5otp, '5, V LIPhysics 322-ElectrodynamicsWinter 2015

Printed Name

Exam 318,2015March

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VECTOR DERTVATIVES

Cunesian. i +. / r !+. / . i i l r =dr. l \ .1.

<t t 0t^ i t t ^

/ ; rL. L l , , \ ^ /af . nr- \(^ ; / ' * (a , , /. j - r f , " - fu,

^ / r r r ' , iJr . \ ^Y- t . . " / '

Sphcr icaf . r1f =d/ i+ rdHA+r\ int) lO6t . l r =.1s;n|bnsd6

t ; ,nJ.rr . v, n ' i - t ' i d- ' .0 '6f,t r JH r \rna 4Q

/r ,p.$c, ,c v.r- -1. I . , , . , , ' ' - ' , ,n, , , - , . I " ' '/ {n ! i ,q | \ |nH ; |o

t t t t . v. , - I f " , """ , , , - ' ' l ,/ \ ,n! L.re da l

I J I a ' - ' , lo . ' f - ' , . , . . , - ' ' ' l ;' ' | , .o nt - u, ' ' ' ' , I Ldr i . , )

L* td,nu, f , , , ; ; ( , I ) - . ,G", : (*" j ,1)- ; ] " , "1,

a2(Dc_u-

F:o_o

t:),(.:r.i_)

tt-l{'..:

C!lindricsl. . l l =. ts i . s tOA +. l . i t t t r=\ ts. tbd.

- i ! ' I i l r I r t^

vr= {+ 9+ 2df r r@ rr :

la

[*r ' '",-[] '

2

Name ___________________________________Student ID ____________ Score_______ last first

, l \ t t ( s rL-T.DA \ I ENT-{L CO \ST{ \TS

.:N8ar lorrCr, \ r r r

/ i : l r IO \ Al

( =.1 l | r . l l l 'nr \

. = 1.6{) , l (J L"C

' , : ( r l l , l ( l lg

iPennr l l i \ i l \ of l rcc \ t r ! ( |

rpcrnrr ihr | r \ ( i f i i .c \ f ! ( ( l

rchrrg. o l the elecrron)

rmi\ \ o l the ele.rro l

SPHERICAL AND CYLI\DRICAL COORI)INATES

Sphc cal

t '= ' i i l 'n \oIr=/nn"!od,

l ' : , ' , r , , r - :t " : ' , " ,1, , , ; .1Ilo=lun i \ t r

( rlindrical

i : . in i r .o.d i - .os ' .*ar i .1" o ii : \ in, \ ,n @i-.o\ , \ ;n r , r i - corodt : c,r , i . rn" P.f .

lli : s in r ! {^or i -sLn ! \urdi +co\di{ : e," ' , .^o i - .o." ' ! r ( ' , i \ in p i0 = \u 'o i - .o 'o i

i : cosdG nn OCr" : \ rn p\ + .osp9i :2

s^ : e,"oi - ' 'n o iC: \ rndi-eo.dii : i

l ,=. \ , - \ ;I

F = 14πε0

qQr2r Coulomb’s Law F =QE+QV×B Lorentz force

E = 14πε0

ρ !r( )r2∫∫∫ r d !τ electric field from continuous charge distribution

3

Name ___________________________________Student ID ____________ Score_______ last first

E•da∫∫ =1ε0Qencl Gauss’ Law (integral) E•∫ dl = 0 (statics)

V r( ) = − E•d l℘

r

∫ Electrostatic potential, and E = −∇V (statics)

∇2V = −ρε0

Poisson’s equation, and ∇2V = 0 Laplace’s equation (regions of no charge)

V (r) = 14πε0

qr and V r( ) = 1

4πε0

ρ !r( )r∫∫∫ d !τ (setting reference point at infinity)

