Solved Problems of Jackson's Electrodynamics 02

PHY 5346 HW Set 1 Solutions – Kimel 2 1.3 In general we take the charge density to be of the form ρ = fr⃗δ, where fr⃗ is determined by physical constraints, such as ∫ ρd 3 x = Q. a) variables: r, θ, φ. d 3 x = dφd cos θr 2 dr ρ = fr⃗δr − R = frδr − R = fRδr − R

∫ ρd 3 x = fR ∫ r 2 drdΩdr − R = 4πfRR 2 = Q → fR = ρr⃗ =

Q 4πR 2

Q δr − R 4πR 2

b) variables: r, φ, z. d 3 x = dφdzrdr ρr⃗ = fr⃗δr − b = fbδr − b

∫ ρd 3 x = fb ∫ dzdφrdrδr − b = 2πfbbL = λL → fb =

λ 2πb

λ δr − b 2πb c) variables: r, φ, z. d 3 x = dφdzrdr Choose the center of the disk at the origin, and the z-axis perpendicular to the plane of the disk ρr⃗ = fr⃗δzθR − r = fδzθR − r where θR − r is a step function. 2 ∫ ρd 3 x = f ∫ δzθR − rdφdzrdr = 2π R2 f = Q → f = πRQ 2 Q ρr⃗ = δzθR − r πR 2 d) variables: r, θ, φ. d 3 x = dφd cos θr 2 dr ρr⃗ = fr⃗δcos θθR − r = frδcos θθR − r ρr⃗ =

∫ ρd 3 x = ∫ frδcos θθR − rdφd cos θr 2 dr = ∫ 0 frr rdrdφ = 2πN ∫ 0 rdr = πR 2 N = Q R

R

where I’ve used the fact that rdrdφ is an element of area and that the charge density is uniformly distributed over area. Q ρr⃗ = δcos θθR − r πR 2 r

PHY 5346 HW Set 1 Solutions – Kimel 3. 1.4 Gauss’s Law:

∫ E⃗ ⋅ da⃗ =

Q enclosed 0 Q a) Conducting sphere: all of the charge is on the surface σ = 4πa 2 Q E4πr 2 = 0, r < a E4πr 2 =  0 , r ∑ a Q ̂ ⃗ = E = 0, r < a E r, r ∑ a 4π 0 r 2

b) Uniform charge density: ρ =

4 3

Q πa 3

, r < a, ρ = 0, r ∑ a.

Qr 3 Qr →E= , r
E4πr 2 =

c) ρ = A r n Q = 4π ∫ r 2 drAr n = 4πAa n+3 /n + 3 → ρ =

n + 3Q n r ,r
ρ = 0, r ∑ a E4πr 2 =

Q n + 3Q 4πr n+3 /n + 3 → E = n+3 4π 0 r 2 4π 0 a Q E= , r∑a 4π 0 r 2

1. n = −2. E=

Q , 4π 0 ra

r
r n+3 a n+3

, r
E=

Q , r∑a 4π 0 r 2

2. n = 2. Qr 3 , 4π 0 a 5 Q E= , 4π 0 r 2 E=

r
PHY 5346 HW Set 1 Solutions – Kimel 4. 1.5 qe −αr 1 + αr2 4π 0 r , ∇ 2 = 12 ∂ r 2 ∂ ∂r r ∂r −αr α −αr ∂ r2 ∂ e r + 2e ∂r ∂r

φr⃗ = ρ ∇ 2 φr⃗ = −  0 ∇2φ = Using ∇ 2

1 r

= −4πδr⃗ ∇2φ = =

q 4π 0

q 1 4π 0 r 2

q 1 ∂ 1 − α 2 r 2 e −αr −αr −αr 2 ∂ −αre + e r 2 4π 0 r 2 ∂r ∂r r 2 −αr −αr 2 −αr 3 −αr − α2 e −αr + α er + αe 2 − 4πδr⃗ − α er + α e 2 r r 3 qα −αr = − 10 qδr⃗ − e 8π

qα 3 −αr ρr⃗ = qδr⃗ − e 8π That is, the charge distribution consists of a positive point charge at the origin, plus an exponentially decreasing negatively charged cloud.

PHY 5346 HW Set 2 Solutions – Kimel ⃗ ⋅ da ⃗= 2. 1.8 We will be using Gauss’s law ∫ E a) 1) Parallel plate capacitor

From Gauss’s law E =

σ 0

=

Q A 0

φ 12 = W=

0 2

∫ E2d3x =

Q 1 4π 0 r 2

∫ a Edr = b

0 2

W= 1 8 0 3) Cylindrical conductor

2

Ad

A 0 φ 12 d

Q2 = 1 d= 1 2 0 A 2 0

2

A 0 φ 12 d

2

A

φ d = 1  0 A 12 d 2 2

, a < r < b.

Q 4π 0

Q 4π 0

→Q=

Q A 0

0 2  W = 0 ∫ E 2 d 3 x =  0 E Ad = 2 2 2) Spherical capacitor

From Gauss’s law, E =

φ 12 d

=

Q enc 0

2

1 Q b − a → Q = 4π 0 baφ 12 4 π 0 ba b − a

∫ a r −2 dr = b

∫ a 4π r rdr 4 b

2

4π 0 baφ 12 b−a

π

= 2

0 2

Q 4π 0

2

−4π −b + a ba

φ 212 b − a = 2π 0 ba ba b − a

=

2 1 Q b − a 8 0 π ba

λL

From Gauss’ s law, E2πrL = φ 12 = W=

0 2

=

0

Q 0

∫ a Edr = b

0 2

∫ E2d3x =

Q 2π 0 L Q 2π 0 L 2π 0 Lφ 12

W= 0 2

b) w = E 1) wr =

0 2

Q A 0

2) wr =

0 2

Q 1 4π 0 r 2

1 4π 0

ln

b a

L

∫a

b

2

Q dr = ln ba r 2π 0 L

2 b 1 Q ln b 2πL ∫ rdr = 2 a 4π 0 L a r

2

ln ba

= π 0 L

φ 212 ln ba

2

3) wr =

0 2

Q 2π 0 Lr

2

2 1 Q 2 0 A 2

=

2

2

=

=

0 < r < d, = 0 otherwise.

Q2 1 2 32 0 π r 4

Q2 1 8 0 π 2 L 2 r 2

, a < r < b, = 0, otherwise

, = 0 otherwise.

PHY 5346 HW Set 2 Solutions – Kimel 3. 1.9 I will be using the principle of virtual work. In the figure below, Fδl is the work done by an external force. If F is along δl (ie. is positive), then the force between the plates is attractive. This work goes into increasing the electrostatic energy carried by the electric field and into forcing charge into the battery holding the plates at constant potential φ 12 .

Conservation of energy gives Fδl = δW + δQφ 12 or ∂Q F = ∂W + φ 12 ∂l ∂l From problem 1.8, a) Charge fixed. 1) Parallel plate capacitor dQ

2

φ2 dQ → W = 1  0 A A 0 = d Q2 W = 1  0 A 12 , φ 12 = A 0 d 2 2 0 A 2 d 2 Q ∂Q = 0, F = ∂W = (attractive) 2 ∂l ∂l 0A 2) Parallel cylinder capacitor φ 12 = λ0 ln da , a = a 1 a 2 λQ W = 1 Qφ 12 → F = ∂W = 1 Q λ0 ∂ ln da  = 1 (attractive) 2 2 2 0d ∂l ∂d b) Potential fixed 1) Parallel plate capacitor Using Gauss’s law, Q =

φ 12 A 0 , d

∂ Q = φ 12 A 0 ∂l d2

2 φ2 φ 212 A 0 1  0 A φ 12 = 1  0 A → F = − 1  0 A 12 + = 2 2 2 d2 d2 d2 2) Parallel cylinder capacitor  Lφ W = 1 Qφ 12 , and Q = 0 d 12 2 ln a 

Qd 0A 2

d

2

=

1 Q2 2 0 A

so  Lφ 2 φ 212 W = 1 0 d 12 , ∂W = − 1  0 L 2 2 ln a  ∂l ln 2 da d ∂Q ∂l

= 0L

φ 12 ln 2 da d

φ 212 φ 212 φ 212 1 L F = − 1 0L +  L = 0 0 2 2 ln 2 da d ln 2 da d ln 2 da d

PHY 5346 HW Set 2 Solutions – Kimel 4. 2.1 We will work in cylidrical coordinate, (ρ, z, φ, with the charge q located at the point ⃗ d = dẑ , and the conducting plane is in the z = 0 plane. Then we know from class the potential is given by

Ez = −

q ∂ 4π 0 ∂z

Ez =

q q − ⃗ ⃗ x−⃗ d x+⃗ d

1 4π 0

φx⃗ =

1 z − d 2 + ρ 2 z−d z − d 2 + ρ 2

q 4π 0

z+d z + d 2 + ρ 2

−

3/2

1 z + d 2 + ρ 2

−

1/2

1/2

3/2

a) σ =  0 E z z = 0 σ = 0 Plotting

q 4π 0 −1

1 2 +r 2

3 2

−d −d + ρ 2 2

+d +d + ρ 2

−

3/2

=−

3/2

2

q 2πd 2

1 2

3 2

π2d4 1 +

ρ2 d2

1 + 2

ρ d

gives 1

2

r

3

4

5

0 -0.2 -0.4 -0.6 -0.8 -1

b) Force of charge on plane ⃗= F

q2 ̂ 1 q−q −ẑ  = 1 z 2 4π 0 2d 4 × 4π 0 d 2

c) 2

F = w = 0 E2 = σ2 = 1 2 0 2 0 A 2 F = 2π

q2 8 0 π 2 d 4

q 2πd 2

ρ

β

∫0

−

1+

ρ2 d2

3

1 12 +

dρ = 2π

ρ d

2

q2 8 0 π 2 d 4

=

3 2

1 d2 4

1 8 0 =

q2

q2 1 4 × 4π 0 d 2

3

d) W=

β

∫ d Fdz =

q2 4 × 4π 0

β

∫d

q2 dz = 4 × 4π 0 d z2

e) W= 1 1 2 4π 0

> i,j,iΦj

qiqj q2 =− 2 × 4π 0 d xj | |x⃗i − ⃗

Notice parts d) and e) are not equal in magnitude, because in d) the image moves when q moves. f) 1 Angstrom = 10 −10 m, q = e = 1. 6 × 10 −19 C.

W=

−19 q2 e =e = e 1. 6 × 10−10 9 × 10 9 V = 3.6 eV 4 × 4π 0 d 4 × 4π 0 d 4 × 10

PHY 5346 HW Set 2 Solutions – Kimel 5. 2.2 The system is described by

a) Using the method of images φx⃗ =

1 4π 0

q q′ + y′ | y| |x⃗ − ⃗ |x⃗ − ⃗

with y ′ = ay , and q ′ = −q ay b) σ = − 0 ∂n∂ φ|x=a = + 0 ∂x∂ φ|x=a 2

σ = 0

1 ∂ 4π 0 ∂x

q q′ + 1/2 1/2 x 2 + y 2 − 2xy cos γ x 2 + y ′2 − 2xy ′ cos γ a 1−

y2 a2

σ = −q 1 4π y 2 + a 2 − 2ay cos γ 3/2 Note y2 q induced = a 2 ∫ σdΩ = −q 1 a 2 2πa 1 − 2 4π a

∫ −1 1

dx , where x = cos γ 3/2 y 2 + a 2 − 2ayx

q 2 q induced = − aa 2 − y 2  = −q 2 aa 2 − y 2  c) qq ′ q 2 ay 1 = , the force is attractive, to the right. 4π 0 a 2 − y 2  4π 0 y ′ − y 2 d) If the conductor were fixed at a different potential, or equivalently if extra charge were put on the conductor, then the potential would be q′ q φx⃗ = 1 + +V 4π 0 |x⃗ − ⃗ y| y′ | |x⃗ − ⃗ and obviously the electric field in the sphere and induced charge on the inside of the sphere would remain unchanged. |F| =


a) Given the potenial for a line charge in the problem, we write down the solution from the figure, λ 4π 0

R2 R2 R2 R2 − ln − ln + ln 2 2 2 xo  x o1  x o2  x o3  2 x⃗ − ⃗ x⃗ − ⃗ x⃗ − ⃗ x⃗ − ⃗ x o3  2 , so φ T |y=0 = 0 x o2  2 = x⃗ − ⃗ x o1  2 , x⃗ − ⃗ Looking at the figure when y = 0, x⃗ − ⃗ x o  2 = x⃗ − ⃗ 2 2 2 2 x o3  , so φ T |x=0 = 0 x o1  = x⃗ − ⃗ x o2  , x⃗ − ⃗ Similarly, when x = 0, x⃗ − ⃗ x o  = x⃗ − ⃗ On the surface φ T = 0, so δφ T = 0, however, ∂φ T ∂φ T δφ T = δx = 0 → = 0 → Et = 0 ∂x t t ∂x t b) We remember ∂φ y0 y0 − σ = − 0 T = −λ π 2 2 ∂y x + x 0  2 + y 20 x − x 0  + y 0 where I’ve applied the symmetries derived in a). Let y0 y0 σ/λ = −1 π x − x  2 + y 2 − x + x  2 + y 2 0 0 0 0 This is an easy function to plot for various combinations of the position of the original line charge (x 0 , y 0 . φT =

ln

c) If we integrate over a strip of width Δz, we find, where we use the integral x0 ∞ y0 ∫ 0 x ∓ x 1 2 + y 2 dx = 12 π ± 2 arctan y0 0 0 ∞ ∞ ΔQ −1 x 0 ΔQ = ∫ σdxΔz → = ∫ σdx = −2λ π tan y0 Δz 0 0 and the total charge induced on the plane is −∞, as expected. d) Expanding

ln

R2 R2 R2 − ln − ln x − x 0  2 + y − y 0  2 x − x 0  2 + y + y 0  2 x + x 0  2 + y − y 0  2 to lowest non-vanishing order in x 0 , y 0 gives xy 16 2 y x 2 0 0 x + y 2  so xy 4λ y x φ → φ asym = π 2 0 0 0 2 x + y 2  This is the quadrupole contribution.

