Instructor Solution Manual Electromagnetics Third Edition Edward J. Rothwell, Michael J. Cloud

SOLUTION MANUAL

Solution Manual to Accompany the Third Edition of

Electromagnetics CRC Press, 2018 E.J. Rothwell and M.J. Cloud March 8, 2018

Chapter 1 1.1. The charge density is given by ρ(r, t) = ρs (θ, φ, t)δ(r − r0 ).

The total charge on the sphere is Z ρ(r, t) dV (V = all of space) Q(t) = V Z 2π Z π Z ∞ = ρs (θ, φ, t)δ(r − r0 ) r2 sin θ dr dθ dφ 0 0 0 Z ∞

Z 2π Z π 2 = ρs (θ, φ, t) δ(r − r0 ) r dr sin θ dθ dφ 0 0 0 Z 2π Z π 2 = r0 ρs (θ, φ, t) sin θ dθ dφ.

0

0

1.2. The charge density is given by ρ(r, t) = ρs (r, θ, t)δ (ρ[φ − φ0 ]) = ρs (r, θ, t)δ (r sin θ[φ − φ0 ]) δ(φ − φ0 ) = ρs (r, θ, t) r sin θ since δ(ax) = δ(x)/a. The total charge on the half-plane is Z Q(t) = ρ(r, t) dV (V = all of space) V Z 2π Z π Z ∞ δ(φ − φ0 ) 2 = ρs (r, θ, t) r sin θ dr dθ dφ r sin θ 0 0 0 Z πZ ∞ Z 2π = ρs (r, θ, t) r dr dθ δ(φ − φ0 ) dφ 0 Z0 π Z0 ∞ = ρs (r, θ, t) r dr dθ.

0

0

1.3. The charge density is given by ρ(r, t) = ρs (φ, z, t)δ(ρ − ρ0 ).

The total charge on the cylinder is Z Q(t) = ρ(r, t) dV (V = all of space) V Z ∞ Z 2π Z ∞ = ρs (φ, z, t)δ(ρ − ρ0 ) ρ dρ dφ dz −∞ 0 0 Z ∞

Z ∞ Z 2π = ρs (φ, z, t) δ(ρ − ρ0 ) ρ dρ dφ dz −∞

0

0

Z ∞ Z 2π = ρ0

ρs (φ, z, t) dφ dz. −∞

0

1

1.4. The charge density is given by ρ(r, t) = ρs (ρ, z, t)δ (ρ[φ − φ0 ]) = ρs (ρ, z, t)

δ(φ − φ0 ) . ρ

The total charge on the half-plane is Z Q(t) = ρ(r, t) dV (V = all of space) V Z ∞ Z 2π Z ∞ δ(φ − φ0 ) ρs (ρ, z, t) ρ dρ dφ dz = ρ −∞ 0 0 Z ∞Z ∞ Z 2π = ρs (ρ, z, t) dρ dz δ(φ − φ0 ) dφ −∞ 0 0 Z ∞Z ∞ = ρs (ρ, z, t) dρ dz. −∞

0

1.5. (a) Write J = ρ̂Jρ (ρ, t).

Apply the continuity equation to the stationary cylinder ρ = a, 0 ≤ z ≤ L: Z Z Z Z d d L 2π a πLa4 dQ = ρ dV = ρ0 ρ2 e−βt ρ dρ dφ dz = −β ρ0 e−βt , dt dt dt 0 0 2 0 I Z L Z 2π dQ = − J · dS = − Jρ (a, t)ρ̂ · ρ̂a dφ dz = −2πLaJρ (a, t), dt S 0 0 Jρ (a, t) = ρ0

∴

βa3 −βt e .

4

Hence

βρ3 −βt e .

4 Equating this to volume charge density times velocity, J(ρ, t) = ρ̂ρ0

ρ̂ρ0

βρ3 −βt e = ρ0 ρ2 e−βt v, 4

we obtain v = ρ̂

βρ .

4

(b) Satisfaction of the first equation follows from 4

1 ∂ ρ −βt 1 ∂ ∇·J= (ρJρ ) = β e ρ0 = βρ0 ρ2 e−βt , ρ ∂ρ ρ ∂ρ 4 and

∂ρ ∂ 2 −βt = ρ0 ρ e = −βρ0 ρ2 e−βt . ∂t ∂t For the second equation, we first calculate 2 1 ∂ ρ β 1 ∂ (ρvρ ) = ∇·v = β = . ρ ∂ρ ρ ∂ρ 4 2 2

But

Dρ ∂ρ = + v · ∇ρ where Dt ∂t

so

∇ρ = ρ̂

∂ 2 −βt ρ0 ρ e = 2ρ̂ρ0 ρe−βt , ∂ρ

Dρ βρ β = −βρ0 ρ2 e−βt + ρ̂ · 2ρ̂ρ0 ρe−βt = − ρ0 ρ2 e−βt Dt 4 2

as desired.

1.6. (a) Z 2π Z π Z 2 Q= 0

0

4r2 cos2 θ δ(θ − π/4)r2 sin θ dr dθ dφ = 56.9 C.

0

(b) Z 2 Z 2π Z ∞ Q= 0

0

4 cos2 φ δ(ρ − 2)ρ dρ dφ dz = 50.3 C.

0

(c) Z 2Z ∞Z ∞ Q= 0

0

4z 3 δ(x)δ(y) dx dy dz = 16 C.

0

(d) Z 2 Z 2π Z ∞

Z Q=

4ρzδ(ρ − 3)ρ dρ dφ dz = 144π C.

ρ dV = 0

(e)

5(z + 2)δ(z)δ(ρ − 3)ρ dρ dφ dz = 30π C.

ρ dV = −∞

1.7.

0

Z ∞ Z πZ ∞

Z Q=

0

0

0

∂ρ = −16x2 e−4t , ∂t

∇ρ = x̂8xe−4t .

∂ρ dρ = + v · ∇ρ dt ∂t = −16x2 e−4t + x̂Ax · x̂8xe−4t = 8x2 e−4t (A − 2).