Eabove −Ebelow =σε0n and

∂Vabove∂n

−∂Vbelow∂n

=1ε0σ boundary conditions

Dabove⊥ −Dbelow

⊥ =σ f

V b( )−V a( ) =WQ W =

12

qii=1

n

∑ V ri( ) and W =12

ρV dτ∫∫∫

Q =CV capacitor

∇2V = −1ε0ρ Poisson’s equation ∇2V = 0 Laplace’s equation

V (r) = 14πε0

1rn+1

( !r )n Pn (cosα)ρ( !r )∫∫∫0

∞

∑ d !τ multipole expansion

p = !r∫∫∫ ρ(!r )d !τ

p = qi!ri

1

n

∑ Vdip (r ) = 1

4πε0

p• rr2 dipole moment

Edip =

p4πε0r

3 (2cosθ r + sinθθ )p

Fmag = I d l∫ ×B

K = dI / d⊥ J = dI / da⊥ surface and volume currents

∇•J+ dρdt

= 0 conserved current (continuity equation for current)ß

4

Name ___________________________________Student ID ____________ Score_______ last first

B r( ) = µ04π

I dl '× rr− r ' 2

∫ B r( ) =µ04π

K× rr− r ' 2

da '∫ B r( ) =µ04π

I J× rr− r ' 2

∫ dτ '

B•d l = µ0Ienc∫ static B =∇×A magnetic vector potential. ∇•A = 0 “Coulomb gauge” convention ∇2A = −µ0J (for Coulomb gauge)

A r( ) = µ04π

J r '( )r− r '∫ dτ ' A r( ) = µ0

4πK r '( )r− r '∫ da ' A r( ) = µ0

4πI

r− r '∫ dl '

A•d l =Φm∫

Babove⊥ −Bbelow

⊥ = 0 Babove|| −Bbelow

|| = µ0K Babove −Bbelow = µ0K× n

Habove|| −Hbelow

|| = µ0K f × n

Aabove −Abelow = 0 ∂Aabove

∂n−∂Abelow

∂n= −µ0K

m = I n da∫∫ = Ia Adip r( ) =µ04πm× rr2

N =m×B F =∇ m•B( ) U = −m•B Jb =∇×M Kb =M× n

H•d l∫ = If,encl Habove

⊥ −Hbelow⊥ = −(Mabove

⊥ −Mbelow⊥ ) Habove

|| −Hbelow|| =K f × n

µ = µ0 1+ χm( )

J =σE V = IR P =VI = I 2R JP =∂P∂t polarization current

ε = E•d∫ ε = −dΦM

dt

5

Name ___________________________________Student ID ____________ Score_______ last first

Φ2 =M I1 Φ = LI ε2 = −MdI1dt ε = −L

dIdt W =

12LI 2

∇•1µ0E×B

#

$%

&

'(+

∂∂t

ε02E 2 +

12µ0

B2#

$%

&

'(= −E•J Poynting’s theorem

dWdt

= −ddt

ε02E 2 +

12µ0

B2"

#$

%

&'dτ∫∫∫ −

1µ0E×B

"

#$

%

&'∫∫ • n da Poynting’s theorem

dFi = Tij daj dFidτ i

=∂Tij∂x j

Tij = ε0 EiEj −12δijE

2"

#$

%

&'+

1µ0

BiBj −12δijB

2"

#$

%

&'

ddt

Pi mech( )+Pi fields( )( ) = Tij∫∫ aj ddτP fields( ) = µ0ε0S

ddτL fields( ) = r×µ0ε0S

∂2 f∂z2

−1V 2

∂2 f∂t2

= 0 λ = 2π / k T = 2π / kV ν =1/T ω = 2πν = kV

eiθ = cosθ + isinθ

vacuum ∇2E−µ0ε0

∂2E∂t2

= 0 ∇2B−µ0ε0

∂2B∂t2

= 0 c =1/ µ0ε0

E z, t( ) = E0e

i kz−ωt( ) B z, t( ) = B0ei kz−ωt( ) E0 = cB0

u =12ε0E0

2 S =

12cε0E0

2k I = S p =12cε0E0

2k

lossless ∇2E−µε ∂

2E∂t2

= 0 ∇2B−µε ∂

2B∂t2

= 0 V =1/ µε = c / n n = µε /µ0ε0

u = 12εE 2 +

1µB2

!