+ ln

R2 x + x 0  2 + y + y

PHY 5346 HW Set 3 Solutions – Kimel 3. 2.5 a) ∞

∫ r |F|dy =

W=

q2a 4π 0

∞

∫r

dy y3 1 −

2

a2 y2

=

q2a 8π 0 r 2 − a 2 

Let us compare this to disassemble the charges − W′ = − 1 8π 0

∑ i≠j

qiqj |x⃗i −x⃗j |

=

aq 2 r

1 4π 0

1 r 1−

a2

=

r2

q2a >W 4π 0 r 2 − a 2 

The reason for this difference is that in the first expression W, the image charge is moving and changing size, whereas in the second, whereas in the second, they don’t. b) In this case W=

∞

∫r

|F|dy =

q 4π 0

∞

∫r

∞ 2y 2 − a 2  Qdy 3 − qa ∫ r 2 2 2 dy y2 yy − a 

Using standard integrals, this gives W=

1 4π 0

q2a qQ q2a − − r 2 2 2 2r 2r − a 

On the other hand qQ + ar q q2a qQ aq 2 aq 2 − = 1 − 2 − r 2 2 2 2 r 4π 0 r − a  r r − a  The first two terms are larger than those found in W for the same reason as found in a), whereas the last term is the same, because Q is fixed on the sphere. − W′ =

1 4π 0

PHY 5346 HW Set 3 Solutions – Kimel 4. 2.6 We are considering two conducting spheres of radii r a and r b respectively. The charges on the spheres are Q a and Q b . a) The process is that you start with q a 1 and q b 1 at the centers of the spheres, and sphere a then is an equipotential from charge q a 1 but not from q b 1 and vice versa. To correct this we use the method of images for spheres as discussed in class. This gives the iterative equations given in the text. b) q a 1 and q b 1 are determined from the two requirements β

β

j=1

j=1

> q a j = Q a and > q b j = Q b As a program equation, we use a do-loop of the form n

>q a j = j=2

−r a q b j − 1 d b j − 1

and similar equations for q b j, x a j, x b j, d a j, d b j. The potential outside the spheres is given by n

φx⃗ =

1 4π 0

> j=1

n

q b j q a j +> ̂ ⃗ ⃗ x − x a jk x − d b jk̂ j=1

This potential is constant on the surface of the spheres by construction. And the force between the spheres is q a jq b k F= 1 > 4π 0 j,k d − x a j − x b k 2 c) Now we take the special case Q a = Q b , r a = r b = R, d = 2R. Then we find, using the iteration equations x a j = x b j = xj x1 = 0, x2 = R/2, x3 = 2R/3, or xj =

j − 1 R j

q a j = q b j = qj qj = q, q2 = −q/2, q3 = q/3, or qj =

−1 j+1 q j

So, as n → β β

β

j=1

j=1

> qj = q >

Q −1 j+1 = q ln 2 = Q → q = ln 2 j

The force between the spheres is q2 −1 j+k F= 1 > 4π 0 R 2 j,k jk 2 − j−1 − k−1 j k Evaluating the sum numerically

2

=

2 1 q 4π 0 R 2

jk > −1 j + k 2 j+k

j,k

2 2 1 0. 0739 1 q 0. 0739 = 1 Q 2 2 4π 0 R 4π 0 R ln 2 2 Comparing this to the force between the charges located at the centers of the spheres Q2 Fp = 1 4π 0 R 2 4 Comparing the two results, we see F = 4 1 2 0. 0739F p = 0. 615F p ln 2 On the surface of the sphere

F=

β

φ=

1 4π 0

> j=1

qj q = 4π 0 R R − xj

β

>−1 j+1 j=1

β

Notice So φ =

1 = 1+1

>−1 j+1 j=1

Q q Q 1 C = 1 = → = 2 ln 2 = 1. 386 4π 0 2R C 4π 0 2 ln 2R 4π 0 R


a) The Green’s function, which vanishes on the surface is obviously Gx⃗, ⃗ x′ =

1 1 − x′ | x ′I | |x⃗ − ⃗ |x⃗ − ⃗

where ⃗ x ′I = x ′ î + y ′ ̂ − z ′ k̂ x ′ = x ′ î + y ′ ̂ + z ′ k̂ , ⃗ b) There is no free charge distribution, so the potential everywhere is determined by the potential on the surface. From Eq. (1.44) ∂Gx⃗, ⃗ x′ ′ φx⃗ = − 1 ∫ φx⃗′  da ′ 4π S ∂n Note that n̂ ′ is in the −z direction, so 2z ∂ Gx⃗, ⃗ x ′ |z ′ =0 = − ∂ Gx⃗, ⃗ x ′ |z ′ =0 = − 3/2 ∂z ′ 2 ∂n ′ x − x  + y − y ′  2 + z 2 So a 2π φx⃗ = z V ∫ ∫ 2π 0 0

ρ ′ dρ ′ dφ ′ 2 x − x ′  + y − y ′  2 + z 2

3/2

where x ′ =ρ ′ cos φ ′ , y ′ = ρ ′ sin φ ′ . c) If ρ = 0, or equivalently x = y = 0, a 2π a ρdρ ρ ′ dρ ′ dφ ′ φz = z V ∫ ∫ = zV ∫ 3/2 3/2 2 ′2 2 2π 0 0 ρ + z 0 ρ + z 2 φz = zV −

z − a 2 + z 2  z a + z  2

2

= V 1−

z a + z 2  2

d) a 2π φx⃗ = z V ∫ ∫ 2π 0 0

ρ ′ dρ ′ dφ ′ ⃗−ρ ⃗ ′ 2 + z2 ρ

3/2

⃗, then ρ ⃗⋅ρ ⃗ ′ = ρρ ′ cos φ ′ In the integration choose the x −axis parallel to ρ a 2π ρ ′ dρ ′ dφ ′ φx⃗ = z V ∫ ∫ 2π 0 0 ρ 2 + ρ ′2 − 2ρ ⃗⋅ρ ⃗ ′ + z 2 3/2 Let r 2 = ρ 2 + z 2 , so a 2π ρ ′ dρ ′ dφ ′ φx⃗ = z V3 ∫ ∫ 3/2 ⃗⋅ρ ⃗′ 2π r 0 0 ρ ′2 −2ρ 1+ r2

We expand the denominator up to factors of O1/r , ( and change notation φ ′ → θ, ρ ′ → α, r 2 → β12  4

φx⃗ = z V3 2π r

∫0 ∫0 a

2π

αdαdθ 1+

⃗⋅α ⃗ α 2 −2ρ r2

3/2

where the denominator in this notation is written 1 3/2 1 + β α − 2ρα cos θ 2

2

or, after expanding, a 2π φx⃗ = z V3 ∫ αdα ∫ 1 − 3 β 2 α 2 + 3β 2 ρα cos θ + 15 β 4 α 4 − 15 β 4 α 3 ρ cos θ + 15 β 4 ρ 2 α 2 cos 2 θ dθ 2π r 0 2 8 2 2 0 Integrating over θ gives a φx⃗ = z V3 ∫ α 2π + 15 β 4 α 4 π − 3β 2 α 2 π + 15 β 4 ρ 2 α 2 π dα 2π r 0 4 2 Integrating over α yields φx⃗ = z V3 5 β 4 πa 6 − 3 a 4 β 2 π + 15 a 4 β 4 ρ 2 π + πa 2 4 8 2π r 8 or

φx⃗ =

Va 2 2 2ρ + z 2  3/2

a2 1− 3 + 5 2 4 ρ + z 2  8

a 4 + 3ρ 2 a 2 2 ρ 2 + z 2 

PHY 5346 HW Set 4 Solutions – Kimel 1. 2.8 The system is pictured below

a) Using the known potential for a line charge, the two line charges above give the potential ′ φr⃗ = 1 λ ln rr = V, a constant. Let us define V ′ = 4π 0 V 2π 0 Then the above equation can be written r ′ 2 = e Vλ′ or r ′2 = r 2 e Vλ′ r 2 ⃗ , the above can be written Writing r ′2 = ⃗r − R 2

R ⃗r + ẑ V′ e λ − 1

V′

R2e λ V′ e λ − 1 2

=

The equation is that of a circle whose center is at −ẑ

R e

V′ λ

−1

, and whose radius is a =

b) The geometry of the system is shown in the fugure.

Note that d = R + d1 + d2 with R

d1 = e

V ′a λ

−1

R

, d2 = e

−V ′b λ

−1

and ′ Va

−V ′b

2λ 2λ a = Re , b = Re −V ′b V ′a e λ − 1 e λ − 1

Forming

V′

Re 2λ e

V′ λ

−1

2

d −a −b = 2

2

2

R

R+ e

V ′a λ

R

+

−1

e

−V ′b λ

−

−1

2

′ Va

Re 2λ V ′a e λ − 1

−

Re e

−V ′b λ

−V ′b 2λ

2

− 1

or R2 e d2 − a2 − b2 =

e

V ′a λ

V ′a −V ′b λ

− 1e

+1 −V ′b λ

− 1

Thus we can write e

d2 − a2 − b2 = 2ab

V ′a −V ′b λ

+1

V ′a

2e 2λ e

= e

−V ′b 2λ

V ′a −V ′b 2λ

+e 2

− V ′a −V ′b 2λ

= cosh

V ′a − V ′b 2λ

or V a − V b = 1 cosh −1 d 2 − a 2 − b 2 λ 2ab 2π 0 Q/L 2π 0 λ Capacitance/unit length = C = = = −1 d 2 −a 2 −b 2 Va − Vb L Va − Vb cosh 2ab

′

c) Suppose a << d , and b << d , and a = ab , then 2π 0 2π 0 C = = 2 1−a 2 +b 2 /d 2 d −1 d 2 −a 2 −b 2 L −1 cosh cosh ′2 2a 2a ′2 2

2

2

cosh −1

2

d 2 1 − a 2 + b 2 /d 2  2a ′2

d 2 1 − a 2 + b 2 /d 2  2a ′2

= e

2π 0 L C

2

= 2π 0 L C

+ negligible terms if 2π 0 L >> 1 C

or ln

d 2 1 − a 2 + b 2 /d 2  a ′2

= 2π 0 L C

or C = L Let us defind α 2 = a 2 + b 2 /d 2 , then 2π 0 C = 2 1−α 2 d L ln a ′2

ln

2π 0 d 2 1−a 2 +b 2 /d 2 a ′2

= 2π  0d 2 + 2π 2 0d 2 α 2 + Oα 4  ln a ′2 ln a ′2

The first term of this result agree with problem 1.7, and the second term gives the appropriate correction asked for. d) In this case, we must take the opposite sign for d 2 − a 2 − b 2 , since a 2 + b 2 > d 2 . Thus 2π 0 C = −1 a 2 +b 2 −d 2 L cosh 2a ′2 If we use the identiy, lnx + x 2 − 1  = cosh −1 x, G.&R., p. 50., then for d = 0

C = L ln in agreement with problem 1.6.

2π 0 2 −b 2 + a2ab

a 2 +b 2 2ab

=

2π 0 ln ab

PHY 5346 HW Set 4 Solutions – Kimel 2. 2.9 The system is pictured below

a) We have treated this probem in class. We found the charge density induced was σ = 3 0 E 0 cos θ We also note the radial force/unit area outward from the surface is σ 2 /2 0 . Thus the force on the right hand hemisphere is, using x = cos θ 1 ⃗ = 1 3 0 E 0  2 2πR 2 ∫ x 3 dx = 1 3 0 E 0  2 2πR 2 /4 = 9 π 0 E 20 R 2 F z = 1 ∫ σ 2 ẑ ⋅ da 2 0 2 0 2 0 4 0 An equal force acting in the opposite direction would be required to keep the hemispheres from sparating. b) Now the charge density is σ = 3 0 E 0 cos θ + Fz =

1 2 0

Q Q Q = 3 0 E 0 x + = 3 0 E 0 x + 2 2 4πR 12π 0 E 0 R 2 4πR

∫ σ 2 ẑ ⋅ da⃗ =

Q 1 3 E  2 2πR 2 ∫ 1 x x + 0 0 2 0 0 12π 0 E 0 R 2

2

dx

Thus 1 F z = 9 π 0 E 20 R 2 + 1 QE 0 + Q2 2 4 32 0 πR 2 An equal force acting in the opposite direction would be required to keep the himispheres from separating.

PHY 5346 HW Set 4 Solutions – Kimel 3. 2.10 As done in class we simulate the electric field E 0 by two charges at ∞

σ = 3 0 E 0 cos θ a) This charge distribution simulates the given system for cos θ > 0. We have treated this probem in class. The potential is given by 3 φx⃗ = −E 0 1 − a3 r cos θ r Using σ = − 0 ∂ φ|surface ∂n We have the charge density on the plate to be 3 σ plate = − 0 ∂ φ|z=0 =  0 E 0 1 − a 3 ∂z ρ σ plate 1 For purposes of plotting, consider  0 E 0 = 1 − x 3 1 0.8 0.6 0.4 0.2

0

2

4

x 6

8

σ boss = 3 0 E 0 cos θ = For plotting, we use

σ boss 3 0 E 0

= cos θ

10

1 0.8 0.6 0.4 0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

b) q = 3 0 E 0 2πa 2 ∫ xdx = 3π 0 E 0 a 2 1

0

c) Now we have φr⃗ = where q ′ = −q ad , d ′ =

1 4π 0 a2 d

q′ q + ⃗r − ⃗ ⃗r − ⃗ d d′

q q′ − ⃗r + ⃗ ⃗r + ⃗ d d′

.