We have dρ/dt = 0 when A = 2.

1.8.

∂ρ = −4βr2 e−βt , ∂t

∇ρ = r̂8re−βt .

dρ ∂ρ = + v · ∇ρ = −4βr2 e−βt + r̂2r · r̂8re−βt = 4r2 (4 − β)e−βt . dt ∂t We have dρ/dt = 0 when β = 4.

3

Chapter 2 2.1. Start with (2.99):

1 P̄ − 0 Ī · E + L̄ · B. P= c

(1)

Put H = µB0 − M:

1 P= P̄ − 0 Ī · E + L̄ · [µ0 H + µ0 M], c

1 P̄ − 0 Ī · E + [µ0 L̄] · H = P − [µ0 L̄] · M. c

(*)

Next examine (2.100):

1 M = −M̄ · E − cQ̄ − Ī · B. µ0

(2)

Put H = µB0 − M:

1 M = −M̄ · E − cQ̄ − Ī · [µ0 H + µ0 M], µ0 M̄ · E + (µ0 cQ̄ − Ī) · H = −M − (µ0 cQ̄ − Ī) · M. Equations (*) and (**) are the constitutive relations between (E, H) and (P, M).

2.2. Examine the transformed Gauss’ law (2.77): γ

∂Dx v ∂Dx ∂Dy ∂Dz −γ 2 + + = ρ. ∂x0 c ∂t0 ∂y 0 ∂z 0

Substitute from (2.74): ∂Hy ∂Dx ∂Dx ∂Hz − = Jx − γv +γ 0 .

0 0 0 ∂y ∂z ∂x ∂t This gives ∂Dx v ∂Hz v ∂Hy v v 2 ∂Dx ∂Dy ∂Dz − + + J − γ + + = ρ, x ∂x0 c2 ∂y 0 c2 ∂z 0 c2 c2 ∂x0 ∂y 0 ∂z 0

∂ v2 ∂ v ∂ v v γDx 1 − 2 + 0 Dy − 2 Hz + 0 Dz + 2 Hy = ρ − 2 Jx , ∂x0 c ∂y c ∂z c c

∂ ∂ v ∂ v v D + γ D − H + γ D + H x y z z y = ργ − γ 2 Jx , 0 0 2 0 2 ∂x ∂y c ∂z c c ∂ 0 ∂ ∂ v Dx + 0 Dy0 + 0 Dz0 = ργ − γ 2 Jx , 0 ∂x ∂y ∂z c γ

or

v ∇0 · D0 = ρ0 = ργ − γ 2 Jx . c

So cρ0 = γ(cρ − βJx ) where β = v/c.

Generalizing, we obtain cρ0 = γ(cρ − β · J).

4

(**)

2.3. The solutions all follow Example 2.4.

2.4. (a) Examine the inequality c2 B 2 > E 2 . We have 2 .

E 2 = Ek2 + E⊥

2 , B 2 = Bk2 + B⊥

In a moving frame we have the following.

(B 0 )2 = B0 · B0 = B0k · B0k + B0⊥ · B0⊥ , B0k = Bk ,

B0⊥ =

So 2 B0⊥ · B0⊥ = γ 2 B⊥ −2

γ (cB⊥ − β × E⊥ ). c

γ2 γ2 2 , B⊥ · (β × E⊥ ) + 2 β 2 E⊥ c c

Also, (E 0 )2 = E0 · E0 = E0k · E0k + E0⊥ · E0⊥ , E0k = Ek ,

E0⊥ = γ(E⊥ + β × cB⊥ ).

So 2 2 . + 2γ 2 cE⊥ · (β × B⊥ ) + γ 2 c2 β 2 B⊥ E0⊥ · E0⊥ = γ 2 E⊥

Thus, the inequality is for a moving frame 2 2 2 2 . + 2γ 2 cE⊥ · (β × B⊥ ) + γ 2 c2 β 2 B⊥ > Ek2 + γ 2 E⊥ − 2γcB⊥ · (β × E⊥ ) + γ 2 β 2 E⊥ c2 Bk2 + c2 γ 2 B⊥

Use E⊥ · (β × B⊥ ) = −B⊥ · (β × E⊥ ) to write 2 2 (1 − β 2 ).

(1 − β 2 ) > Ek2 + γ 2 E⊥ c2 Bk2 + c2 γ 2 B⊥

Since 1 − β 2 = 1/γ 2 , 2 2 , ) > Ek2 + E⊥ c2 (Bk2 + B⊥

or c2 B 2 > E 2 , which is true, so c(B 0 )2 > (E 0 )2 . (b) Similar to part (a).

2.5. (a) Setting E · B = (Ek + E⊥ ) · (Bk + B⊥ ) = Ek Bk + E⊥ · B⊥ = 0, we get Ek Bk = −E⊥ · B⊥

(1)

2 2 cB 2 > E 2 =⇒ c(Bk2 + B⊥ ) > Ek2 + E⊥ .

(2)

We also have From Problem 2.4 we know that 2

2

2

2

0 0 c(Bk0 + B⊥ ) > Ek0 + E⊥ .

0 2 > 0 and (2) is true. To have E0 = 0 we need Thus if E0 = 0 then Bk0 2 + B⊥

5

(A) E0k = Ek = 0 =⇒ Ek = 0.

(B) E0⊥ = γ(E⊥ + β × cB⊥ ) = 0.

Note that (B) is satisfied if β × cB⊥ = −E⊥ for some β. Examine 2.

(C) E⊥ · (β × cB⊥ ) = cβ · (E⊥ × B⊥ ) = E⊥

From (2), since Ek = 0, E⊥ · B⊥ = 0 =⇒ E⊥ ⊥ B⊥ ⊥ β.