"#

$

%& S =

1µE×B I =

12εVE0

2

θi =θr n1 sinθi = n2 sinθt R = Ir / Ii T = It / Ii

E0rE0i ⊥

=

n1µ1cosθi −

n2µ2cosθt

n1µ1cosθi +

n2µ2cosθt

E0tE0i ⊥

=2 n1µ1cosθi

n1µ1cosθi +

n2µ2cosθt

6

Name ___________________________________Student ID ____________ Score_______ last first

E0rE0i ||

=−n2µ2cosθi +

n1µ1cosθt

n2µ2cosθi +

n1µ1cosθt

E0tE0i ||

=2 n1µ1cosθt

n2µ2cosθi +

n1µ1cosθt

tanθB ≅ n2 / n1 sinθc = n2 / n1

lossy ∇2E−µσ ∂E

∂t−µε

∂2E∂t2

= 0 ∇2B−µσ ∂B

∂t−µε

∂2B∂t2

= 0

k 2 = µεω 2 + iµσω d =1/ Im k

Re k =ω εµ2

1+ σεω

!

"#

$

%&2

+1'

(

))

*

+

,,

1/2

Im k =ω εµ2

1+ σεω

!

"#

$

%&2

−1(

)

**

+

,

--

1/2

poor conductor Re k = µε ω Im k =µεσ2

good conductor Re k = Im k =ωσµ2

y+γ y+ω02y = e

mE(t) E(t) = E0e

iωt y(t) = y0eiωt y0 =

e /mω02 −ω 2( )− iγω

E0

ε = ε0 +Nfe2

m1

ω02 −ω 2( )

− iγω

dilute n =cωRe k =1+ 1

2ε0Nfe2

mω02 −ω 2( )

ω02 −ω 2( )

2+γ 2ω 2

α = 2 Imk =1+ ω

cε0Nfe2

mγω

ω02 −ω 2( )

2+γ 2ω 2

plasma σ =Nfe2

m1

γ − iω dilute plasma σ =Nfe2

m1−iω ω p =

Nfe2

ε0m

dilute plasma k2 =

1c2

ω 2 −ω p2{ }

7

Name ___________________________________Student ID ____________ Score_______ last first

B^SIC EQUATTONS OF ELECTnODYMIT|I(XI

Marr!.L F4dda

v.n=lpo ""= -#v.B=0v x a = 611aoeo9E

j .D= pr

O, ."=-#v.l=0vxH=1ra19

Ardf,lry n Br

Dcltido l

I D=foD+P

I n= l r -mha.!d.b

Ldrdr b... Lt

F=4(E+rxE)

E!.r!/, M@nttd..nd For.t

Lituar twalial

I P=.or.E. D =.8I

I v=,-n ' n=lr

E=-iv -+. B=vxA

I t l _ | _\Et tsy: u = i ! \eoE'r -R')dt

Lt '.EnMt P = €o /(E x E) /r

Pttyrtting v.ctor: 5 - 11n x rltto

Lmotfonrutai P = #c'a|

8

NrDC Strd.Dt ID Sc!rc

- I. (2S pohts tot l) Wryc!gigawEtts ( I ot wans)

-d ,i"1,:iliT;:ton' A laser beam in vacuum has povrcr 20

r. (6 p$) Find the magnitude fid dirccrion of the poynting vcctor.1s) = ?P*to I wans \t(t ^^iL L6

@=25' lo ' - ' " ' r " rv/a>(N 0t pe((dN Af peorh64r, at

b. (9 pts) Find thc peak values of tbe E r;::{#)\rr:T&ryt .7,':.oo=6/c. m'ltt

7"ro1ff

; glH'*: HLff;T,H'i',91-T'- l'"1't to"- or rcfractiou r'6' Assumingoroe E rna n ncjas iulllil.ffild:,::ffi'"tr$ffi?ffi f;l*ffi*;1 ls 5k4u€A' av l-.tt€ F4crae G, HedtcEEo = 2,2-tor/6"= l .- l * ,6t Vorrs/)Sr,,rc€ (57 I s o^/ ctr 17. 6<Ot *.EoHo lS

U N cFt.r tuoij , H{* C€ 4 , '(",if," b" )/ NC a€Asq O,/ (E , ,;bo.7,3 \ / i^6.LL T6stA

9

tr. (25 poirb foht) Wrve! h cotductoB. Consider I I MHz plade wave in vacuumincidcnr oo a thick slab. o-f *nn". Co-nry.^a.a good conducor,.no*r"g;,il, ;o youcar sssumc ib conductivity is 6 x IOt(t/Oym

NrDGlost .frrsl

c. (5 pts) What's rhc wavelength ir thc coppcr?