−q σ = − 0 ∂ φ|r=a = 4π ∂r

q ind = 2πa 2

q ind =

−

−qa 2 d 2 − a 2  2a

2 2 q ind = −1 q d − a 2 d

d 2 − a 2  d 2 − a 2  − 3 3 a⃗ a−⃗ d a+⃗ d a⃗ d 2 − a 2  d 2 − a 2  − 3 3 a⃗ a−⃗ d a+⃗ d a⃗

−q 4π

∫0

1 da

1 − d−a

1

2d − d2 − a2

1 + 1 − d+a 2 a +d 2

2 2 a + d2

= −q 1 −

dx

1 a + d2 2

d 2 − a 2  d a2 + d2

PHY 5346 HW Set 4 Solutions – Kimel 4. 2.11 The system is pictured in the following figure:

a) The potential for a line charge is (see problem 2.3) φr⃗ = λ ln rr∞ 2π 0 Thus for this system φr⃗ =

τ ln 2π 0

r∞ ⃗ ⃗r − R

+

τ ′ ln 2π 0

To determine τ ′ and R ′ , we need two conditions: I) As r → ∞, we want φ → 0, so τ ′ = −τ. II) φr⃗ = b  = φr⃗ = −b  or ′ ln b − R R−b

′ = ln b + R b+R

or b − R′ R−b ′ This is an equation for R with the solution

=

b + R′ b+R

2 R′ = b R

The same condition we found for a sphere. b)

r∞ ⃗′ ⃗r − R

r2 +

φr⃗ = τ ln 4π 0

b4 R2

2

− 2r bR cos φ

r 2 + R 2 − 2rR cos φ

as r → ∞ φr⃗ =

2 τ ln 1 − 2b cos φ/rR 4π 0 1 − 2R cos φ/r

=

τ ln 1 − 2 b 2 − R 2  cos φ rR 4π 0

Using ln1 + x = x − 1 x 2 + 1 x 3 + Ox 4  3 2 1 b 2 − R 2  cos φ φr⃗ = − τ 2π 0 rR c) σ = − 0 ∂ φ|r=b = − τ ∂ ln 4π ∂r ∂r τ 2πb

σ= where y = R/b. Plotting σ/

τ 2πb

gy = g2, g4 = −

3 5−4 cos φ

b4 R2

2

− 2r bR cos φ

r 2 + R 2 − 2rR cos φ

r=b

1−y y + 1 − 2y cos φ 2

2

1−y 2 y 2 +1−2y cos φ

=

r2 +

, for y = 2, 4, gives

1 − y2 y 2 + 1 − 2y cos φ

, − 17−815cos φ

0.5

1

1.5

2

2.5

3

0 -0.5 -1 -1.5 -2 -2.5 -3

d) If the line charges are a distance d apart, then the electric field at τ from τ ′ is, using Gauss’s law ′ E= τ 2π 0 d The force on τ is τLE, ie, ′ 2 F = ττ L = − τ L , and the force is attractive. 2π 0 d 2π 0 d


a) Notice from the figure, Φρ, −φ = Φρ, φ; thus from Eq. (2.71) in the text, β

Φρ, φ = a 0 + ∑ a n ρ n cosnφ n=1

∫ −π/2 Φb, φ = 2πa 0 = πV1 + πV2 → a 0 = 3π/2

V1 + V2 2

Using

∫ −π/2 cos mφ cos nφdφ = δ nm π 3π/2

Applying this to Φ, only odd terms m contribute in the sum and m−1 2V 1 − V 2  am = −1 2 πmb m Thus 2V 1 − V 2  i m ρ m e imφ Φρ, φ = V 1 + V 2 + Im ∑ π 2 mb m m odd

Using 2

∑ m odd

x m = ln 1 + x m 1 − x

and Im lnA + iB = tan −1 B/A we get V 1 − V 2  V + V2 Φρ, φ = 1 + tan −1 π 2 as desired. b) σ = − 0 ∂ Φρ, φ|ρ=b ∂ρ

ρ

2 b cos φ 1−

ρ2 b2

σ = − 0 σ = −2 0 :

ρ

2 b cos φ

V 1 − V 2  ∂ tan −1 π ∂ρ

1−

ρ2 b2

ρ=b

b +ρ V1 − V2 V − V2 bcos φ 4 |ρ=b = − 0 1 π 2 2 4 2 2 2 πb cos φ b − 2b ρ + ρ + 4ρ b cos φ 2

2


a) As suggested in the text and in class, we will superpose solutions of the form (2.56) for the two sides with Vx, y, z = V. 1) First consider the side Vx, y, a = z : ∞

∑ A nm sinα n x sinβ m y sinhγ nm z

Φ 1 x, y, z =

n,m=1

with α n = sine functions,

nπ a

, βm =

mπ a

π a

, γ nm =

n 2 + m 2 . Projecting out A nm using the orthogonality of the

16V sinhγ nm anmπ 2 where both n, and m are odd. (Later we will use n = 2p + 1, m = 2q + 1 2) In order to express Φ 2 x, y, z in a form like the above, we make the coordinate transformation x ′ = y, y ′ = x, z ′ = −z + a So Φ 2 x, y, z = Φ 1 x ′ , y ′ , z ′  = Φ 1 y, x, = z + a A nm =

Φx, y, z = Φ 1 x, y, z + Φ 2 x, y, z b) Φ a , a , a  = 16 ⋅2 V 2 2 2 π

∞

∑ p,q=0

−1 p+q 2p + 12q + 1 cosh γ nm

where I have used the identity sinhγ nm a = 2 sinh ∞

Let fp, q ≡ ∑ p,q=0

−1 p+q 2p+12q+1 cosh

2p+1 2 +2q+1 2

γ nm a γ a  cosh nm  2 2 π 2

a 2

(p,q)

fp, q

Error

Sum

0,0

0.213484

4.4%

.214384

1,0

−0. 004641

2.13%

0.20974

0,1

−0. 004641 0.013% 0.20510

1,1 0.0002835 0.015% 0.20539 The first three terms give an accuracy of 3 significant figures. Φ a , a , a  = 16 ⋅ 0. 20539 V = 0. 332 96V 2 2 2 π2 Φ av  a , a , a  = 2 V = 0. 333. . . . V 2 2 2 6 c) σx, y, a = − 0 ∂ Φ|z=a ∂z σx, y, a = − 162 0 V π ∞

coshγ nm a − 1 = − 162 0 V ∑ sinα n x sinβ m y sinhγ nm a π n,m odd ∞

σx, y, a = − 162 0 V ∑ sinα n x sinβ m y tanh π n,m odd

γ nm a 2


1

∫ 0 Pl xdx The problem is symmetric around the z axis so φr, θ =

∑A l r l + B l r −l−1 Pl cos θ l

The A l and B l are determined by the conditions 1)

∫ −1 φa, xPl xdx =

2 A a l + B l a −l−1  2l + 1 l

∫ −1 φb, xPl xdx =

2 A b l + B l b −l−1  2l + 1 l

1

2) 1

Solving these two equations gives 1 1 2l + 1 a l+1 ∫ φa, xP l xdx − b l+1 ∫ φb, xP l xdx Al = 2l+1 2l+1 −1 −1 2a −b  B l = a l+1 2l + 1 2

∫ −1 φa, xPl xdx − A l a 2l+1 1

Using

∫ −1 φa, xPl xdx = V ∫ 0 Pl xdx 1

1

∫ −1 φb, xPl xdx = V ∫ −1 Pl xdx = V−1 l ∫ 0 Pl xdx 1

0

1

So Al =

1 2l + 1 l+1 l+1 l Va − b −1 P l xdx ∫ 0 2a 2l+1 − b 2l+1  1 B l = a l+1 2l + 1 V ∫ P l xdx − A l a 2l+1 2 0

Note that

∫ 0 Pl xdx =

1 ∫ 1 P l xdx 2 −1 1 for l even. For even l ∑ 0, ∫ P l xdx = 0. Thus we have 1

0

∫ 0 P0 xdx = 1; ∫ 0 P1 xdx = 1

1

1, 2

∫ 0 P3 xdx = − 18 1

and 3 7 A0 = V , A1 = Va 2 + b 2 , A 3 = − Va 4 + b 4  2 4a 3 − b 3  16a 7 − b 7  B 0 = 1 Va − 1 Va = 0 2 2 3 B1 = 3 a2V − Va 2 + b 2 a 3 = 3 Va 2 b 2 b3 + a 3 4 4 4a 3 − b 3  −a + b 7 B3 = − 7 a4V − a7 − Va 4 + b 4  16 16a 7 − b 7 

3 3 = − 7 Va 4 b 4 b 7 + a 7 16 −a + b 1

As b → ∞, only the B l terms (and A 0  survive. Thus using the general expression for ∫ P l xdx 0 given by (3.26) 3 4 φr, θ = V P 0 x + 3 a2 P 1 x − 7 a4 P 3 x +. . . . . 2 r 8 r 2 Let’s now solve the problem neglecting the outer sphere (since b → ∞ using the Green’s function result (2.19) this integral give, for cosθ = 1 φr, θ = V 1 − ρ 2  2

1 − 1 − ρ

1 1 + ρ 2 

with ρ = a/r. Expanding the above, 2 4 φr, θ = V ar + 3 a2 − 7 a4 +. . . . . . 2 r 8 r 2 Comparing with our previous solution with x = 1, we see the Green’s function solution differs by having a B 0 term and by not having an A 0 term. All the other higher power terms agree in the series. This difference is due to having a potential at ∞ in the original problem.

PHY 5346 HW Set 5 Solutions – Kimel 4. 3.4 Slice the sphere equally by n planes slicing through the z axis, subtending angle Δφ about this axis with the surface of each slice of the pie alternating as ±V. φr, θ, φ =

> A lm r l Y ml θ, φ l,m

so A lm = 1l a

∫ dΩY ml θ, φ ∗ φa, θ, φ

Symmetries: A l−m = −1 m A lm  ∗ φr, θ, φ + 2Δφ = φr, θ, φ where Δφ = 2π 2n Thus m = ±n, and integral multiples thereof φ−r⃗ = −φr⃗, n = 1 φ−r⃗ = φr⃗, n > 1 Since PY ml θ, φ = −1 l Y ml θ, φ Then l is odd for n = 1; l is even for n > 1 Thus we only have contributions of l ≥ n. Using A lm = 1l ∫ dΩY ml θ, φ ∗ φa, θ, φ a The integral over φ can be done trivially, since the integrand is just e −imφ leaving the desired answer in terms of an integral over cosθ. n = 1 case: I am going to keep only the lowest novanishing terms, involving A 11 and A 1−1 . 1 1 ∗ 1 φ = rA 11 Y 11 + A 1−1 Y −1 1  = rA 11 Y 1 + A 11 Y 1   = 2r ReA 11 Y 1 

Y 11 = − A 11 = − 1a

3 V 8π

3 1 − x 2  1/2 e iφ 8π π

∫ −1 1 − x 2  1/2 dx ∫ 0 e −iφ dφ − ∫ π 1

2π

e −iφ dφ

A 11 = 2iπ a φ = 2r Re

2iπ a

3 V 8π

−

3 V 8π

3 sin θe iφ 8π

= 3r Vsin θ sin φ 2a

From the figure

we see sin θ sin φ = cos θ ′ So φ = 3r Vcos θ ′ = V 3 ar P 1 cos θ ′  +. . . . . . 2a 2 The other terms, for l = 2, 3, can be obtained in the same way in agreement with the result of (3.36)

PHY 5346 HW Set 6 Solutions – Kimel 2. 3.10 This problem is described by

a) From the class notes Φρ, z, φ =

∑A nν sin νφ + B nν cos νφI ν nν

nπρ L

sin nπz L

where 1 A nν = 2 πL I ν nπb L

∫0 ∫0 L

2π

Vφ, z sin nπa  sinνφdφdz, L

∫0 ∫0 L

1 B nν = 2 πL I ν nπb L

2π

Vφ, z sin nπa  cosνφdφdz, L

∫0 ∫0 L

1 B nν = 1 πL I ν nπb L

2π

Vφ, z sin nπa dφdz, L

ν∑0 νΦ0 ν=0

Noting π 2

∫−

3π 2

π 2

sin νφdφ − ∫ π sin νφdφ

=0

2

we conclude A nν = 0. Similarly, noting π 2

∫−

3π 2

π 2

cos νφdφ − ∫ π cos νφdφ 2

=

4−1 m , 2m + 1

m = 0, 1, 2, . . . .

where I’ve recognized that ν must be odd, ie, ν = 2m + 1. Also L 2 dz = , l = 0, 1, 2, . . . . . ∫ 0 sin nπz L 2l + 1π where again I’ve recognized that n must be odd, ie, n = 2l + 1. Thus 16−1 m V B nν = 2 nπb π I 2m+1  L 2l + 12m + 1 b) Now z = L/2,

L ∑∑ b, L ∑∑ ρ. Then from the class notes

I 2m+1 

2l + 1πρ 2L

2l + 1πρ 1  L Γ2m + 2

m+1

Also 2l + 1π L

sin

= −1 l

so Φρ, z, φ =

ρ b

l+m

V ∑ π 2 2l16−1 + 12m + 1 l,m

2m+1

cos2m + 1φ

Using ∞

tan

−1x

=

∑ 1lx + 1 −1 l 2l+1

l=0

π = tan −1 1 = 4

∞

∑ 2l−1 +1 l

l=0

so Φρ, z, φ = 4V π

−1 ∑ 2m +1 m

m

ρ b

2m+1

cos2m + 1φ

Remembering from problem 2.13 that

∑ m

−1 m 2m + 1

ρ b

2m+1

cos2m + 1φ = 1 tan −1 2

we find −1 Φρ, z, φ = 2V π tan

which is the answer for problem 2.13.

2

ρ b

1−

cos φ ρ2 b2

2

ρ b

1−

cos φ ρ2 b2

PHY 5346 HW Set 6 Solutions – Kimel 3. 4.1

q lm =

⃗d 3 x = > q i r ll Y m∗ ∫ r l Y m∗ l θ, φρx l θ i , φ i  i

Using Y m∗ l θ, φ =

2l + 1l − m! m P l xe −−mφ = N ml P ml xe −−mφ 4πl + m!

From the figure we get q lm = a l N ml P ml 0q1 − −1 m 1 − i m  = 0, for m even, so m = 2n + 1, n = 0, 1, 2, . . . q lm = 2qa l N ml P ml 01 − −1 n i b) The figure for this system is

Since the sum of the charges equals zero, l ≥ 1. m m m∗ l m q lm = qa l Y m∗ l x = 1, φ + Y l x = −1, φ = qa N l P l 1 + P l −1 From the Rodrigues formula for P ml x, we see P ml ±1 = 0, for m ≠ 0. So q lm = qa l N 0l 1 + −1 l P l 1 Thus l is even, but l ≠ 0

q lm = 2qa l N 0l c) Using the fact that N 0l =

2l+1 4π

and Y 0l =

β

Φx⃗ =

x >2qa l  Prll+1

≈

l=2

2l+1 4π

Pl

2qa 2 P 2 x = 0 on x-y plane) r3 2

qa Φx⃗ = − 3 r

Let us plot Φx⃗/−q/a, ie,

1  ar  3

=

1 x3

8 7 6 5 4 3 2 1 0 0.5

1

1.5

2

x

2.5

3

The exact answer on the x-y plane is −q Φx⃗ = a So let’s plot

2 − x 1 x3

,

2 x

2 x 1+ −

1 x2

−q = a

3

1 x

− 3 4

1 x

5

+ 5 8

2 x 1+

1 x2

0.3 0.25 0.2 0.15 0.1 0.05 0 1

2

where the smaller is the exact answer.