So β is in the direction of E⊥ × B⊥ and has magnitude E⊥ /cB⊥ from (C). So β=

E⊥ E⊥ × B⊥ 1 · = 2 (E⊥ × B⊥ ) cB⊥ E⊥ B⊥ cB⊥

will make E0 = 0 if E · B = 0 and c2 B 2 > E 2 . (b) If c2 B 2 < E 2 , then (1) is satisfied by Bk = 0 and we need B0⊥ = 0 to have B0 = 0. Note that B0⊥ = 0 =⇒ cB⊥ − β × E⊥ = 0 =⇒ cB⊥ = β × E⊥ .

Examine B⊥ · (cB⊥ ) = B⊥ · (β × E⊥ ) = β · (E⊥ × B⊥ ).

So β is in the direction E⊥ × B⊥ and has magnitude β= Thus β=

2 cB⊥ cB⊥ . = E⊥ B⊥ E⊥

cB⊥ E⊥ × B⊥ c · = 2 (E⊥ × B⊥ ) E⊥ E⊥ B⊥ E⊥

will make B0 = 0 if E · B = 0 and c2 B 2 > E 2 .

2.6. Lab frame: F = QE = Q(Ek + E⊥ ).

Inertial frame: F0 = QE0 + Qv × B0 .

Use E0 = E0k + E0⊥ = Ek + γ(E⊥ + β × cB⊥ ), γ B0 = B0k + B0⊥ = Bk + (cB⊥ − β × E⊥ ), c to get γ F0 = QEk + QγE⊥ + Qγβ × cB⊥ + Qv × Bk + Qγv × B⊥ − Q v × (β × E⊥ ). c Now use

v × (β × E⊥ ) = β × (β × E⊥ ) = β(β · E⊥ ) − E⊥ (β · β) = −β 2 E⊥ c 6

to write F0 = QEk + QγE⊥ (1 + β 2 ) + 2Qγv × B⊥ + Qv × Bk

1 + 2Qγv × B⊥ + Qv × Bk . = Q(Ek + E⊥ ) + QγE⊥ 1 + β 2 − γ Therefore

h i p F0 = F + Q γE⊥ (1 + β 2 − 1 − β 2 ) + 2γv × B⊥ + v × Bk .

2.7. We have D0 = 0 E0 ,

B0 = µ0 H0 ,

γ β × H⊥ = 0 Ek + 0 γE⊥ + 0 γcβ × B⊥ , c γ Bk + γB⊥ − β × E⊥ = µ0 Hk + µ0 γH⊥ − µ0 γcβ × D⊥ . c Dk + γD⊥ +

Cross β into (1): β × Dk + γβ × D⊥ +

γ β × (β × H⊥ ) = 0 β × Ek + 0 γβ × E⊥ + 0 γcβ × (β × B⊥ ). c

Use β × (β × A⊥ ) = β(β · A⊥ ) − A⊥ β · β = −β 2 A⊥ to write γβ × D⊥ =

γ 2 β H⊥ + 0 γβ × E⊥ − 0 γcβ 2 B⊥ . c

Substituting into (2), i hγ γ 0 0 0 2 0 0 2 Bk + γB⊥ − β × E⊥ = µ Hk + µ γH⊥ − µ c β H⊥ + γβ × E⊥ − γcβ B⊥ , c c Bk + γB⊥ − µ0 0 γc2 β 2 B⊥ =

γ β × E⊥ − µ0 0 cγβ × E⊥ + µ0 Hk − µ0 γH⊥ + γµ0 β 2 H⊥ . c

We attempt to write this as B = Bk + B⊥ = AH + BHk + Cβ × E. Equate k components to get Bk = µ0 Hk .

Equate ⊥ components to get γB⊥ (1 − β 2 n2 ) = or B⊥ =

γ (1 − n2 )β × E⊥ − µ0 γ(β 2 − 1)H⊥ c

2 1 1 − n2 0 β −1 β × E − µ H⊥ . ⊥ c 1 − β 2 n2 1 − β 2 n2

7

(1) (2)

So 1 − β2 1 1 − n2 H⊥ + β × E⊥ 2 2 1−β n c 1 − β 2 n2 = AH + BHk + Cβ × E

B = Bk + B⊥ = µ0 Hk + µ0

= A(Hk + H⊥ ) + BHk + Cβ × (Ek + E⊥ ) = (A + B)Hk + AH⊥ + Cβ × E⊥ .

We have A + B = µ0 ,

A = µ0

1 − β2 , 1 − β 2 n2

hence

C=

B = µ0 − A = µ0 β 2

1 − n2 .

1 − β 2 n2

Hk = β̂(β̂ · H) =

ββ ·H β2

1 1 − n2 , c 1 − β 2 n2

Writing

we get B = AH + BHk + Cβ × E

B ββ 0 A Ī + 0 2 · H − Ω × E (Ω = Cβ) =µ µ0 µ β

n2 − 1 1 − β2 n2 − 1 β Ī − = µ0 ββ · H − ×E 1 − β 2 n2 1 − β 2 n2 1 − n2 β 2 c = µ0 Ā · H − Ω × E where

1 − β2 n2 − 1 Ā = Ī − ββ 1 − β 2 n2 1 − β2

and Ω=

n2 − 1 β .

1 − n2 β 2 c

By following similar steps we can show that D = 0 Ā · E + Ω × H. 2.8. Start with D0 = 0 E0 and B0 = µ0 H0 . Substitute (2.44)–(2.46) to get v×H = 0 (E + v × B), c2 v×E B− = µ0 (H − v × D). c2 D+

Cross v into (1): v×D=−

v × (v × H) + 0 v × E + 0 v × (v × B). c2 8

(1) (2)

Substitute this into (2):

v × (v × H) v×E 0 0 − µ0 0 v × E − µ0 0 v × (v × B) B= +µH−µ − c2 c2

v × (v × H) v × (v × B) v×E 0 0 2 0 − c2 µ0 0 (1 − µ c ) + µ H + . = 2 2 c c c2 Since n2 = c2 µ0 0 , we have v × (v × B) B + n2 = µ0 c2 Note that

v × (v × H) v×E 2 H+ − (n − 1).

