left-hand-side of Faraday’s law becomes We can use the Taylor expansion to approximate @Ey E~ d~s = x y (11.4.4) · @x Tuesday,I April 24, 18 13 13-8 Assume that x and y are very small such that time derivative of z-component of magnetic field is nearly uniform over area element Rate of change of magnetic flux on right-hand-side of Eq. (11.4.1) is d @B B~ dA~ = z x y (11.4.5) dt · @t ZZ Equating two sides of Faraday’s Law and dividing through by area x y @E @B y = z @x @t (11.4.6) Eq. (11.4.6) indicates that at each point in space time- varying B-field is associated with spatially varying E-field Tuesday, April 24, 18 14 "Ey Ey (x + !x) = Ey (x) + !x +!. (13.4.3) "x Then left-hand-side of Faraday’s law becomes ! ! #Ey "" E ! ds # $x.$y . (13.4.4) #x We assume that !x and !y are very small such that the time derivative of the z - component of the magnetic field is nearly uniform over the area element. Then the rate of change of magnetic flux on the right-hand-side of Eq. (13.4.1) is given by d ! ! $B ! ## B " dA = ! z %x %y . (13.4.5) dt $t Equating the two sides of Faraday’s Law and dividing through by the area !x!y yields !E !B y = " z . (13.4.6) !x !t Eq. (13.4.6) result indicates that at each point in space a time-varying magnetic field is associated with a spatially varying electric field. Second condition on relationship between electric and magnetic fields The second condition on the relationship between the electric and magnetic fields may be maydeduced be by deduced using the Ampereby using-Maxwell Ampere-Maxwell equation: equation d ~ ! ! d ! ! ~ ~ B d~s = µ0iincB!d+s =µµ0"✏0 E!dAE A (11.4.7)(13.4.7) "# 0 0 ## C · dtdt S · I Z ConsiderConsider a rectangulara rectangular loop in theloop xy- inplane xy depicted - plane in Figure depicted 13.4.3, with a unit normal nˆ = ˆj. ! Evaluating lineFigure integral 13.4.3 of Spatial magnetic variation fieldof the magneticaround fieldclosed B path 13-9 B~ d~s = B (x) z B (x + x) z (11.4.8) · z z Tuesday, April I24, 18 15 Use Taylor expansion to approximate @B B (x + x)=B (x)+ z x + (11.4.9) z z @x ··· Left-hand-side of Maxwell-Ampere law becomes @B B~ d~s = z x z (11.4.10) · @x I Assuming that x and z are very small such that time derivative of y -component of electric field is nearly uniform over area element Rate of change of electric flux on right-hand-side of Eq.(11.4.7) is d @Ey µ0 ✏0 E~ dA~ = µ0 ✏0 x z (11.4.11) dt · @t ZZ Equating two sides of Maxwell-Ampere law and dividing by x z yields @Bz @Ey = µ ✏ (11.4.12) @x 0 0 @t Eq. (11.4.12) indicates that at each point in space time-varying E-field is associated to spatially varying B-field Tuesday, April 24, 18 16 Eq. (11.4.6) and Eq. (11.4.12) are coupled differential equations To uncouple them first take another partial derivative of Eq. (11.4.6) with respect to x @2E @ @B @ @B y = z = z Eq. (11.4.13) @x2 @x @t @t @x ⇣ ⌘ ⇣ ⌘ Assumed that field B z is sufficiently well behaved such that partial derivatives are interchangeable @ @B @ @B z = z @x @t @t @x Eq. (11.4.14) ⇣ ⌘ ⇣ ⌘ Substitute Eq. (11.4.12) into Eq. (11.4.13) One-dimensional wave equation @2E @2E y = µ ✏ y Eq. (11.4.15) @x2 0 0 @t2 By a dimensional analysis 1 quantity has same dimensions as speed squared µ0✏0 Tuesday, April 24, 18 17 Repeat argument to find a one-dimensional wave equation satisfied by z-component of magnetic field ☛ taking @/@x of Eq. (11.4.12) 2 @ Bz @ @Ey @ @Ey Eq. (11.4.16) = µµ0✏0 = µµ0✏@0 @x2 0E0 @x @t 0 0 @t @x Substitute Eq. (11.4.6) into Eq. (11.4.16)⇣ yielding⌘ a one-dimensional wave equation satisfied by z -component of magnetic field 2 2 @ Bz @ Bz = µ0✏0 Eq. (11.4.17) @x2 0E0 @t2 General form of a one-dimensional wave equation is given by @2 (x, t) 11 @2 (x, t) = Eq. (11.4.18) @x2 v⌫22 @t2 where ⌫ v is speed of