3 x

4

5

1 x

7

− 35 64

1 x

9

+. . . .

PHY 5346 HW Set 7 Solutions – Kimel 2. 4.2 We want to show that we can obtain the potential and potential energy of an elementary diplole: ⃗ p⋅⃗ x Φx⃗ = 1 4π 0 r 3 ⃗ 0 ⃗⋅E W = −p from the general formulas Φx⃗ = W=

1 4π 0

⃗ d x ∫ ρx x′ | |x⃗ − ⃗ ′

3 ′

∫ ρx⃗Φx⃗d 3 x

using the effective charge density ⃗⋅⃗ ρ eff = −p ∇δx⃗ where I’ve chosen the origin to be at ⃗ x0. ′ ⃗ ⃗ ⋅ ∇ δx⃗′ d 3 x ′ −p 1 Φx⃗ = 1 ∫ =− 1 ⃗ p ⋅ ∫⃗ ∇ 4π 0 4π 0 x′ | x′ | |x⃗ − ⃗ |x⃗ − ⃗ Φx⃗ =

δx⃗′ d 3 x ′

p⋅⃗ x 1 ⃗ 3 4π 0 r

Similarly, W=

∫ ρx⃗Φx⃗d 3 x = − ∫ ⃗p ⋅ ⃗∇δx⃗Φx⃗d 3 x = ⃗p ⋅ ∫ δx⃗∇⃗ Φx⃗d 3 x = −p⃗ ⋅ E⃗0

PHY 5346 HW Set 7 Solutions – Kimel 3. 4.7 a) Since ρ does not depend on φ, we can write it in terms of spherical harmonics with m = 0. First note Y 02 =

5 4π

3 1 − sin 2 θ − 1 2 2

or sin 2 θ = − 2 4π Y 02 + 4π 2 Y 00 3 5 3 Thus only the m = 0, l = 0, 2 multipoles contribute. 2 4π β 2 1 2 −r 3 = 1 q 00 = ∫ 0 r 64π r e dr = 2 34π 8π 3 2 π q 20 = − 2 3 Φx⃗ =

1 4π 0

4π 5

β

∫0 r4

1 r 2 e −r dr = − 2 3 64π

Y0 Y0 4πq 00 r0 + 4πq 20 23 5r Φx⃗ =

1 4π 0

=

4π 45 = −3 5 5 4π π P 4π q 00 r0 +

1 4π 0

4π q P 2 5 20 r 3

P0 − 6 P2 r r3

b) Φx⃗ =

1 4π 0

⃗ d x ∫ ρx x′ | |x⃗ − ⃗ ′

3 ′

Using 1 1 = 4π > + 1r > 2l x′ | |x⃗ − ⃗ lm

r< r>

l

m ′ ′ Y m∗ l θ , φ Y l θ, φ

, we see only the l = 0, 2 and m = 0 terms of the expansion contribute in the potential. Next take r ′ > r. ⃗′  ′ 1 ′ ′ ′2 ′ ρx Φx⃗ = 1 4π > r l Y ml θ, φ ∫ Y m∗ dr l θ , φ r dΩ ′l+1 4π 0 2l + 1 r lm Φx⃗ =

1 4π Y 0 4π 2 0 4π 0 3 Φx⃗ = Φx⃗ =

β

∫0

0 1 r 2 e −r rdr + Y 2 r 2 − 2 5 64π 3

0 1 4π Y 0 4π 2 3 + Y 2 r 2 − 2 0 5 4π 0 3 32π 3

1 4π P 2 3 + P 2 r 2 − 2 0 5 4π 0 3 3 32π

1 64π

=

4π 5

β

∫0

4π 5 1 4π 0

1 r 2 e −r 1 dr r 64π 1 64π P0 − r 2 P2 120 4

PHY 5346 HW Set 7 Solutions – Kimel 4. 4.9 a) The system is described by

Since there is azimuthal symmetry, choosing the z-axis through q, Φ out = Φ out = Φ in =

∑ B l r −l−1 Pl +

1 4π 0

l

∑ B l r −l−1 Pl +

1 4π 0

q r>

∑

l

1 4π 0

q x′ | |x⃗ − ⃗ r< r>

l

Pl

l

∑ A l r l Pl +

q r>

l

∑

r< r>

l

Pl

l

Boundary conditions: At the surface, r ′ = d = r > , r = a = r < . 1) Φ out = Φ in |r=a , or 2)  ∂r∂ Φ in k

B l = A l a 2l+1 = ∂r∂ Φ out |r=a , or letting k = 0

∑ lA l a l−1 Pl + l

q a l−1 l d dl =

Pl

=

∑ −l + 1B l a −l−2 Pl + l

∑ −l + 1A l a l−1 Pl + l

q a l−1 l d dl

q a l−1 l d dl

Pl

Pl

or Al =

a1 − kl 1 + kl + 1 d l+1

Bl =

a1 − kla 2l+1 1 + kl + 1 d l+1

4π Remember that P l = 2+1 Y 0l , and substitute the above coefficients into the expansion to get the answer requested by the problem.


a) Since there is azimutal symmetry, Φr, θ =

∑A l r l + B l r −l−1 Pl cos θ l

Also ⃗ Φr, θ ⃗ = −x⃗∇ ⃗ = x⃗E D D r = −x⃗ ∑lA l r l−1 − l + 1B l r −l−2 P l cos θ l

between the spheres,

∫ D r dΩr 2 = Q,

and is independent of r.

Thus A l = 0, B l = 0, l Φ 0 → D r =

∫ D r dΩr 2 = 2πB 0

0 ∫

0 −1

d cos θ +  ∫ d cos θ 1

0

B0 = ⃗ = E

Q 2π 0 1 +

Q 2π 0 1 +

 0

 0

x⃗B 0 r2 = 2πB 0  0 +  = Q



r 2

r̂

b)

∫ D r dA = D r A = σ f A → σ f = D r = x⃗E r σf =

Q 2π 0 1 +

 0

r 2

,

cos θ ≥ 0

σf =

Q , 2π1 + 0 r 2

cos θ < 0

c)

∫ ρ pol dV = σ pol A = ∫ −∇⃗ ⋅ P⃗dV = −PA → σ pol = −P = − 0 χ e E σ pol = −x⃗/ 0 − 1

Q 2π1 + 0 r 2

Notice σ pol + σ f = σ tot =

Q =  0 E, as expected. 2π1 + 0 r 2

PHY 5346 Homework Set 10 Solutions – Kimel 1. 5.1 The system is described by

We want to show φm = −

μ0I Ω 4π

Suppose the observation point is moved by a displacement δx⃗, or equivlently that the loop is displaced by −δx⃗. ⃗ φ m , then ⃗ = −∇ If we are to have B ⃗ δφ m = −δx⃗ ⋅ B Using the law of Biot and Savart,

δφ m = −

μ0I 4π

∮ δx⃗ ⋅

dl⃗′ × ⃗r r3 δφ m = −

δx⃗ × dl⃗′

=−

μ0I 4π

μ0I 4π

∮ r̂ ⋅ δdA r2

∮ ⃗r ⋅

Or, φm = −

μ0I Ω 4π

r3 =−

=−

μ0I δΩ 4π

μ0I 4π

∮ r̂ ⋅

δx⃗ × dl⃗′ r2

PHY 5346 Homework Set 10 Solutions – Kimel 2. 5.2 a) The system is described by

First consider a point at the axis of the solenoid at point z 0 . Using the results of problem 5.1, μ0 NIdzΩ 4π

dφ m = From the figure, Ω=

⃗

∫ r̂ ⋅r 2dA

φm =

=

θ ∫ dArcos 2

= 2πz ∫

ρdρ z = 2π − +1 2 3/2 2 ρ + z  R + z 2 

R

2

0

β μ0 1 NI ∫ z − + 1z 2 2 2 z0 R + z 

Br = −

μ0 NI −z 0 + R 2 + z 20  2

dz =

μ0 NI ∂ −z 0 + R 2 + z 20  2 ∂z 0

=

−z 0 + R 2 + z 20  μ0 NI 2 R 2 + z 20 

In the limit z 0 → 0 Br =

μ0 NI 2

By symmetry, thej loops to the left of z 0 give the same contribution, so B = B l + B r = μ 0 NI H = NI ⃗ is directed along the z axis, so By symmetry, B ⃗ =0 ⃗⋅B δφ m = −δρ ⃗ is directed ⊥ to the z axis. Thus for a given z, φ m is independent of ρ, and consequently if δρ H = NI everywhere within the solenoid.

If you are on the outside of the solenoid at position z 0 , by symmetry the magnetic field must be in the z direction. Thus using the above argument, φ m must not depend on ρ. Set us take ρ far away ⃗ directed along from the axis of the solenoid, so that we can replace the loops by elementary dipoles m the z axis. Thus for any point z 0 we will have a contributions φm 

⃗ ⋅ ⃗r 2 ⃗ ⋅ ⃗r 1 m m + 3 r 32 r1

⃗ ⋅ ⃗r 1 = −m ⃗ ⋅ ⃗r 2 and r 1 = r 2 . Thus where m H=0

PHY 5346 Homework Set 10 Solutions – Kimel 3. 5.8 Using the same arguments that lead to Eq. (5.35), we can write Aφ =

μ0 4π

∫

′

d 3 x cos φ ′ J φ r ′ , θ ′  x′ | |x⃗ − ⃗

Choose ⃗ x in the x − z plane. Then we use the expansion 1 = x′ | |x⃗ − ⃗

r l< m∗ ′ ′ m Y l θ , φ Y l θ, 0 r l+1 >

> 2l4π+ 1 l,m

The cos φ ′ factor leads to only an m = 1 contribution in the expansion. Using 2l + 1 l − m! P m cos θ 4π l + m! l

Y ml θ, 0 = and

l−1! l+1!

=

1 ll+1

, we have on the inside

Aφ =

μ0 4π

′

′

′

J φ r , θ  > ll +1 1 r l P1l cos θ ∫ d 3 x ′ Pl cos θr ′l+1 1

l

which can be written Aφ = −

μ0 4π

> m l r l P1l cos θ l

with ml = −

1 ll + 1

′

′

′

J φ r , θ  ∫ d 3 x ′ Pl cos θr ′l+1 1

A similar expression can be written on the outside by redefining r < and r > .

PHY 5346 Homework Set 9 Solutions – Kimel 1. 5.10 a) From Eq. (5.35) A φ r, θ =

μ0 I 4π a

∫r

′2

dr ′ dΩ ′ sin θ ′ cos ϕ ′ δcos θ ′ δr ′ − a x′ | |x⃗ − ⃗

Using the expansion of 1/|x⃗ − ⃗ x ′ | given by Eq. (3.149), 4 1 = π x′ | |x⃗ − ⃗

β

β

∫ 0 dk coskz − z 

1 I 0 kρ < K 0 kρ >  + > cosmϕ − ϕ ′  I m kρ < K m kρ >  2

′

m=1

We orient the coordinate system so ϕ = 0, and because of the cosϕ ′ factor, m = 1. Thus, A φ r, θ =

μ 0 I 4π 4π a π

β

∫ 0 dk ∫ r ′2 dr ′ d cos θ ′ sin θ ′ δcos θ ′ δr ′ − a coskzI 1 kρ < K 1 kρ >  β μ A φ r, θ = π0 aI ∫ dk coskzI 1 kρ < K 1 kρ >  0

where ρ < ρ >  is the smaller (larger) of a and ρ. b) From problem 3.16 b), 1 = x′ | |x⃗ − ⃗

β

β

> ∫ 0 dke imφ−φ  J m kρJ m kρ ′ e −k|z| ′

m=−β

Note z ′ = 0, and φ = 0, so Aφ =

μ 0 Ia 2

β

∫ 0 dke −k|z|J 1 kρJ 1 ka

PHY 5346 Homework Set 10 Solutions – Kimel 4. 5.16 a) The system is shown in the figure

I shall use the magnetic potential approach and will call inside the sphere region 1 and outside the sphere region 2. φ 1 = φ loop + > A l r l P l l

φ 2 = φ loop + > B l r −l−1 P l l

⃗ φ, and we have the boundary conditions, ⃗ = −∇ where H H 1|| = H 2|| → φ 1 r = b = φ 2 r = b μ 0 ∂ φ 1 r = b = μ ∂ φ 2 r = b ∂r ∂r We are given that b >> a, so θ φ loop = 1 m cos 4π r 2 with m = πa 2 I. (From the form of φ loop, only the l = 1 term contributes.) The boundary conditions give A 1 b 1 = B 1 b −1−1 −

2μ 0 m 2μm + μ0A1 = − − 2μB 1 b −3 3 4πb 4πb 3

So μ − μ 0  A 1 = − 2 m3 4π b 2μ + μ 0 

On the inside, at the center of the loop ⃗ φ loop − ⃗ ⃗ = −∇ H ∇A 1 r cos θ ⃗ φ loop at the center of the loop, which is directed in the z direction. From Eq. (5.40), we are given −∇ H z = μ10 −B θ  − A 1 If μ >> μ 0 A 1 → − 1 m3 4π b and from (5.40), at r = 0 2 3 H z = I + 1 m3 = I + I a 3 = I 1 + a 3 2a 4π b 2a 4 b 2a 2b

PHY 5346 Homework Set 9 Solutions – Kimel 2. 5.18 a) From the results of Problem 5.17, we can replace the problem stated by the system

where I ∗ is equidistant from the interface and is equal to I ∗ = loop is a. Now from Eq. (5.7)

μ r −1 μ r +1

I. The radius of each current

⃗on I = I ∫ dl⃗ × Br⃗ F ⃗ θ = dlB r −θ̂ + dlB θ r̂ ⃗ = dl⃗ × B ⃗ r + dl⃗ × B dl⃗ × B By symmetry, only the z − component survives, so, from the figure ⃗ ⋅ ẑ = dlB r dl⃗ × B

a 4d + a 2 2

2d 4d 2 + a 2

+ dlB θ

So Fz =

2πaI aB + 2dB r θ 4d 2 + a 2

with B r and B θ given by Eqs. (5.48) and (5.49) and cos θ =

2d 4d 2 +a 2

,r=

4d 2 + a 2 , and I → I ∗ .

c) To determine the limiting term, simply let r → 2d and take the lowest non-vanishing term in the expansion of the magnetic flux density. F z = πaI aB r + 2dB θ d μ0I∗ a a F z = πaI a d 4d 2d 2 Fz → −

+ 2d −

μ0I∗ a2 4

1 2d 3

− a 2d

3πμ 0 a 4 I × I ∗ 32 d4

The minus sign shows the force is attractive if I and I ∗ are in the same direction. This same result can be gotten more directly, using

F z = ∇ z mB z  with m = πa 2 I, and (from Eq. (5.64)) Bz =

μ0 4π

2m ∗ z3

with m ∗ = πa 2 I ∗ , and z = 2d Fz =

μ0 2πa 2 I ∗ πa 2 I − 3 4 4π 2d

with agrees with out previous result.