2 c c2

v × (v × B) v2 |B| |B| ∼ c2 c2

and

v2 v × (v × E) |E| |E| ∼ c2 c2

since v 2 /c2 1. Therefore B = µ0 H − (n2 − 1)

v×E . c2

2.9.

B0 − H0 µ0 Bk 1 γ + (cB⊥ − β × E⊥ ) − Hk − γ(H⊥ − β × cD⊥ ). = µ0 µ0 c

M0 =

Equate k components: M0k =

Bk − Hk . µ0

M=

B − H, µ0

But

so M0k = Mk .

Equate ⊥ components: 1 γ (cB⊥ − β × E⊥ ) − γ(H⊥ − β × cD⊥ ) µ0 c

1 γ = γB⊥ − γH⊥ + β × γcD⊥ − E⊥ µ0 µ0 c

B⊥ 1 =γ − H⊥ + γcβ × D⊥ − E⊥ µ0 µ0 c2

B⊥ =γ − H⊥ + γcβ × (D⊥ − 0 E⊥ ). µ0

M0⊥ =

9

But P⊥ = D⊥ − 0 E⊥ , so M0⊥ = γM⊥ + γcβ × P⊥ = γ(M⊥ + β × cP⊥ ).

2.10. (a) Into D0 = 0 E0 + P0 , substitute (2.44) and (2.45): D+v×

H = 0 (E + v × B) + P0 . c2

Next use D = 0 E + P and the fact that B = µ0 (H + M) =⇒ v × B = µ0 v × H + µ0 v × M to get 0 E + P +

v×H = 0 E + 0 µ0 v × H + 0 µ0 v × M + P0 . c2

Therefore P0 =

v×M + P. c2

(b) Into B0 = µ0 (H0 + M0 ), substitute H0 = H − v × D and B0 = B − v × E/c2 : B−

v×E = µ0 (H − v × D) + µ0 M0 . c2

Next use B = µ0 (H + M) and the fact that D = 0 E + P =⇒ v × D = 0 v × E + v × P to get µ0 H + µ0 M −

v×E = µ0 H − µ0 ( 0 v × E + v × P) + µ0 M0 . c2

Therefore µ0 M = −µ0 v × P + µ0 M0 as desired.

2.11. Start with (2.114) and (2.115): P = P0 +

v×M , c2

M = M0 − v × P.

Then v × P = v × P0 +

v × (v × M) c2

and we have M = M0 − v × P0 − or M+

v × (v × M) c2

v2 v̂ × (v̂ × M) = M0 − v × P0 . c2 10

But

v2 v̂ × (v̂ × M) |M| c2

because v 2 /c2 1, so M = M0 − v × P0 .

Now use P0 = 0 χ0e E0 , to write M=

M0 =

χ0m 0 B, µ0 µ0r

χ0m 0 B − 0 χ0e v × E0 . µ0 µ0r

By (2.148) and (2.147) we have B0 = B −

v×E , c2

E0 = E + v × B,

so

v×E χ0m B− − 0 χ0e v × (E + v × B) M= µ0 µ0r c2 0

χm χ0m 0 0 =B − 0 χe v × (v × B) − + 0 χe v × E µ0 µ0r c2 µ0 µ0r

0

χm 1 χ0 0 v × (v × B) 0 − χ −

+ χ = B m 0 e e v × E. µ0 µ0r c2 µ0r But χ0e so

2 v × (v × B) χ0m 0 v v̂ × (v̂ × B) B , =

χ e 2 c2 c µ0r

χ0 M = m0 B − 0 µ0 µ r

χ0m 0 + χe v × E. µ0r

2.12. Start with Pk = P0k .

P = Pk + P⊥ , Then

cP0⊥ = γ(cP⊥ − β × M⊥ ) =⇒ P⊥ = so P = P0k +

1 0 1 P⊥ + β × M⊥ , γ c

1 0 1 P + β × M⊥ . γ ⊥ c

Use (2.150), M⊥ =

1 0 M − β × cP⊥ , γ ⊥

to get 1 1 P = P0k + P0⊥ + β × γ

c

11

1 0 M − β × cP⊥ . γ ⊥

Now rearrange, P + β × (β × P⊥ ) = P0k +

1 0 1 P + β × M0⊥ , γ ⊥ cγ

and use the fact that β × (β × P⊥ ) = −β 2 P⊥ to get P − β 2 P⊥ = P0k +

1 1 β × M0⊥ + P0⊥ . cγ γ

Equating k terms, we get Pk = P0k .

Equating ⊥ terms, we get (1 − β 2 )P⊥ =

1 1 β × M0⊥ + P0⊥ . cγ γ

Next use χ0m 0 χm γ B = [cB⊥ − β × E⊥ ], µ0 µ0r ⊥ µ0 µ0r c P0⊥ = 0 χ0e E0⊥ = 0 χ0e γ[E⊥ + β × cB⊥ ],

M0⊥ =

to get 1 γ χ0m 1 γ χ0m cβ × B − β × (β × E⊥ ) + 0 χ0e E⊥ + 0 χ0e β × cB⊥ ⊥ cγ c µ0 µ0r cγ c µ0 µ0r

χ0m 1 χ0m 0 2 0 + E⊥ 0 χe + 2 β = (β × B⊥ ) c 0 χe + c µ0 µ0r c µ0 µ0r

χ0m χ0m 0 0 2 = (β × cB⊥ ) 0 χe + 2 + E⊥ 0 χe + 2 β c µ0 µ0r c µ0 µ0r

χ0 χ0m 2 0 = (β × cB⊥ ) 0 χ0e + m + E

χ + β ⊥ 0 e µ0r µ0r

(1 − β 2 )P⊥ =

so that

(β × cB⊥ ) E⊥ χ0m χ0m 2 0 0 P⊥ = 0 χe + 0 + 0 χe + 0 β .

1 − β2 µr 1 − β2 µr

Now Pk = P0k = 0 χ0e E0k = 0 χ0e Ek = 0 χ0e (E − E⊥ ).