propagation and ( x, t ) is wave function Ey and Bz satisfy wave equation and propagate with speed 1 v⌫ = Eq. (11.4.19) pµ0 ✏0 Tuesday, April 24, 18 18 1 ! = , (13.4.21) 0 µ c2 0 Comparing Eqs. (13.4.21) and (13.4.20), we conclude that the speed of propagation is exactly equal to the speed of light, 7 12 2 Taking µ0 =4⇡ 10 Tm/A and ✏0 =8.854 10 C/N/m ⇥ 1 8 -1 ⇥ ⌫v = vc==2c = 299792458.997 10m !s m. /s (13.4.22) Maxwell’s Equations predict that E- and⇥ B-fields Thus, we conclude that Maxwell’s Equations predict that electric and magnetic field can propagate through space at the speed propagateof light. The spectrumthrough of space electromagnetic at speed waves of lightis shown in Figure 13.4.4. Electromagnetic spectrum Tuesday, April 24, 18 19 Figure 13.4.4 Electromagnetic spectrum 13.4.1 One-Dimensional Wave Equation We shall now explore the properties of wave functions ! (x,t) that are solutions to the one-dimensional wave equation, Eq. (13.4.18). Many types of physical phenomena can be described by wave functions. We have already seen in the previous section that the plane transverse electric and magnetic fields, E (x,t) and B (x,t) , propagating at the y z speed of light satisfy a one-dimensional wave equations. The transverse displacement of a stretched string, y(x,t) will propagate along a string oriented along the x -direction at s speed dependent on the material properties and tension of the string. In each of these 13-12 Tuesday, April 24, 18 20 Standing electromagnetic waves can be formed by confining the electromagnetic waves within two perfectly reflecting conductors, as shown in Figure 13.4.8. 13.6 Poynting Vector In Chapters 5 and 11 we had seen that electric and magnetic fields store energy. Energy 11.5 Poynting Vector can also be transported by electromagnetic waves that consist of both fields. Consider a Energy can also be transportedplane by electromagnetic wave waves passing through a small volume element of area A and thickness dx , as shown in Figure 13.6.1. Consider plane EM wave passing through small volume element of area A and thickness dx as shown in ☛ Total energy stored in electromagnetic fields in volume element is 2 1 2 B dU = uA dx =(uE + uB) Adx = Figure✏00 E 13.6.1+ ElectromagneticAdx Eq. wave (11.5.1) passing through a volume element 2 E µ0 ⇣ ⌘ The2 total energy stored in the electromagnetic fields in the volume element is given by 1 2 B uE = ✏00 E uB = Eq. (11.5.2) 2 E 2µ0 1 " B2 # dU = uAdx = (u + u )Adx = ! E 2 + Adx , (13.6.1) Because electromagnetic wave propagates with speed ofE lightB c $ 0 % 2 & µ0 ' time it takes for wave to movewhere through volume element is dt = dx/c 2 1 2 B One may obtain rate of change of energy per unit areauE = !0 E , uB = . (13.6.2) 2 2µ 2 0 dU c 2 B S = = ✏0 E + Adt are2 theE 0 energy densitiesµ associated withEq. the (11.5.3) electric and magnetic fields. Because the electromagnetic wave0 propagates with the speed of light c , the amount of time it takes Tuesday, April 24, 18 ⇣ ⌘ 21 for the wave to move through the volume element is dt = dx / c . Thus, one may obtain the rate of change of energy per unit area, denoted by the symbol S , as 2 dU c " 2 B # S = = $ !0 E + % . (13.6.3) Adt 2 & µ0 ' -2 The SI unit of S is [W ! m ]. Recall that the magnitude of the fields satisfy E = cB and c = 1/ µ ! . Therefore Eq. (13.6.3) may be rewritten as 0 0 13-21 2 SI unit of S is [W m ] · Recall that magnitude of fields satisfy E = cB and c =1/pµ0✏0 Therefore Eq. (11.5.3) may be rewritten as 2 2 c 2 B cB 2 EB S = ✏00E + = = c✏00E = Eq. (11.5.4) 2 E µ0 µ0 E µ0 Turn this energy⇣ flow into⌘ a vector by assigning direction as direction of propagation Rate of energy flow per unit area is called Poynting vector S ~ (after British physicist John Poynting) and defined by vector product 1 S~ = E~ B~ Eq. (11.5.5) µ0 ⇥ Plane transverse electromagnetic waves fields E ~ and B~ are perpendicular and magnitude of S~ E~ B~ EB S~ = ⇥ = = S Eq. (11.5.6) | | µ0 µ0 Tuesday, April 24, 18 22 2 2 c " 2 B # cB 2 EB S = $ !0 E + % = = c!0 E = . (13.6.4) 2 & µ0 ' µ0 µ0 We can turn this energy flow into a vector by assigning the direction as the direction of ! propagation. The rate of energy flow per unit area is called the Poynting vector S (after the British physicist John Poynting), and defined by the vector product ! 