=−

3πμ 0 a 4 I × I ∗ 32 d4


⃗ is constant within the cylinder. The The effective volume magnetic charge density is zero, since M ⃗ from Eq. (5.99)) is M 0 , on the top surface and −M 0 on the effective surface charge density (n̂ ⋅ M bottom surface. From the bottom surface the potential is (for z > 0 a ρdρ Φ b = 1 −M 0 2π ∫ = − M0 1/2 2 2 4π 2 0 ρ + z 

a 2 + z 2  − z

By symmetry, the potential from the top surface is (on the inside) Φt = M0 2

a 2 + L − z 2

− L − z

The total magnetic potential is Φ = Φb + Φt = −

M0 2

M a 2 + z 2  − z + 0 2

a 2 + L − z 2

− L − z

So, on the inside of the cylinder, M Hz = − ∂ − 0 2 ∂z

Hz = − while above the cylinder,

M0 2

a 2 + L − z 2

z − a + z 2 

L−z a + L − z 2

a 2 + z 2  − z +

M0 2

2−

2

2

− L − z

Hz = − M0 2

−

z−L a + L − z 2

z + a + z 2  2

2

with a similar expression below the cylinder. ⃗ = μ0 H ⃗ +M ⃗ B Thus inside the cylinder, Bz = μ0 −

z − a + z 2 

L−z a + L − z 2

μ0M0 2

z + a + z 2 

L−z a + L − z 2

μ0M0 2

z − a + z 2 

z−L a + L − z 2

M0 2

Bz =

2−

2

2

2

2

while above the cylinder, Bz =

2

2

First we plot B z in units of a for L = 5a 1 2

1+z 2

1 2

1+z 2

gz =

+

z

5−z 1+5−z 2

−

z

z−5 1+5−z 2

if z < 5 if 5 < z

gz 1 0.8 0.6 0.4 0.2

0

2

And similarly, H z in units of a for L = 5a.

4

z

6

8

10

+ M0

− 12

2−

fz = − 12

−

−

z 1+z 2

5−z 1+5−z 2

+

z

if 5 < z

z−5

1+z 2

if z < 5

1+5−z 2

fz

0.4 0.2

0 -0.2 -0.4

2

4

z

6

8

10


Since the wires are nonpermeable, μ = μ 0 . The system is made of parts with cylindrical symmetry, so we can determine B using Ampere’s law.

∫ B⃗ ⋅ dl⃗ = μ 0 ∫ ⃗J ⋅ da⃗

⃗ ⃗ = μ 0⃗ ∇×B J, or On the outside of each wire,

∫ B⃗ ⋅ dl⃗ = B2πρ = μ 0 I → B out =

μ0I 2πρ

On the inside of each wire

∫ B⃗ ⋅ dl⃗ = B2πρ = μ 0 I ρR 2 , 2

B in =

μ0I ρ with R = a, b 2π R 2

From the right-hand rule, the B from each wire is in the φ̂ direction. From the above figure, using ⃗ is in the ±ẑ direction. Since ⃗ ⃗ =B ⃗, the general expression for the vector potential, we see A ∇×A B z = − ∂ A z → A z = − ∫ B z dρ ∂ρ Thus Az =

−

μ0I 2π

ln

ρ R

+C

− μ4π0 I

=− ρ2 R2

,

μ0I 4π

ln

ρ2 R2

+ 1 on the outside

on the inside

where I’ve determined C = 1/2, from the requirement that A z be continuous at ρ = R. Let l be the

length of the wire. Then we know the total potential energy is given by W= 1 2 Consider the second term

l 2

∫ ⃗J ⋅ A⃗d 3 x =

l ∫J Ada + J Ada a a b b 2

∫ J b Ada b . The system is pictured as

From the figure ⃗ b, ⃗a = ⃗ d+ρ ρ so, since J b =

ρ 2a = d 2 + ρ 2b − 2dρ b cos φ

I πb 2

l 2

∫ J b Ada b =

∫A out ρ a  + A in ρ b ρ b dρ b dφ

l I 2 πb 2

μ I = l I2 0 2 πb 4π

∫

ln

ρ 2b ρ 2a + 1 − b2 a2

ρ b dρ b dφ

b 2 ρ2 μ I ≃ l I 2 0 2π ∫ ln d 2 + 1 − 2b 2 πb 4π b a 0 2 μ I = l I 2 0 2π 1 b 2 1 + 2 ln d 2 2 πb 4π 4 a

The first term

l 2

= l 2

μ0 4π

ρ b dρ b 1 + 2 ln d I 2 a 2

∫ J a Ada a is equal to l 2

∫ J a Ada a =

l 2

μ0 4π

1 + 2 ln d I 2 b 2

Thus W= l 2

μ0 4π

2 1 + 2 ln d I 2 = l L I 2 ab 2 l

or L = μ 0 1 + 2 ln d 2 l ab 4π


Using Ampere’s law in integral form

∫ B⃗ ⋅ dl⃗ = μ 0 I enclosed we get μ0I ρ ,ρ
B= B=

μ0I 1 ,b<ρ a

Now the energy in the magnetic field is given by ( l is the length of the wires) W= 1 2 =

μ0I 2π

1 2μ 0 =

1 2μ 0

2

∫ B⃗ ⋅ H⃗ d 3 x =

l 2π ∫ μ0I 2π

b 0

2

ρ b2

2

1 2μ 0

∫ B2d3x

ρdρ + 2π ∫

lπ 1 + 2 ln a 2 b

μ → L = 0 l 4π

1 + 2 ln a b 2

If the inner wire is hollow, B = 0, ρ < b, so L = μ 0 ln a l 2π b

a b

1 ρ

= l L I2 2 l

2

ρdρ


This problem is very much like 5.26, except the wires are superconducting. We know from section 5.13 that the magnetic field within a superconductor is zero. We will be using W= 1 2

∫ ⃗J ⋅ A⃗d 3 x =

l ∫J Ada + J Ada a a b b 2

Using the same arguments as applied in problem 5.26, μI

− 2π ln

Az =

ρ R

μI

+C

= − 4π ln

0, Thus if we consider the second term l 2

l 2

ρ2 R2

on the inside

∫ J b Ada b ,

∫ J b Ada b =

∫A out ρ a  + A in ρ b ρ b dρ b dφ

l I 2 πb 2

b 2 μI ≃ l I2 2π ∫ ln d 2 ρ b dρ b = l 2 πb 4π 2 0 a

The first term

l 2

+ 0 on the outside

μ 4π

2 ln da I 2

∫ J a Ada a is equal to l 2

∫ J a Ada a =

l 2

μ 4π

2 ln d I 2 b

Thus W= l 2

μ 4π

2 2 ln d I 2 = l L I 2 ab 2 l

so L = l

μ 4π

2 2 ln d ab

Now using the methods of problem 1.6, assuming the left wire has charge Q, and the right wire charge −Q, we find φ 12 =

∫b

d−a

Edr =

Q l

2π

∫b

d−a

Q

2 1 + 1 dr ≃ l ln d r d−r 2π ab

Q

C = l = 2π 2 φ 12 l ln dab Thus L × C = l l

μ 4π

2 2 ln d ab

× 2π = μ 2 ln dab

PHY 5346 Homework Set 12 Solutions – Kimel 1. 6.11 a) Consider the momentum contained in the volume

Δp = ΔtcΔAg F=

Δp = cΔAg Δt

P = F = cg ΔA where I’m using the time averaged quantities. In class we found cg = 1c S = 1  0 |E 0 |2 = u 2 Thus P=u b) We are given S = 1. 4 × 10 3 W/m 2 But we know P = u =

S c

. From Newton’s second law

1. 4 × 10 3 W/m 2 F = F/A = S/c = a= m = 4. 66 × 10 −3 m/s 2 2 8 −3 m/A m/A 3 × 10 m/s × 1 × 10 kg/m In the solar wind, there are approximately 10 × 10 4 protons/m 2 − sec, with average velocity v = 4 × 10 5 m/s. Δp = P = 10 × 10 4 × 4 × 10 5 × 1. 67 × 10 −27 = 6. 68 × 10 −17 N/m 2 ΔtA : 2 −17 F = F/A = P = 6. 68 × 10 N/m = 6. 68 × 10 −14 m/s 2 a= m m/A m/A 1 × 10 −3 kg/m 2

PHY 5346 Homework Set 12 Solutions – Kimel 2. 7.1 I shall apply Eqs.(26), (27), and (28) s0 + s1 = a 1 2

s0 − s1 = a2 2

δ l = δ 2 − δ 1 = sin −1

s3 2a 1 a 2

s0 + s3 = a + 2

s0 − s3 = a− 2

δ c = δ − − δ + = sin −1

s2 2a + a −

a) s 0 = 3, s 1 = −1, s 2 = 2, s 3 = −2 a 1 = 1, a 2 = −2 2 2

δ l = sin −1

a+ =

1 , 2

2 = − 1 π rad 4 5 2

a− =

2

δ c = sin −1 2

= 1. 107 1 rad

1 2

5 2

b) s 0 = 25, s 1 = 0, s 2 = 24, s 3 = 7 a1 =

δ l = sin −1

25 , a = 2 2

s3 2a 1 a 2

= sin −1

a+ =

32 = 4, 2

7 2

δ c = δ − − δ + = sin −1

25 2

25 2

a− =

25 2

= 0. 283 79 rad

s0 − s3 = 3 2

24 24 × 3

= 1 π rad 2

To plot the two cases Re E x ≡ X = cos x, ReE y ≡ Y = rcoxx − δ l , where r = a 2 /a 1 and x = ωt. Case a) cos x, 2 cosx + π4 

1 0.5

-1

-0.8 -0.6 -0.4 -0.2

0.2

0.4

0.6

0.8

1

0.2

0.4

0.6

0.8

1

-0.5 -1

Case b) cos x, cosx − 0. 283 79 1 0.8 0.6 0.4 0.2 -1

0 -0.8 -0.6 -0.4 -0.2 -0.2 -0.4 -0.6 -0.8 -1

PHY 5346 Homework Set 13 Solutions – Kimel 1. 7.2 a) The figure describes the muliple internal reflections which interfere to give the overall reflection and refraction:

For the ij interface I shall use the notation r ij =

E ′0 i = n 2n i + nj E0

R ij =

n −n E ′′0 = ni + nj i j E0

Thus from the figure E ′′0 = E 0 R 12 + r 12 E 0 R 23 r 21 e iφ + r 12 E 0 R 23 R 21 R 23 r 21 e i2φ +. . . . β

E ′′0 = E 0 R 12 + r 12 E 0 R 23 r 21 e iφ >R 21 R 23 e iφ  n=0

E ′′0 = E 0 R 12 +

r 12 r 21 R 23 −iφ e − R 21 R 23 

Similarly E ′0 = E 0 r 12 r 23 + E 0 r 12 R 23 R 21 r 23 e iφ +. . . .

n

E ′0 = E 0

r 12 r 23 1 − R 21 R 23 e iφ

where the phase shift for the internally reflected wave is given by φ=

ωn 2 2d 2π2d = c λ2

Now for a plane wave S i = 1 |E 0i |2 2v i Thus ′′ |E ′′ | R= S = 02 S |E 0 |

2

′ ′ T = vv 13 S = nn 3 S 1 S S

From the above R=

R 212 +

2r 12 r 21 R 23 R 12 cos φ − R 21 R 23  + R 12 r 21 R 23  2 1 + R 21 R 23  2 − 2R 21 R 23 cos φ

T = nn 3 1

r 12 r 23  2 1 + R 21 R 23  2 − 2R 21 R 23 cos φ

Since these two equations are simple functions of φ, which is linearly proportional to the frequency, they are simple functions of frequency which you should plot. b) Since in part a) we used the convention that the incident wave is from the left, I will rephrase this question so that n 1 is are, n 2 is the coating, and n 3 is glass. In this case, we will have n 1 < n 2 < n 3 , and R 21 R 23 < 0. Thus for T to be a maximum, from the above equation cosφ = −1, or φ = π. φ=

2π2d λ =π→d= 2 λ2 4

where λ 2 is the wavelength in the medium = λn 21

PHY 5346 Homework Set 13 Solutions – Kimel 2. 7.4 We have a nonpermeable conducting material, so μ = μ 0 , and we have J = σE, where σ is the conductivity. The following figure describes the system:

The two boundary conditions that we must satisfy for plane waves are E 0 + E ′′0 − E ′0 = 0 kE 0 − E ′′0  − k ′ E ′0 = 0 Or ′ E ′′0 = k − k′ E0 k+k

⃗ . Adding in this term in Maxwell’s equations for a We must take into account the fact that ⃗ J = σE plane wave, we get k= ω c σ k ′2 = μω 2 1 + i ω Thus we can write k′ =

μ ωα + iβ

with α=

σ 2 1 +  ω  +1 2

1/2

1/2

σ 2 1 +  ω  −1 2

β= Thus

1 − μ 0 cα − i μ 0 cβ E ′′0 = E0 1 + μ 0 cα + i μ 0 cβ 1) For a very poor conductor σ is very small, so keeping only first order in σ α=