P = Pk + P⊥ , Thus, P = 0 χ0e E + E⊥

0 χe

1 χ0m 2 (β × cB⊥ ) χ0m 0 −1 + 0 β + 0 χe + 0 .

1 − β2 µr 1 − β2 µr

If v/c 1, then β 2 1 and P = 0 χ0e E +

v c

× cB⊥

χ0m 0 0 χe + 0 . µr

Use v × B⊥ = v × (B − Bk ) = v × B since v × Bk = 0.

Obtain P = 0 χ0e E + 0

χ0 χ0e + m µ0r

A similar set of steps can be used to find M. 12

v × B.

2.13. Start with (2.102): ∇×E=−

∂B . ∂t

Integrate over a surface S: Z

Z (∇ × E) · dS = − S

∂B · dS. S ∂t

Use (2.127): Z

∂B d · dS = dt S ∂t

Z

Z

[v(∇ · B) − ∇ × (v × B)] · dS.

B · dS − S

S

But ∇ · B = 0 from (2.96), so Z Z Z d (∇ × E) · dS = − B · dS − [∇ × (v × B)] · dS. dt S S S Use Stokes’s theorem:

I E · dl = − Γ

or

d dt

Z

I B · dS − Γ

d [E + v × B] · dl = − dt Γ

Z

I

Therefore

I

(v × B) · dl,

S

Z

d E · dl = − dt Γ ∗

B · dS

where

B · dS. S

E∗ = E + v × B.

S

Next, start with (2.103): ∇×

B ∂E = J + JM + JP + 0 . µ0 ∂t

Integrate over a surface S and use Stokes’s theorem: I Z Z 1 ∂E B · dl = [J + JM + JP ] · dS + 0 · dS. µ0 Γ S S ∂t Use (2.127): Z

∂E d · dS = dt S ∂t

Z

Z E · dS −

S

[v(∇ · E) − ∇ × (v × E)] · dS. S

But ∇·E=

ρ + ρP 0

from (2.104) so I Z Z I Z 1 d B · dl = 0 E · dS − (ρ + ρP )v · dS + 0 (v × E) · dl + [J + JM + Jp ] · dS. µ0 Γ dt S S γ S So

or

I

Z Z d [B − µ0 0 v × E] · dl = µ0 0 E · dS + µ0 [J + JM + Jp ] · dS, dt S Γ S I Z Z 1 d E · dS B+ · dl = µ0 J+ · dS + 2 c dt S Γ S 13

where J+ = J + JM + JP − (ρ + ρP )v

B+ = B −

and

1 v × E. c2

Next, start with (2.135): Z

I (n̂ × E) dS = − S

∂B dV. V ∂t

Use (2.132): Z

d ∂B dV = dt V ∂t

So

Z

I B dV −

B(v · n̂) dS.

V

S

Z I d B dV + B(v · n̂) dS, (n̂ × E) dS = − dt V S S I Z d [(n̂ × E) − (v · n̂)B] dS = − B dV. dt V S

I

or

Finally, start with (2.136): I Z Z ∂E (n̂ × B) dS = µ0 [J + JM + JP ] dV + µ0 0 dV. S V V ∂t Use (2.132): Z

∂E d dV = dt V ∂t

So

I E dV −

V

E(v · n̂) dS. S

Z 1 d E dV, c2 dt V S V S

I Z Z 1 1 d (n̂ × B) + 2 (v · n̂)E dS = µ0 [J + JM + JP ] dV + 2 E dV. c c dt V S V

I

(n̂ × B) dS +

or

Z

1 c2

I

Z

E(v · n̂) dS = µ0

[J + JM + JP ] dV +

2.14. Start by integrating (2.141) over a time-changing surface to get Z Z Z ∂B (∇ × E) · dS = − Jm · dS − · dS. S S S ∂t Use (2.127): Z

d ∂B · dS = dt S ∂t

Z

Z B · dS −

S

Z v(∇ · B) · dS +

S

∇ × (v × B) · dS S

where ∇ · B = ρm by (2.143). Hence, by Stokes’s theorem, Z Z Z I ∂B d · dS = B · dS − ρm v · dS + (v × B) · dl. dt S S ∂t S Γ Substitute this into (A) and use Stokes’s theorem again: I Z Z Z I d B · dS + ρm v · dS − (v × B) · dl.

E · dl = − Jm · dS − dt S Γ S S Γ 14

(A)

So

I (E + v × B) · dl = − Γ

d dt

Z

Z

(Jm − ρm v) · dS.

B · dS − S

S

Hence (2.37) is changed by the addition of the third term. Since (2.142) contains no magnetic source terms, (2.138) is unchanged.

Next, integrate (2.141) over a time-changing volume: Z Z Z ∂B Jm dV − (∇ × E) dV = − dV. V ∂t V V Apply the curl theorem: I

Z (n̂ × E) dS = −

S

Z Jm dV −

V

∂B dV. V ∂t

(B)

Use (2.132): Z

d ∂B dV = dt V ∂t

Z

I B dV −

V

B(v · n̂) dS. S

So (B) becomes I

Z

d (n̂ × E) dS = − Jm dV − dt S V

or

I

Z

I B(v · n̂) dS

B dV + V

Z

S

d [n̂ × E − (v · n̂)B] dS = − Jm dV − dt S V

Z B dV. V

Thus, (2.139) is modified by the addition of the first term on the right-hand side. Since (2.103) contains no magnetic sources, (2.140) is unchanged.

2.15. Consider Figure 2.5, where the sources are magnetic charge ρm and magnetic current Jm .

Integrate (2.142) over V , with J = 0: Z Z Z Z ∂D n̂1 × H1 dS + n̂2 × H2 dS + n̂3 × H dS = dV. S1 S2 S3 V ∂t Choose δ = k∆ so that most of the source lies within V . As ∆ → 0, we have S1 = S2 , n̂1 = −n̂2 = n̂12 , and V → 0. Thus Z ∂D dV → 0 V ∂t and

Z n̂12 × (H1 − H2 ) dS = 0 S1

or n̂12 × (H1 − H2 ) = 0.