1 ! ! S = E!B . (13.6.5) µ0 ! ! For our plane transverse electromagnetic waves, the fields E and B are perpendicular, ! and the magnitude of S is ! ! ! E ! B EB S = = = S . (13.6.6) µ µ 0 0 As an example, suppose the electric field associated with a plane sinusoidal ! ˆ electromagnetic wave is E = E0 cos(kx "!t)j . The corresponding magnetic field is ! As an example, suppose electricˆ field associated with a plane sinusoidal B = B0 cos(kx "!t)k , and the direction of propagation is the positive x -direction. The electromagnetic wave is E~ = E cos(kx !t)ˆ| Poynting vector is then 0 Corresponding magnetic field is B~ = B cos(kx !t) kˆ 0 and direction of !propagation1 is positive x -direction E B ˆ ˆ 0 0 2 ˆ S = (E0 cos(kx !"t)j) # (B0 cos(kx !"t)k) = cos (kx !"t)i (13.6.7) Poynting vector is thenµ ☛ Eq. (11.5.7) µ 0 0 ~ 1 ˆ E0B0 2 S = (E0 cos(kx ! !t)ˆ|) (B0 cos(kx !t) k)= cos (kx !t)ˆı Asµ0 expected, S points⇥ in the direction of waveµ0 propagation (Figure 13.6.2). As expected ☛ S~ points in direction of wave propagation Tuesday, April 24, 18 23 Figure 13.6.2 Electric and magnetic fields, and the Poynting vector for a plane wave on the plane x = 0 . The intensity of the wave, I , is defined as the time-average of S , and is given by E B E B E 2 cB2 I = S = 0 0 cos2 (kx !"t) = 0 0 = 0 = 0 , (13.6.8) µ 2µ 2cµ 2µ 0 0 0 0 13-22 Intensity of wave I is defined as time-average of S E B E B E2 cB2 I = S = 0 0 cos2(kx !t) = 0 0 = 0 = 0 Eq. (11.5.8) h i µ0 h i 2µ0 2cµ0 2µ0 2 1 recall ☛ cos (kx !t) = Eq. (11.5.9) h i 2 To relate intensity to energy density we first note equality between electric and magnetic energy densities 2 2 2 2 B (E/c) E ✏00E Eq. (11.5.10) uB = = = 2 = E = uE 2µ0 2µ0 2c µ0 2 Time-averaged energy density of wave is then 2 2 ✏00 2 1 2 B0 u = uE + uB = ✏00 E = E E0 = B = Eq. (11.5.11) h i h i E h i 2 µ0 h i 2µ0 Comparing Eqs. (11.5.8) and Eq. (11.5.11) we can conclude that intensity is related to average energy density by I = S = c u Eq. (11.5.12) h i h i Tuesday, April 24, 18 24 11.6 Momentum and Radiation Pressure An electromagnetic wave transports not only energy but also momentum and hence can exert a radiation pressure on a surface due to absorption and reflection of momentum When a plane electromagnetic wave is completely absorbed by a surface momentum transferred is related to energy absorbed by U p = complete absorption Eq. (11.6.1) c (We shall not prove this result as it involves a more complicated description of energy and momentum stored in electromagnetic fields) If EM wave is completely reflected by a surface such as a mirror 2 U p = complete reflection Eq. (11.6.2) c Tuesday, April 24, 18 25 For a wave that is completely absorbed time-averaged radiation pressure (force per unit area) is given by F 1 dp 1 dU P = h i = = Eq. (11.6.3) A A dt Ac dt Because time-averaged rate thatD energyE deliveredD Eto surface is dU = S A dt h i Eq. (11.6.4) D E Substitute Eq. (11.6.4) into Eq. (11.6.3) yielding S P = h i complete absorption Eq. (11.6.5) c If radiation is completely reflected radiation pressure is twice as great as case of complete absorption 2 S P = h i complete reflection Eq. (11.6.6) c Tuesday, April 24, 18 26 Tuesday, April 24, 18 27