β=

σ ω

σ = 2ω

≈1

1/2

σ 2 1 +  ω  −1 2

2) For the case of a very good conductor, α≈

1/2

σ 2 1 +  ω  +1 2

≈

σ 2ω

>> 1, so 2 μ 0 ωδ 2

2ω

=

1 ωδ μ 0 

σ = 1 2ω ωδ μ 0 

β≈

where I have used (5.165) to relate the conductivity to the skin depth. σ= 1− E ′′0 = E0 1+

c ωδ c ωδ

R=

c − i ωδ δ− = c + i ωδ δ+

E ′′0 E0

c ω c ω

2 μ 0 ωδ 2

− i ωc 2 ω ≈ −1 + ω c 1 + i δ = −1 + c 1 − iδ + i ωc

2

= −1 + δω/c 2 +

ωδ c

2

≈ 1 − 2δω/c

PHY 5347 Homework Set 1 Solutions – Kimel 1. 8.3

x

z y a) ∇ 2t + γ 2 ψ = 0,

ψ = E z TM or ψ = H z TE

As in class, we will use cylindrical coordinates, and assume ψρ, φ = RρQφ We get the two equations ∂ 2 Qφ = −m 2 Qφ with solns Qφ = e ±imφ , m = 0, 1, 2, . . . ∂φ 2 d 2 Rx + 1 dRx + 1 − m 2 x dx x2 dx 2

Rx

Bessel eqn.

with regular solutions J m x, and singular solution (which we reject as nonphysical) N m x. Here x = γρ. Solutions: TM: BC: J m x mn  = 0, and E z ρ, φ = E 0 J m γ mn ρe ±imφ ,

m = 0, 1, 2, . . . . ; n = 1, 2, 3, . . . ; γ mn = x mn /R

Lowest cutoff frequencies: ω mn = Using the results of Jackson, p. 114,

γ mn = x mn R μ μ

x 0n = 2. 405, 5. 52, 8. 654, . . . x 1n = 3. 832, 7. 016, 10. 173, . . . . x 2n = 5. 136, 8. 417, 11. 620, . . . . TE: BC: J ′m x ′mn  = 0, and E z ρ, φ = E 0 J m γ ′mn ρe ±imφ ,

m = 0, 1, 2, . . . . ; n = 1, 2, 3, . . . ; γ ′mn = x ′mn /R

Lowest cutoff frequencies: γ ′mn x ′mn = R μ μ

ω mn = Using the results of Jackson, p. 370,

x ′0n = 3. 832, 7. 016, 10. 173, . . . x ′1n = 1. 841, 5. 331, 8. 536, . . . . x ′2n = 3. 054, 6. 706, 9. 970, . . . . From the above we see the lowest cutoff frequency is the TE mode ω ′11 = 1. 841K, with K = 1/ R μ The next four lowest cutoff frequencies are: ω 01 = 2. 405K = 1. 31ω ′11 ′ = 3. 054K = 1. 66ω ′11 ω 21

ω ′01 = 3. 832K = 2. 08ω ′11 ω 11 = 3. 832K = 2. 08ω ′11 b) From Eq. (8.63) in the text 1/2

βλ 

ξ λ + η λ ωωλ

ω ω2 1 − ω λ2

2

For TM modes, η λ = 0, and for TE mode, ξ λ + η λ is of order unity. So for comparison purposes, I’ll take 1/2

β 11 TE = f 1 x =

2 1 + 1. 841 2 x

x 1−

1.841 2 x2

1/2

β 01 TM = f 2 x =

x 1−

where I’ve expressed the functions in terms of x = ω/K.

2.405 2 x2

1/2

1+

1.841 2 x2

x 1− 1.841 2

2

x

14 12 10 8 6 4 2 0 2

4

x6

8

10

1/2 x 1− 2.405 2

2

x

14 12 10 8 6 4 2 0

3

4

5

6x

7

8

9

10


a) TM: ∇ 2t + γ 2 ψ = 0;

ψ|B = 0; E z = ψx, ye ±ikz−iωt ; B z = 0

Since we have a node along y = x, then we just take the antisymmetrized version for the square waveguide, developed in class, ie, nπy mπy nπx ψx, y = E 0 sin mπx a  sin a  − sin a  sin a  Again 2 γ 2mn = cπ2 m 2 + n 2 , m, n = 1, 2, 3. . . . , but m ≠ n a

TE: ∇ 2t + γ 2 ψ = 0; Now the BC require

∂ψ | ∂n B

∂ψ | = 0; H z = ψx, ye ±ikz−iωt ; E z = 0 ∂n B

= 0, but using a 45 0 rotation of coordinates, we see ∂ = ∂n

1 2

− ∂ + ∂ ∂x ∂y

Thus the combination nπy mπy nπx ψx, y = H 0 cos mπx a  cos a  + cos a  cos a  satisfies the above BC on the diagonal, as you can see by direct substitution. 2 γ 2mn = cπ2 m 2 + n 2 , m, n = 0, 1, 2, 3. . . . , but m ≠ n = 0 a

b) The lowest cutoff freq. are: TM: ω 12 or ω 21 .

TE: ω 01 or ω 10 . From Eq. (8.63) in the text

β 12 TM 

β 01 TE 

ω 1 − ω 212 /ω 2

ω 1 − ω 212 /ω 2

1/2

1/2

1+

ω 201 ω2

For the square wave guide, we don’t have the antisymmetrization, but the formulas for the cutoff frequencies are the same without the present restrictions on m and n. So for the square guide, the cut off frequencies are TM: ω 11 TE: ω 01 (as before)

PHY 5347 Homework Set 2 Solutions – Kimel 1. 8.5 a)

R

L For the TM modes, we saw in class the resonance frequencies are TM:

ω mnp =

1 μ

x 2mn p2π2 + R2 L2

p = 0, 1, 2, . . . . m = 0, 1, 2, . . . . n = 1, 2, 3, . . . .

TE:

ω ′mnp =

1 μ

p2π2 x ′2 mn + L2 R2

p = 1, 2, 3, . . . . m = 0, 1, 2, . . . . n = 1, 2, 3, . . . .

Thus ω mnp 1 μ

ω ′mnp 1 μ

=

x 2mn p2π2 + 2 R L2

=

x ′2 p2π2 mn + R2 L2

The lowest four frequencies are (in these units) ω 010 = 2. 405 ω 110 = 3. 832 ω ′111 =

1. 841 2 + π 2

R L

2

ω ′211 =

3. 054 2 + π 2

R L

2

2. 405, 3. 832, 1. 841 2 + π 2 x 2 , 3. 054 2 + π 2 x 2

10 8 6 4 2

0

0.2

0.4

0.6

0.8

1 x

1.2

1.4

1.6

1.8

2

where x = R/L. The answer is ”No.” ω ′111 and ω 010 cross when 1. 841 2 + π 2 x 2 = 2. 405 or x = 0. 492 58. For frequencies smaller than this cross over frequency, ω ′111 is lowest, whereas for larger frequencies, ω 010 is lowest.

PHY 5347 Homework Set 3 Solutions – Kimel 3. 9.1 a)

ρx⃗, t = qδzδy − sin ω 0 tδx − d cos ωt To illustrate the equivalence of the two methods, I’ll consider the lowest two moments. n = 0 : Qt = n=1:⃗ pt =

∫ ρx⃗, td 3 x = q = Reqe −i0⋅ωt 

∫ ρx⃗, tx⃗d 3 x = qdî cos ωt +

sin ωt = Reqdî + i e −i1⋅ωt

So we identify ⃗ p = qdî + i  as the quantity to be used in Jackson’s formulas. Arbitrary n: The n’th multipoles will contribute with maximum frequencies of ω n = nω. b) The proof that we can write β

ρx⃗, t = ρ 0 x⃗ + ∑ Re2ρ n x⃗e −inωt n=1

with ρ n x⃗ = 1τ

τ

∫ 0 ρx⃗, te inωt dt

was presented in lecture and will not be repeated here. c) We have already calculated the n = 0, 1 moments by the method of part a). Now we compute these moments by the method of part b). n=0: ρ 0 x⃗ = 1τ Q= n=1:

∫ ρ 0 x⃗d 3 x =

q τ

τ

∫ 0 qδzδy − sin ω 0 tδx − d cos ωt dt τ

∫ 0 dt ∫ d 3 xδzδy − sin ω 0 tδx − d cos ωt = q

ρ 1 x⃗ = 1τ ⃗ px⃗ =

∫ d 3 xx⃗2ρ 1 x⃗ = =

as before.

τ

∫ 0 qδzδy − sin ω 0 tδx − d cos ωt e iωt dt

2qd τ

τ

2q τ

τ

∫ 0 dt ∫ d 3 xx⃗δzδy − sin ω 0 tδx − d cos ωt

∫ 0 dte iωt î cos ωt +

sin ωt = qdî + i 

PHY 5347 Homework Set 3 Solutions – Kimel 3. 9.2 First consider a rotating charge which is at an angle α at time t = 0.

Compared to the lecture notes for this problem, where we assumed α = 0, we should let ωt → ωt + α. Thus using the result developed in class, we can write for this problem 1 Q α t = Re

3 qd 2 2

i

0

i −1 0 0

0

e −i2a e −i2ωt

0

From the figure Q tot t = Q α1 t + Q α2 t + Q α3 t + Q α4 t 1 = Re

3 qd 2 −e −i π2 + e −i 3π2 − e −i 5π2 + e −i 7π2 2

1 Q tot

0

i −1 0 0

Thus from the class notes

i 0

0

i −1 0 0

= 3 qd 2 4i 2

i

0

0

0

e −i2ωt

dP = c 2 Z 0 k 6 qd 2 3 2 dΩ 1152π 2 P=

2

161 − cos 4 θ =

c2Z0k6 qd 2 3 360π 2

2

16 =

And, of course, the frequency of the radiation is 2ω.

1 c 2 Z k 6 q 2 d 4 1 − cos 4 θ 0 32π 2 1 c2Z0k6q2d4 10π


Since the problem has azimuthal symmetry, we can expand Vr⃗, t (in the radiation zone) in terms of Legendre polynomials: Vr⃗, t =

∑ b l tr −l−1 Pl cos θ l

Using the orthogonality of the Legendre polynomials, the leading term of the expansion in the radiation zone will be the l = 1 term. 1 b 1 t = 3 R 2 ∫ xVr⃗, tdx = 3 VR 2 cos ωt 2 2 −1

So, Vr⃗, t = with ⃗ p=

3 2

⃗ p ⋅ r̂ p ⋅ r̂ −iωt 3 VR 2 cos ωt /r 2 = ⃗ cos ωt = Re e 2 2 r r2

VR 2 ẑ , which should be used in the radiation formulas developed in lecture. dP = c 2 Z 0 k 4 |p ⃗ |2 sin 2 θ dΩ 32π 2 2 4 ⃗ |2 P = c Z 0 k2 8π = 1 c 2 Z 0 k 4 |p 3 12π 32π

with ⃗ p=

3 2

VR 2 ẑ .

PHY 5347 Homework Set 4 Solutions – Kimel 1. 9.11 We are working with small sources in the radiation zone.

From the notes and Eqs (9.170) and (9.172), Q lm = M lm = − 1 l+1

3 ∫ r l Y m∗ l ρd x

⃗ ∫ r l Y m∗ l ∇⋅

⃗r × ⃗ J d3x

a) Electric Dipole Radiation: Q 1m = =

3 ∫ rY m∗ 1 ρd x

∫ rδxδy−qδz − a cos ω 0 t − qδz + a cos ω 0 t + 2qδz Y m∗ 1 dxdydz

0 0 m∗ = −qa cos ω 0 tY m∗ 1 0, φ + Y 1 π, φ = −qaδ m0 cos ω 0 tY 1 0 + Y 1 π = 0

b) Magnetic Dipole Radiation: Since the particles move in an orbit with no area, ⃗r × ⃗ J = 0 → M 1m = 0 c) Electric Quadrupole Radiation: Q 2m =

3 ∫ r 2 Y m∗ 2 ρd x

= −qδ m0 a 2 cos 2 ω 0 tY 02 θ = 0 + a 2 cos 2 ω 0 tY 02 θ = π = −qδ m0 a 2 Y 02 0cos 2ω 0 t + 1 where I have used cos 2 ω 0 t = cos 2ω 0 t + 1/2. Thus the Fourier Series decomposition of this moment yields terms with frequency 0, and 2ω 0 . The first term does not contribute to radiation, and the second can be written

Q 20 t = Re−2qa 2 Y 02 0e −2iω 0 t so Q 20 = −2qa 2 Y 02 0 is the quantity that is used in the radiation formulas of Jackson. Using Eq. (9.151) dP 2, 0 = Z 0 |a2, 0|2 X ⃗ 20 dΩ 2k 2

2

and from Eq. (9.169) a2, 0 = where I have used Y 02 0 =

ck 4 i5 × 3 5 4π

3 Q 20 = ck 4 2 i5 × 3

P2, 0 =

5 4π

. Thus |a2, 0|2 =

dP 2, 0 = Z 0 dΩ 2k 2

3 −2qa 2  2

1 c2k8q2a4 30π

1 c2k8q2a4 30π

15 sin 2 θ cos 2 θ 8π

=

1 Z k 6 c 2 q 2 a 4 sin 2 θ cos 2 θ 0 32π 2

2π Z k 6 c 2 q 2 a 4 ∫ 1 1 − x 2 x 2 dx = 2π Z k 6 c 2 q 2 a 4 × 4 0 0 15 32π 2 32π 2 −1 P2, 0 =

1 Z k6c2q2a4 60π 0

PHY 5347 Homework Set 5 Solutions – Kimel 1. The system is described by

and is azimuthally symmetric Rθ = R 0 1 + βtP 2 cos θ ; βt = β 0 cos ωt; kR << 1 Q=

∫ ρr 2 drdφd cos θ = 2π ∫ −1 d cos θρ ∫ 0 1

Rθ

r 2 dr

1 = 2π ρ ∫ R 30 1 + 3βP 1 P 2 d cos θ + Oβ 2  = 4π ρR 30 → ρ = 3 3 Q 3 3 4πR 0 −1

where I’ve used the fact that 1 = P 0 . Since the system is azimuthally symmetric, Q lm = δ m0 Q l0 . Q lm = 2πρδ m0 ∫ Using Y 0l =