Strdalt lD - Scor.

r. (5 pts) Wbat's rhc wsve's skin deprh in thc copper?A =t / rn? = t k 'uG 42,/z

a iL --- 5 - t / -L21f D6 6xtoz \1r to-1 )

== 6f. zt *tb. (5 pts) Whst's rhc ravelengh in vacuum?7 =c/€ = 3*ros/tre -_ 3@.-t

V -n/P.Q =2TA=o,9 \ rn

d. (5 pb) Whrt's thc wave's propogation velocity (pb8sc velocity) in vacuum?

V-C

G. (5 pts) Whst's rhc wsve's propagation velocity @hasc vctocity) in thc coppcr?

V='1*E = V€ = O,Lt qn, tot = qoo, n/,

l0

Name Studetrt ID Score

lII. (25 points total) Stress tensor. Consider E andB field ofthe form E=Ei, B =0.

a. (10 pts) Find the components ofthe Maxwell Stress Tensor; rite them below:

,.-.2-c

b. (15 pts) Now consider the flat inlerface between a z-ero-field region and rhe fieldregion E = Ei . For various orientations ofthe interface relative to the E field, shownbelow, hnd the magnitude ofthe pressure (or tension, or shear) on the interface andindicate on each figure the direction ofthe pressue (or tension, or shear).

Zo

-')

IIt .o I =-r i

irtJF< -a4* -

)Fv)4t

.LL

2-

TE^J k dt\t

.-=

'v,t-c2/ , -" /uL

Pe€>tu8€

5*€nal1

Name ______________________________________ Student ID _______________ Score________ last first

Physics 322A, Winter 2015 Exam 3, page ED-UWA322A151T-E3(EPW,WAB)Sol.doc

IV. [25 points total] This question consists of three independent parts; A, B, and C.

A. [5 pts] For each of the following plane waves, identify which direction the wave is propagating. Briefly explain your reasoning for each wave. Assume k, l, α, and ω are positive constants.

Eo exp[i(kx −ωt)] Eoeαz exp[i(−ky+ωt)]

To find the direction of propagation, we can imagine tracking a location of fixed phase on the wave. Thus in the first case, as time increases, x must also increase to keep phase constant, so the wave is travelling in the +x direction. Similarly, the second wave is travelling in the +y direction.

−Eo exp[−i(kx +ωt)] Eo exp[i(kx − ly−ωt)] The third wave is travelling in the –x direction, for as time increases x must decrease to keep the phase constant. The fourth wave is travelling partly in the +x direction and partly in the –y direction, so altogether it is travelling in the (kx-ly)/sqrt(k2 + l2) direction.

B. [6 pts] Each of the following diagrams shows the fields (to scale) on either side of the boundary between two linear media. There are no free currents at the boundary, but there may be free charge.

E1

B2

B1

E2

Case 1 Case 2

E1

B2

B1

E2

Case 3

E1

B2

B1

E2

For each diagram, state whether or not the fields shown are physically possible. Explain your reasoning for each case.

Case 1 is not possible (the perpendicular components of B are not continuous, implying that there is magnetic charge at the boundary). Case 2 is also not possible (the parallel components of E are not continuous, implying there is a discontinuity in B at the boundary, which in turn implies that there is free current at the boundary). Case 3 is possible; the parallel components of E are continuous, as are the perpendicular components of B. The perpendicular components of E are not continuous, but that is okay since there may be free charge at the boundary.

C. A plane wave is traveling through a linear medium. The plane wave has wavelength λ, period T, amplitude E

!"o, and is traveling in the (x + y) direction.

i. [4 pts] What direction could this wave be polarized? Explain.

It could be polarized in any direction perpendicular to the direction of propagation; for instance, in +z.

ii. [3 pts] How fast is this wave travelling?

The speed of a wave is defined as wavelength over frequency, or λ/T.

iii. [7 pts] Write the complex waveform that describes this plane wave, including the polarization. Explain.

Eo exp[i(2πλ{ x2+y2}− 2π

Tt)] z

The angular frequency is 2π/T, and the wave vector is 2π /λ. The wave is travelling in the (x+y) direction, so the relative sign between the two terms in the exponent must be negative, and since the wave vector has equal components in both directions, each has a factor of 1/sqrt(2).