Next integrate (2.141) over V : Z Z n̂1 × E1 dS + S1

Z ∂B n̂2 × E2 dS + n̂3 × E dS = −Jm − dV. ∂t S2 S3 V Z

15

Choose δ = k∆ so that most of the source lies within V . As ∆ → 0, we have S1 = S2 , n̂1 = −n̂2 = n̂12 , and V → 0. So Z Z Jm dV n̂12 × (E1 − E2 ) dS = − V

S1

Z δ/2

Z =−

Jm dS dx S1 −δ/2 Z δ/2

=−

Z Jms dS.

f (x, ∆) dx −δ/2

But

S1

Z δ/2 lim

∆→0 −δ/2

so

f (x, ∆) dx = 1,

Z

Z n̂12 × (E1 − E2 ) dS = −

Jms dS

S1

S1

or n̂12 × (E1 − E2 ) = −Jms .

Next integrate (2.143) over V : Z Z B1 · dS + S1

Z

Z

B2 · dS +

S2

B · dS =

ρm dV.

S3

V

As ∆ → 0, we have S1 = S2 , n̂1 = −n̂2 = n̂12 . Therefore Z Z (B1 − B2 ) · n̂12 dS = ρm dV S1

V

Z

Z δ/2

=

ρm dS dx S1 −δ/2 Z δ/2

=

Z

f (x, ∆) dx −δ/2

ρms dS S1

Z =

ρms dS S1

and we obtain (B1 − B2 ) · n̂12 = ρms .

Integration of (2.144) with ρ = 0 gives (D1 − D2 ) · n̂12 = 0.

2.16. Consider a surface carrying an electric current Js as shown in Figure 2.6. We begin with (2.148) where V is the volume V1 shown in Figure 2.6. Since V1 does not contain Js , we have Z Z Z Z d [n̂ × H − (v · n̂)D] dS + [n̂ × H + (v · n̂)D] dS = J dV + D dV. (A) dt V1 S1 S10 V1 16

Apply the same formula to the region V2 : Z Z Z [n̂ × H + (v · n̂)D] dS = [n̂ × H − (v · n̂)D] dS +

d J dV + dt V2

S20

S2

Z D dV.

(B)

V2

Add (A) and (B): Z Z [n̂ × H − (v · n̂)D] dS −

Z Z d D dV − [n̂10 × H + (v · n̂10 )D] dS J dV − dt V1 +V2 S10 V1 +V2 Z − [n̂20 × H + (v · n̂20 )D] dS = 0.

S1 +S2

S20

If δ is very small, then S1 + S2 = S, V1 + V2 = V , and n̂10 = −n̂20 = n̂12 . Then Z Z Z Z Z d [n̂×H+(v· n̂)D] dS − J dV − (v· n̂12 )(D1 −D2 ) dS. D dV = n̂12 ×(H1 −H2 ) dS + dt V V S S10 S10 (C) Now apply (2.148) to the volume region V that intersects the surface. This region does contain the surface current Js .

Z Z Z Z d [n̂ × H + (v · n̂)D] dS − J dV − D dV = Js dS. (D) dt V S V S10 Consistency between (C) and (D) requires that n̂12 × (H1 − H2 ) + (v · n̂12 )(D1 − D2 ) = Js .

This is the desired boundary condition. Application of (2.147) to a surface containing a magnetic surface current Jm will, through an analogous set of steps, lead to the second boundary condition.

2.17. Start with (2.252) and (2.253):

∂ ∂ ¯ ∇ + ζ̄ · E = − µ̄ · H − Jm , ∂t ∂t

∂ ∂ ¯ ∇ − ξ̄ · H = ¯ · E + J. ∂t ∂t

Define a new dyadic ¯ −1 by ¯ −1 · ¯ = ¯ · ¯ −1 = Ī.

Then (2.252) becomes

∂E ∂ −1 ¯ = ¯ · ∇ − ξ̄ · H − ¯ −1 · J. ∂t ∂t Next, differentiate (2.252) with respect to time:

∂ ∂E ∂2 ∂Jm ¯ ∇ + ζ̄ · = − 2 (µ̄ · H) − . ∂t ∂t ∂t ∂t Lastly, substitute (*) to get

∂ ∂ ∂2 ∂ ∂Jm −1 ¯ ¯ ¯ ∇ + ζ̄ · ¯ · ∇ − ξ̄ + 2 µ̄ · H = ∇ + ξ̄ · ¯ −1 · J − . ∂t ∂t ∂t ∂t ∂t For an anisotropic medium we have B = µ̄ · H and D = ¯ · E, so ξ̄ = ζ̄ = 0. Then we get

∂2 −1 ¯ ¯ ¯ · ¯ −1 · J − ∂Jm . ∇ · ¯ · ∇ + 2 µ̄ · H = ∇ ∂t ∂t 17

(*)

2.18. Let D(r, t) = (r)E(r, t),

B(r, t) = µ(r)H(r, t).

Maxwell’s equations: ∇ × E = −µ

∂H , ∂t

∇×H=J+

∂E , ∂t

∇ · ( E) = ρ.