2l+1 4π

1 −1

dxY 0l ∫

Rθ 0

r l+2 dr =

2πρδ m0 l+3

0 ∫ −1 dxR l+3 0 1 + l + 3βP 2 Y l 1

P l and 1 = P 0 , Q lm =

2πρδ m0 l+3

2l + 1 R l+3 2δ + l + 3β 2 δ 0 l0 4π 2l + 1 l2

Notice that the l = 0 term is time independent and thus does not contribute to the radiation. Next consider the l = 2 term. Q 20 t = 2 π ρ 5 R 50 β = ρ = 3 3 Q 2 π ρ 5 R 50 β = 5 4πR 0 5 Q 20 t = Re

3 R 2 Qβ 0 e −iωt 0 20π

3 R 2 Qβt 0 20π

3 R 2 Qβ 0 0 20π

Q 20 =

dP2, 0 Z ⃗ 20 = 02 |a2, 0|2 X dΩ 2k a E 2, 0 = ⃗ 20 X

2

ck 4 i5 × 3

3 Q 2 20

= 15 sin 2 θ cos 2 θ 8π

dP2, 0 ck 4 = Z 02 dΩ 2k 5 × 3

3 Q 2 20

2

× 15 sin 2 θ cos 2 θ 8π

2 4 ck 4  Z Z 2 1 1 0 |ck Q 20 | 0 2 = sin θ cos θ = π 160 k 2 160 k 2

= P=

2

2

3 20π

R 20 Qβ 0

π

2

sin 2 θ cos 2 θ

9 Z 0 k 6 c 2 R 40 Q 2 β 20 sin 2 θ cos 2 θ 3200π 2

1 9 9 Z k 6 c 2 R 40 Q 2 β 20 × 2π ∫ 1 − x 2 x 2 dx = Z 0 k 6 c 2 R 40 Q 2 β 20 × 2π × 4 2 0 15 3200π 3200π 2 −1

P=

3 Z k6c2R4Q2β2 0 0 2000π 0

PHY 5347 Homework Set 5 Solutions – Kimel 2. The system is described by

It = I 0 cos ωt = ReI 0 e −iωt ⃗ Jt = 1a Itδr − aδcos θφ̂ where I determined the normalization constant

1 a

⃗=I by the condition ∫ ⃗ J ⋅ da

⃗ Jt = Re Ia0 δr − aδcos θφ̂ e −iωt

→⃗ J = Ia0 δr − aδcos θφ̂

⃗ and E ⃗ in the radiation zone given by Eq.(9.149). Since this We use the general expression for H system has no net charge density and there is no intrisic magnetization, the the expansion coefficients in these equations are given by a E l, m =

a M l, m =

k2 i ll + 1 k2 i ll + 1

∫ Y m∗ l ik ⃗ ∫ Y m∗ l ∇⋅

⃗r ⋅ ⃗ J j l krd 3 x

⃗r × ⃗ J j l krd 3 x

a) ⃗r ⋅ ⃗ J = 0 in the first equation, so there is no electric multipole radiation. In spherical coordinates ⃗r × ⃗ J = −aJθ̂ ⃗ in spherical coordinates given in the back of the book, Using the formulas for ⃗ ∇⋅A ⃗ ∇ ⋅ ⃗r × ⃗ J = − 1 ∂ J sin θJ = − cos θ J − ∂ J sin θ ∂θ sin θ ∂θ The first term does not contribute, because cos θ = 0, while the second term can be written, using the chain rule,

⃗ ∇ ⋅ ⃗r × ⃗ J = sin θ

∂ J ∂ cos θ

The problem has azimuthal symmetry, so m = 0. Realizing derivatives of δ −functions are defined by integration by parts, a M l, m =

δ m0 k 2 i ll + 1

∫ Y 0∗l

sin θ

∂ J j krd 3 x = l ∂ cos θ

ik 2 ll + 1

∫

∂ sin θY 0∗  Jj l krd 3 x l ∂ cos θ

a M l, 0 =

i2πk 2 I 0 a 2 j ka ∂ sin θY 0 | l l cos θ=0 a ∂ cos θ ll + 1

a M l, 0 =

i2πk 2 I 0 a 2 j l ka1 − x 2  1/2 d Y 0 x| x=0 a dx l ll + 1

Since Y 0l x is either an even or odd polynomial in x, then only odd l contribute to a M l, 0. This ⃗ and E ⃗ in the radiation zone are known through determines the expansion coefficients, and thus H Eq.(9.149). The power distribution is given by Eq. (9.151) b) From our previous answers, we see a E l, m = 0, and that the lowest magnetic multipole contribution is a M 1, 0. 2 1/2 a M 1, 0 = i2πk I 0 aj 1 ka1 − x 2  d Y 01 x|x=0 dx 2

Using j 1 ka → ka ; 3

d Y 0 x| = dx 1 x=0

a M 1, 0 = i2πk 3 I 0 a 2

M l0 =

ik 3 3

1 = ik 3 2 M l0 24π 3 1 24π

i2πk 3 I 0 a 2 2

3 4π

=

3 I πa 2 4π 0

Note that you would get the same answer, if you used Eq. (9.172) directly. From Eq. (9.151) dP = Z 0 dΩ 2k 2

3

2πk Fa

2

1 24π

2

3 sin 2 θ = 1 Z 0 k 4 I πa 2  2 sin 2 θ 0 8π 32π 2

If we compare this result with the one that we get for an elementary magnetic dipole, which is given by Eq. (9.23) ⃗ /c, with the substitution ⃗ p→m

dP = 1 Z 0 k 4 |m ⃗ |2 sin 2 θ dΩ 32π 2 Thus we may identify ⃗ | = I 0 πa 2 |m as would be expected.

PHY 5347 Homework Set 6 Solutions – Kimel 1. 10.1 a) Let us first simplify the expression we want to get for the cross section. Using n̂ 0 = ẑ , dσ ̂ , n̂ , n̂  = k 4 a 6 5 − |̂ 0 ⋅ n̂ |2 − 1 |n̂ ⋅ ẑ × ̂ |2 − ẑ ⋅ n̂ 0 dΩ 0 0 4 4 Orienting the system as

and using ̂ 0 = α 0 x̂ + β 0 , with |α 0 |2 + |β 0 |2 = 1 n̂ = cos θẑ + sin θx̂ then dσ ̂ , n̂ , n̂  = k 4 a 6 5 − |α |2 sin 2 θ − 1 |β 0 |2 sin 2 θ − cos θ 0 4 4 dΩ 0 0 Using the result for the perfectly conducting sphere Eq. (10.14) dσ n̂ , ̂, ̂ 0 , n̂ 0  = k 4 a 6 ̂ ∗ ⋅ ̂ 0 − 1 ẑ × ̂  ⋅ n̂ × ̂ ∗  0 2 dΩ

2

Using ̂ ⊥ = dσ n̂ , ̂ ⊥ , ̂ 0 , n̂ 0  = k 4 a 6 β 0 − 1 β 0 cos θ 2 dΩ

2

2

= k 4 a 6 |β 0 |2 1 − 1 cos θ 2

Similarly ̂ || = x̂ ′ dσ n̂ , ̂ , ̂ , n̂  = k 4 a 6 α cos θ − 1 α 0 || 0 0 dΩ 2 0 By definition

2

= k 4 a 6 |α 0 |2 cos θ − 1 2

2

dσ ̂ , n̂ , n̂  = dσ n̂ , ̂ ⊥ , ̂ 0 , n̂ 0  + dσ n̂ , ̂ || , ̂ 0 , n̂ 0  dΩ dΩ dΩ 0 0 = k 4 a 6 |β 0 |2 1 − 1 cos θ 2

2

+ |α 0 |2 cos θ − 1 2

2

which simplifies to dσ ̂ , n̂ , n̂  = k 4 a 6 5 − |α |2 sin 2 θ − 1 |β |2 sin 2 θ − cos θ 0 4 0 4 dΩ 0 0 using |α 0 |2 + |β 0 |2 = 1, and cos 2 θ = 1 − sin 2 θ. b) If ̂ 0 is linearly polarized making an angle φ with respect to the x axis, then ̂ 0 = α 0 x̂ + β 0

= cos φx̂ + sin φ , so α 0 = cos φ, β 0 = sin φ

Then from part a) dσ ̂ , n̂ , n̂  = k 4 a 6 5 − |α 0 |2 sin 2 θ − 1 |β 0 |2 sin 2 θ − cos θ 4 4 dΩ 0 0 = k 4 a 6 5 − cos 2 φ sin 2 θ − 1 sin 2 φ sin 2 θ − cos θ 4 4 Using cos2φ = cos 2 φ − sin 2 φ, this expression simplifies to dσ ̂ , n̂ , n̂  = k 4 a 6 5 1 + cos 2 θ − 3 sin 2 cos 2φ − cos θ 8 dΩ 0 0 8 as desired.

PHY 5347 Homework Set 7 Solutions – Kimel 1. 11.3 Let us just focus on the 0, 1 component transformation, since the 2, 3 components remain unchanged, if we take the relative velocities between Lorentz frames to be along the x direction. We want to relate a single Lorentz transformation to two sequential transformations as described by

Thus we require A = A2A1 where A is a Lorentz transformation. Rewritten explicitly, the above equation reads γ

−βγ

−βγ

γ

=

=

γ2

−β 2 γ 2

γ1

−β 1 γ 1

−β 2 γ 2

γ2

−β 1 γ 1

γ1

γ2γ1 + β2γ2β1γ1

−γ 2 β 1 γ 1 − β 2 γ 2 γ 1

−γ 2 β 1 γ 1 − β 2 γ 2 γ 1

γ2γ1 + β2γ2β1γ1

So γ = γ2γ1 + β2γ2β1γ1 βγ = γ 2 β 1 γ 1 + β 2 γ 2 γ 1 βγ γ β γ + β2γ2γ1 β1 + β2 β= γ = 2 1 1 = γ2γ1 + β2γ2β1γ1 1 + β2β1 Or v = v 1 +v 1vv22 1 + c2 as required.


From the class notes, and Eq. (11.31), we know u || =

u ′|| + v vu ′|| c2

1+

u⊥ =

;

u ′⊥ γ1 +

vu ′|| c2



and dt = dt ′ γ1 +

vu ′||  c2

Thus taking the differential of the first equation above and using the second equation for dt, du || ≡ a || = dt

1 +

vu ′|| c2

a ′|| − u ′|| + v cv2 a ′|| vu ′||

γ1 +

c2

1−

=

3

1 +

3/2 v2 c2 vu ′|| 3  c2

Similarly, du ⊥ ≡ a = ⊥ dt

vu ′||

1+

c2

a ′⊥ − u ′⊥

γ 2 1 +

vu ′|| c2

v c2

a ′||

3

This is equal to the expression we want to prove, du ⊥ ≡ a = ⊥ dt

1− 1 +

v2 c2 vu ′|| c2

3

v × a ⃗′ × ⃗ a ′⊥ + ⃗ u′  c2

since the BAC - CAB theorem shows that v × a ⃗′ × ⃗ a ′⊥ + ⃗ u′  = c2

1+

vu ′|| c2

a ′⊥ − u ′⊥ v2 a ′|| c

a ′||

PHY 5347 Homework Set 7 Solutions – Kimel 3. 11.15 ⃗ =β ⃗ = 0. Then ⃗ || ẑ , so β ⃗⋅E ⃗⋅B From Eq. (11.149), it is clear that we should take β ⃗+β ⃗ ⃗×B E ′ = γE ⃗ ′ = γB ⃗−β ⃗ ⃗×E B The vectors in parentheses should make the same angle wrt the x axis θ ′ if they are to be parallel. This can best be seen from the figure,

From the figure ⃗ ′ = γE 0 î − β2E 0  sin θî + β2E 0 cos θ̂  E ⃗ ′ = γcos θ2E 0 î + sin θ2E 0 ̂ − βE 0 ̂  B Thus tan θ ′ =

2β cos θ 2 sin θ − β = 1 − 2β sin θ 2 cos θ

2 cos θ ⋅ 2β cos θ − 1 − 2β sin θ ⋅ 2 sin θ − β = 0 or 2β 2 sin θ − 5β + 2 sin θ = 0 This quadratic equation has the solution

β=

1 5− 4 sin θ

25 − 16 sin 2 θ

where I’ve chosen the solution which give β = 0 if θ = 0. If θ << 1, then β → 0, and the original fields are parallel. If θ → π/2 then β =

1 4

5 − 3 = 1/2. γ =

2 3

⃗ ′ = 0 + Oθ − π ̂ E 2 ⃗′ = B ⃗ ′ = γ2E 0 ̂ − 1 E 0 ̂  = γE 0 3 ̂ B 2 2 So in these two limits, the fields are parallel to the x and y axes, respectively.

PHY 5347 Homework Set 8 Solutions – Kimel 2. 12.1 (a) q L = − 1 mu α u α − c u α A α 2

(invariant Lagrangian)

Show this Lagrangian gives the correct eqn. of motion, ie, Eq. (12.2) du α = e F αβ u β mc dτ τ2

The Action is A = ∫ Ldτ. τ1

δA = 0 yields the Lagrange equations of motion d ∂L − ∂L = 0 dτ ∂u α ∂x α d ∂L = −m d u α − q ∂A α dx μ c ∂x μ dτ dτ dτ ∂u α ∂L = − q u ∂ α A β c β ∂x α So

d ∂L dτ ∂u α

−

∂L ∂x α

= 0 yields q q q m d u α = c u β ∂ α A β − c ∂ μ A α u μ = c u β ∂ α A β − ∂ β A α  dτ

or d u α = q F αβ u β mc dτ

PHY 5347 Homework Set 8 Solutions – Kimel 3. 12.2 (a) L ′ = L + d λt, ⃗ x dt x  − λt 1 , ⃗ x = 0 → L and L ′ yield the same Euler-Lagrange Eqs. of Mot. δ ∫ L ′ − Ldt = δλt 2 , ⃗ t2

t1

where the last equality follows from the fact that variation at the end points is zero since the end points are held fixed. (b) For simplicity of notation in this part, I’m going to set c = 1. ⃗ − eφ with A μ = φ, A ⃗ ⃗⋅A L = −m 1 − u 2 + eu If A α → A α + ∂ α Λ, then φ → φ + ∂0Λ ⃗ →⃗ A a−⃗ ∇Λ where the minus sign in the second equation should be noticed. Thus ⃗ − eφ − eu ⃗⋅A ⃗⋅⃗ L → −m 1 − u 2 + eu ∇Λ − e∂ 0 Λ Now μ d Λt, ⃗ x = ∂ μ Λt, ⃗ x ∂x = ∂ Λ + ⃗ ∇Λ ⋅ ⃗ u dt ∂x ∂t ∂t

Or ⃗ − eφ − e d Λt, ⃗ ⃗⋅A L → −m 1 − u 2 + eu x dt By the argument of part (a), this Lagrangian gives the same equations of motion as the original Lagrangian.