Curl of Faraday’s law: ∇ × (∇ × E) = −

∂ ∂ ∇ × (µH) = − [µ∇ × H − H × ∇µ] ∂t ∂t

∂ ∂E ∇(∇ · E) − ∇ E = − µ J+ − H × ∇µ ∂t ∂t ∂J ∂ 2 E ∂H = −µ − µ 2 + × ∇µ ∂t ∂t ∂t ∂2E 1 ∂J − µ 2 − (∇ × E) × ∇µ = −µ ∂t ∂t µ 2

∇ · ( E) = ρ

∇ · E + E · ∇ = ρ ρ ∇ ∇·E= −E·

=⇒ =⇒

∂J ∂2E ∇µ − µ 2 − (∇ × E) × ∂t ∂t µ

∇ ∇µ ∂J ρ ∂2E − (∇ × E) × =µ +∇ ∇2 E − µ 2 + ∇ E · ∂t

µ ∂t

∴

∇

ρ ∇ −E·

− ∇2 E = −µ

2.19. We have from Ampere’s law and Faraday’s law ∂Ey ∂Hx = 0 , ∂z ∂t ∂Ey ∂Hx = µ(z) . ∂z ∂t

(*) (**)

We differentiate (*) with respect to t and (**) with respect to z to give, ∂ 2 Ey ∂ 2 Hx = 0 , ∂z∂t ∂t2

∂ 2 Ey ∂ 2 Hx ∂µ(z) ∂Hx = µ(z) + . ∂z 2 ∂z∂t ∂z ∂t

Substituting for ∂ 2 Hx /∂z∂t and using (**) gives the wave equation for Ey : ∂ 2 Ey ∂ 2 Ey 1 ∂µ(z) ∂Ey − − µ(z) = 0.

0 ∂z 2 µ(z) ∂z ∂z ∂t2 To find the wave equation for Hx we differentiate (**) with respect to t and (*) with respect to z to give, ∂ 2 Ey ∂ 2 Ey ∂ 2 Hx ∂ 2 Hx =

, = µ(z) .

0 ∂z 2 ∂z∂t ∂z∂t ∂t2 18

Substituting for ∂ 2 Ey /∂z∂t gives ∂ 2 Hx ∂ 2 Hx − µ(z) = 0, 0 ∂z 2 ∂t2 which is the desired wave equation.

2.20. For µ(r) = µ0 and (r) = (z) we have ∇µ = 0, Setting J = 0 and ρ = 0 we get

∂2E 1 ∂ 2 ∇ E − µ 2 + ∇ E · ẑ = 0, ∂t ∂z

∇ = x̂

∂ . ∂z

∂2H 1 ∂ ∇ H − µ 2 − (∇ × H) × ẑ = 0. ∂t ∂z 2

Now use

1 ∂ 1 ∂ ∇ E · ẑ = ∇ Ez ∂z ∂z

1 ∂ 1 ∂ + ∇Ez = Ez ∇ ∂z ∂z

∂ 1 1 ∂ 1 ∂ + ∇ + ∇Ez = Ez ∇ ∂z ∂z ∂z " # 1 ∂ 1 ∂2 1 ∂ 2 + = ẑEz − ∇Ez ∂z 2 2 ∂z ∂z to get the wave equations " # ∂2E 1 ∂2 1 ∂ 2 1 ∂ ∇ E − µ 2 + ẑEz − 2 + ∇Ez = 0, 2 ∂t ∂z

∂z ∂z 2

∇2 H − µ

∂ 2 H 1 ∂ + ẑ × (∇ × H) = 0. ∂t2 ∂z

If E(r, t) = ŷEy (z, t) and H(r, t) = x̂Hx (z, t) then

∂Hx ∂Hx ẑ × (∇ × H) = ẑ × ŷ = −x̂ , ∂z ∂z

∇Ez = 0, and

∇2 E = ŷ

∂ 2 Ey , ∂z 2

∇2 H = x̂

∂ 2 Hx . ∂z 2

With these the wave equations become ∂ 2 Ey ∂ 2 Ey − ŷµ = 0, ∂z 2 ∂t2 ∂ 2 Hx 1 ∂ ∂Hx ∂ 2 Hx x̂ − x̂µ − x̂ = 0, ∂z 2 ∂t2 ∂z ∂z ŷ

which are identical to (2.247) and (2.245) from Example 2.7.

19

2.21. We want to show that (2.265) ∇2 E − µ0 0

∂ ∂2E 1 = ∇(ρ + ρP ) + µ0 (J + JM + Jp ) 2 ∂t 0 ∂t

is equivalent to (2.259) ∂2E ∂J = −∇ × Jm − µ ∂t2 ∂t when D = E, B = µH, and Jm = 0. The equivalent sources are ∇ × (∇ × E) + µ

Jp =

∂P , ∂t

JM = ∇ × M,

ρP = −∇ · P,

where P = D − 0 E, Therefore JP = and

∂ ( E − 0 E) = 0 ∂t

JM = ∇ ×

M=

B − H. µ0

∂E ∂E − 0 ∂t ∂t

µ µ H −∇×H= ∇×H−∇×H µ0 µ0

since the material is homogeneous. Also, ρP = −∇ · D + 0 ∇ · E. Substitute the equivalent sources into (2.265):

∂2E 1 ∂ µ ∂E ∂E ∇ E − µ0 0 2 = ∇(ρ − ∇ · D + 0 ∇ · E) + µ0 J+ ∇×H−∇×H+ − 0 ∂t 0 ∂t µ0 ∂t ∂t

∂ µ ∂E ∂E ∂E µ ∂D = ∇(∇ · E) + µ0 J+ J+ −J− + − 0 ∂t µ0 µ0 ∂t ∂t ∂t ∂t 2 2 ∂ E ∂J ∂ D = ∇(∇ · E) − µ0 0 2 + µ +µ 2 . ∂t ∂t ∂t 2

So −∇2 E + ∇(∇ · E) + µ or ∇ × (∇ × E) + µ

∂2E ∂J = −µ 2 ∂t ∂t

∂2E ∂J = −µ 2 ∂t ∂t

which is (2.259).

Next we substitute the equivalent sources into (2.266) and use

B µ µ ∂D ∇× = ∇×H= J+ µ0 µ0 µ0 ∂t

20

to get

∂E ∂2B 1 ∂D − 0 ∇ B − µ0 0 2 = −µ0 ∇ × J + ∇ × B − ∇ × H + ∂t µ0 ∂t ∂t

µ µ ∂E ∂D ∂D ∂E = −µ0 ∇ × J + J + −J− + − 0 µ0 µ0 ∂t ∂t ∂t ∂t ∂ ∂ = −µ∇ × J − µ (∇ × E) − µ0 0 (∇ × E). ∂t ∂t 2

So ∇2 B = −µ∇ × J + µ

∂2B , ∂t2

∇ × ∇ × B = µ∇ × J − µ ∇ × ∇ × H + µ

∂2B , ∂t2

∂2H = ∇ × J, ∂t2

which is (2.260).