PHY 5347 Homework Set 9 Solutions – Kimel 1. 12.3 ⃗ 0 along z, ⃗ v 0 along y a) Take E

Generally ⃗ dp v ×B ⃗+ ⃗ ⃗ ; =e E c dt

dE = ev⃗ ⋅ E ⃗ dt

⃗ = 0, and E ⃗ = E 0 ẑ Since B dp z = eE 0 dt dp y =0 dt The initial condition ⃗ p0 = mv 0 and the above equations show the subsequent motion is in the y − z plane. Consistent with this initial condition, we have p y t = mv 0 ; p z t = eE 0 t Et =

⃗ p 2 tc 2 + m 2 c 4 =

m 2 v 20 c 2 + m 2 c 4 + ceE 0 t 2 =

ω 20 + ceE 0 t 2

Using ⃗ p = mγv⃗ = E2 ⃗ v c v y t =

yt = where ρ ≡

ω0 ceE 0

. So

mv 0 c 2 ceE 0

p y t = Et/c 2

∫0 t

mv 0 c 2 ω 20 + ceE 0 t 2

mv 0 c dt = sinh −1 t/ρ eE 2 2 0 ρ +t

mv 0 c tceE sinh −1  ω 0 0  eE 0

yt =

Eq. (1)

Similarly p z t = Et/c 2

v z t =

eE 0 tc 2 ω 20 + ceE 0 t 2

Thus 2 zt = eE 0 c ceE 0

∫0 t

tdt =c ρ2 + t2

ρ 2 + t 2  − ρ

Eq.(2)

b) From Eq. (1) t=

ω0 eE y eE sinh mv00 c  = ρ sinhky, with k = mv 0c 0 ceE 0

Then from Eq.(2) z = cρ Let us plot

sinh 2 ky + 1 − 1

Eq.(3)

sinh 2 x + 1 − 1 20 18 16 14 12 10 8 6 4 2 0

1

2

x

3

4

5

For small t : t/ρ << 1, and ky << 1. Thus we can taylor expand Eq.(3) and get z = cρk 2 y 2 /2 which is quadratic in y giving a parabolic shape. For large t : t/ρ >> 1, and we see the sinh term dominates in Eq.(3) and we get z which is an exponential shape.

cρe ky 2

PHY 5347 Homework Set 9 Solutions – Kimel 2. 12.5 a) The system is described by

⃗ and E ⃗ . We want E ⃗ ′⊥ = 0 = γE + u⃗ × B ⃗ . Background: particle having m, e. Choose ⃗ u ⊥ to B c ⃗ = −E ⃗ ; now B ⃗× ⃗ ⃗ = cE ⃗×B ⃗ . Using BAC - CAB on the lhs of the equation gives Thus u⃗c × B u×B ⃗ ⃗ ⃗ u = c E×B 2 . Thus from the figure, B

⃗ ⃗ ⃗ u = c E ×2 B = c E ẑ B B Then, using Eq. 11. 149 ⃗ || = 0; E

⃗ ′⊥ = 0; E

⃗ ′⊥ = 1 B ⃗ ′ = 0; B ⃗ B || γ =

1−

u c

2

⃗ B

So ⃗ ′⊥ = B

1−

E B

2

⃗ = B

B 2 − E 2 B̂ 2 B2

Now from the class notes, ′ ⃗′ = ⃗ du ⃗ B ′ , where ω ⃗ B ′ = eB′ , where in this case E ′ is the energy of the particle. u′ × ω ′ dt E

I’ll choose the same boundary conditions as in class, decribed in the figure.

So ⃗ u ′⊥ = ω B ′ acosω B ′ t ′ ̂ 3 − sinω B ′ t ′ ̂ 1 where u ′⊥ t = 0 = ω B ′ a (ie, the BC determine a. ⃗ x ′ t ′  = u ′|| t ′ ̂ 2 + a̂ 3 sin ω B ′ t ′ + ̂ 1 cos ω B ′ t ′  Consider the inverse Lorentz transformation between the frames, γ

βγ 0 0

ct ′

βγ

γ

0 0

z′

x

0

0

1 0

x′

y

0

0

0 1

y′

ct z

=

or t = γt ′ + βγz ′ /c = γt ′ + βγa sin ω B ′ t ′ /c ≡ ft ′  → t ′ = f −1 t So zt = βγcf −1 t + γa sin ω B ′ f −1 t xt = a cos ω B ′ f −1 t ′ −1 f t yt = u ||0

⃗ field alone. Then the b) If |E| > |B|, one can transform to a frame where the field is a static E solution is as we found in section 12.3 of the text, with the above transformation taking you to the unprimed frame.

PHY 5347 Homework Set 9 Solutions – Kimel 3. 12.14 a) We are given L = − 1 ∂ α A β ∂ α A β − 1c J α A α 8π which can be rewritten L = − 1 ∂ β A α ∂ β A α − 1c J α A α 8π Using the Euler-Lagrange equations of motion, ∂β

∂L − ∂L = 0 ∂A α ∂∂ β A α 

Noting ∂L = − 1 ∂βAα 4π ∂∂ β A α  ∂L = − 1 J α c ∂A α The Euler-Lagrange equations of motion are α ∂ β ∂ β A α  = ∂ β ∂ β A α  = 4π c J

or α ∂ β ∂ β A α − ∂ α A β + ∂ α A β  = ∂ β F βα + ∂ β ∂ α A β = 4π c J

If we assume the Lorentz gauge, ∂ β A β = 0, then the above reduces to α ∂ β F βα = 4π c J

Maxwell’s equations, given by Eq. (11.141). b) Eq. 12. 85 gives −

1 F F αβ  − 1 J α A α c 16π αβ

The term in parentheses can be written F αβ F αβ = 2∂ α A β ∂ α A β − 2∂ α A β ∂ β A α  + 2A β ∂ β ∂ α A α The last term vanishes if we choose the Lorentz gauge, and the second term is of the form of a 4-divergence. Thus the Lagrangian of this problem differs from the usual one, of Eq. (12.85) by a 4-divergence ∂ α A β ∂ β A α . The 4-divergence does not change the euations of motion since the fields vanish at the limits of integration given by the action. Using the generalized Gauss’s theorem or by integrating by parts, we

see the 4-divergence gives zero contribution to the action.

PHY 5347 Homework Set 10 Solutions – Kimel 2. 14.2 Background. In the nonrelativistic approximation the Lienard-Wiechert potenials are ⃗ ⃗ x⃗, t = eβ |ret A R

φx⃗, t = e |ret , R

Let us assume that we observe the radiation close enough to source so R/c << 1 and t ′ ≅ t. Then in the radiation zone ⃗ = −n̂ ∂ × A ⃗ = eβ̇ × n̂ ⃗ = n̂ ∂ × A ⃗ =⃗ B ∇×A ′ cR ∂R c∂t ⃗ =B ⃗ × n̂ E dPt ⃗ = c RB 4π dΩ

2

2 = e β̇ × n̂ 4πc

2

=

e 2 |v̇ |2 sin 2 θ 4πc 3

where θ is the angle between n̂ and β̇ (assuming here the particle is moving linearly) 2 Pt = 2e 3 |v̇ |2 3c

Let the time-average be defined by 〈ft〉 ≡ 1τ

τ

∫ 0 ftdt

Then dPt dΩ

=

e 2 sin 2 θ |v̇ |2 4πc 3

2 〈Pt〉 = 2e 3 |v̇ |2 3c

a) Suppose ⃗ xt = ẑ a cos ω 0 t. Then v̇ = |v̇ |2

= aω 20  2 1τ

d2z dt 2

= −aω 20 cos ω 0 t

τ

∫ 0 cos 2 ω 0 tdt = aω 20  2 /2

So dPt dΩ

=

e 2 aω 2  2 sin 2 θ 0 8πc 3

2 2 〈Pt〉 = e 3 aω 20  3c

b) Suppose ⃗ xt = Rî cos ω 0 t +

sin ω 0 t. Then

v̇ t = −Rω 20 î cos ω 0 t +

sin ω 0 t k̂

î n̂ × v̇ =

sin θ cos φ

sin θ sin φ

cos θ

−Rω 20 cos ω 0 t

−Rω sin ω 0 t

0

2 dPt = e 3 Rω 20  2 cos 2 θ sin 2 ω 0 t + cos 2 ω 0 t + sin 2 θ sin 2 ω 0 t + φ  dΩ 4πc

dPt dΩ 〈Pt〉 =

=

2 e 2 Rω 2  2 1 + cos θ 0 2 4πc 3

2 e 2 Rω 2  2 2π ∫ 1 1 + x  dx = 2e 2 Rω 2  2 0 0 2 −1 3c 3 4πc 3

PHY 5347 Homework Set 11 Solutions – Kimel 2. 14.12 a) From Jackson, Eq (14.38) dP = e 2 dΩ 4πc

⃗ × β̇ n̂ × n̂ − β ⃗ 5 1 − n̂ ⋅ β

2

Using azimuthal symmetry, we can choose n̂ in the x-z plane.

From the figure n̂ = cos θẑ + sin θx̂ ⃗t ′  = − a ω 0 sin ω 0 t ′ ẑ = −β sin ω 0 t ′ ẑ β c aω 2 β̇ t ′  = − c 0 cos ω 0 t ′ ẑ = −ω 0 β cos ω 0 t ′ ẑ ⃗ × β̊ = 0 Using β ⃗ × β̇ n̂ × n̂ − β

2

= n̂ × β̇

2

and n̂ × β̇ =

ω 0 β sin θ cos ω 0 t ′

So 2 4 sin 2 θ cos 2 ω 0 t ′ dP t ′  = e cβ dΩ 4πa 2 1 + β cos θ sin ω 0 t ′  5

Defining φ = ω 0 t ′ ,

e 2 cβ 4 1 〈 dP t ′ 〉 = dΩ 4πa 2 2π

∫0

2π

sin 2 θ cos 2 φ dφ 1 + β cos θ sin φ 5

Or, doing the integral, 4 + β 2 cos 2 θ e 2 cβ 4 〈 dP 〉 = sin 2 θ 2 7/2 2 2 dΩ 32πa 1 − β cos θ c)

4+.05 2 cos 2 θ 7/2 1−.05 2 cos 2 θ

sin 2 θ 5 4 3 2 1

0

4+.95 2 cos 2 θ 7/2 1−.95 2 cos 2 θ

0.5

1

1.5

2

2.5

3

0.5

1

1.5

2

2.5

3

sin 2 θ 300 250 200 150 100 50 0

PHY 5347 Homework Set 12 Solutions – Kimel 1. 16.1 It’s useful to apply in this case the Virial Theorem, familiar from classical mechanics: 〈T〉 = 1 〈 dV 〉r 2 dr If V = ar n , then 〈T〉 = n 〈V〉 2 In our case V =

1 2

kr 2 , with k = mω 20 , so n = 2 and 〈T〉 = 〈V〉

Or, 〈 dV 〉 = E r dr We are given dE = − τ 〈 dV m dr dt

2

〉

This can be rewritten dE = − τ kE m dt So τ

E = E 0 e − m kt = E 0 e −τω 0 t = E 0 e −Γt 2

Similarly, ⃗ τ 〈 1 dV 〉L dL ⃗ = −m r dr dt But

1 dV r dr

= mω 20 , so L = L 0 e −τω 0 t = L 0 e −Γt 2

PHY 5347 Homework Set 12 Solutions – Kimel 2. 16.2 V = −, q = −e dE = − τ 〈 dV m dt dr

2

〉

If V = ar n , then the Virial theorem tells us 〈T〉 = n 〈V〉 2 In the present case, n = −1, so 2 E = 〈T〉 + 〈V〉 = 1 〈V〉 = − Ze 2r 2

dV = Ze 2 dr r2 Now dE = − τ 〈 dV m dr dt

2

〉

gives d 1 = 2Ze 2 τ dr rt mr 4 t or r 2 dr = −2Ze 2 τdt/m But τ =

2 e2 3 c3m

, so r 2 dr = −3Zcτ 3 τt

Integrating both sides gives r 3 t = r 30 − 9Zcτ 3 τt b) At this point, for simplicity of notation, I’m going to take c =  = 1. Then from problem 14.21, 1 = 2 e 2 Ze 2  4 m T 3 n5 We are given 2 r = n a0 Z

Where a 0 =

1 me 2

= Bohr radius, and τ =

2 e2 3 m

Z − dn = − Z dr = Z 32 Zτ 2 = Z 3 dt 2a 0 n dt 2a 0 n r 2a 0 n a0n2

2

2 Z 2 em 3

2

4 6 = 2 Zm e 5 3 n

in agreement with the result of problem 14.21. c) From part b

But rt =

n 2f a 0 Z

, r0 =

t=

r 30 − r 3 t 9Zτ 2

−

n 2f a 0 Z

n 2i a 0 Z

t=

n 2i a 0 Z

3

9Zτ

3

n6 − n6 = 1 a 30 i 4 2 f 9 Z τ

2

In our present case Z = 1, so t= 1 9

1 me 2

3

n 6i − n 6f 2 e2 3 m

2

= 1 e −10 n 6i − n 6f  4m

In these units, (from the particle data book) MeV −1 = 6. 6 × 10 −22 s. e 2 = α = 1/137, and m = 207 ×. 511MeV. t=

1 × 6. 6 × 10 −22 s n 6i − n 6f  = 7. 53 × 10 −14 n 6i − n 6f s 4 × 207 ×. 5111/137 5

For the cases desired,

t 1 = 7. 53 × 10 −14 10 6 − 4 6 s = 7. 5 × 10 −8 s t 2 = 7. 53 × 10 −14 10 6 − 1 6 s = 7. 530 0 × 10 −8 s Just as a check on working with these units, notice 2 1 τ = 2 em = 2 6. 6 × 10 −22 s = 6. 29 × 10 −24 s 3 . 511 × 137 3

in agreement with what we found before.

Solved Problems of Jackson's Electrodynamics 02

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