2.22. Eliminating the third term, we obtain from (2.284) H(z, t) =

H0 − Ω z z H0 Ω z z e v f t− + ev f t + .

2 v 2 v

To plot this, let = 81 0 and µ = µ0 (which implies v = 3.33 × 107 m/s), σ = 2 × 10−4 S/m, and f (t) = rect(t/τ ) with τ = 1 µs. This gives the plot shown in Figure 1. We see that this plot is very similar to Text Figure 2.8. The pulse shape is mostly determined by the exponential factors, and these factors also affect the amplitudes of the pulses in a similar way to Text Figure 2.8. The most important contribution from the third term in (2.364) is to create the wake or “tail” seen in Text Figure 2.8.

2.23. (a) Let b = b(t) = vt be the radius of the expanding surface. The charge density is ρ(r, t) =

Q Q δ(r − b) = δ(r − vt).

4πb2 4π(vt)2

For a Gaussian surface of radius r, Gauss’s law reads ( I 0, r < b = vt, D · n̂ dS = Q, r > b = vt.

S So

( 0, r < b = vt, 4πr Dr (r, t) = = QU (r − vt) Q, r > b = vt.

2

and we have E(r, t) = r̂

Q U (r − vt).

4π 0 r2

21

2.0

t=0 t = -0.5 µs

H(z,t)/H 0

1.5

t = -8 µs t = -4 µs

1.0 0.5 0.0

-0.5 0

100

200

300

400

z-position (m)

2.0

t=0

H(z,t)/H 0

1.5

t = 0.5 µs

1.0

t = 4 µs

0.5

t = 8 µs

0.0 -0.5 0

100

200

300

400

z-position (m)

Figure 1 (b) By (2.300) we have 1 1 ∂ψ ∇ψ(r, t) = − r̂ 0 0 ∂r

1 ∂ Q vt − r = − r̂ U (r − vt) 0 ∂r 4πr vt

Q ∂ vt − r vt − r =− r̂ U (r − vt) + δ(r − vt) 4π 0 ∂r vtr vtr

Q 1 −r − vt + r =− r̂ U (r − vt) 4π 0 vt r2 Q = r̂ U (r − vt) 4π 0 r2

E(r, t) = −

for r > vt. (c) By the definition of volume current we have J = ρv =

Q Q δ(r − vt)[vr̂] = r̂ δ(r − vt).

4π(vt)2 4πvt2

22

Compare this to (2.298):

∂ψ ∂ ∂Er J=∇ = (∇ψ) = −r̂ 0 ∂t ∂t ∂t

Q ∂ U (r − vt) = −r̂ 0 ∂t 4π 0 r2 Q ∂ U (r − vt) = −r̂ 4πr2 ∂t Q = r̂ vδ(r − vt) 4πr2 Q = r̂ vδ(r − vt) 4π(vt)2 Q = r̂ δ(r − vt).

4πvt2 2.24. (a) Under the assumptions stated, Gauss’s law reads ∇ · E = 0. We therefore set ∇·E=

∂ ∂ [A(x + y) cos(ωt)] + [B(x − y) cos(ωt)] = 0 ∂x ∂y

and find that A = B. (b) Since

∂Ey ∂Ex ∂ ∂ ∇ × E = ẑ − = ẑ [B(x − y) cos(ωt)] − [A(x + y) cos(ωt)] = 0, ∂x ∂y ∂x ∂y Faraday’s law shows that B is constant.

2.25. (a) Use Faraday’s law: −µ0

π π ∂H π = ∇ × E = −x̂E0 β sin x sin(ωt − βz) + ẑE0 cos x cos(ωt − βz), ∂t a a a

and integrate with respect to t: H(r, t) = −x̂

π π E0 E0 π β sin x cos(ωt − βz) − ẑ cos x sin(ωt − βz). ωµ0 a ωµ0 a a

(b) Since J = 0, Ampere’s law reads ∇ × H = 0

∂E . ∂t

Substituting E and H from above, we find that q β = ω 2 µ0 0 −

π 2 a .

2.26. (a) D = E = ŷ30 0 x2 y 2 e−αt U (t).

23

(b) Z t D(r, t) = 0 E(r, t) + 0

χe (r, t − t0 )E(r, t0 ) dt0

−∞ 2 −αt

= ŷ10 0 x e

Z t

0

ŷ30 0 x2 y 2 U (t − t0 )U (t0 )e−αt dt0

U (t) + −∞

= ŷ10 0 x2 e−αt U (t) +

Z t

0

ŷ30 0 x2 y 2 U (t)e−αt dt0 0

e−αt 1 2 −αt 2 2 = ŷ10 0 x e U (t) + ŷ30 0 x y U (t) . − α α

(c) D = ¯ · E = (3 0 x̂x̂ + 2 0 x̂ŷ + 4 0 ẑŷ) · [ŷ10x2 e−αt U (t)] = x̂20 0 x2 e−αt U (t) + ẑ40 0 x2 e−αt U (t). (d) D = (E + β∇ × E) = 7 0 [ŷ10x2 e−αt U (t) + ẑ200xe−αt U (t)] = 70 0 xe−αt U (t) [ŷx + ẑ20].

2.27.

I

Z 2π Z π

I

∗

J · dS = S

(J − ρv) · n̂ dS = S

0

−ρ0 v0 r̂ · r̂R2 (t) sin θ dθ dφ

0

= −4πρ0 v0 R2 (t) = −4πρ0 v03 t2 .

Z 2π Z π Z R(t)

Z ρ dV = V

∴

4 4 ρ0 r2 sin θ dr dθ dφ = πR3 (t) = πv03 t3 , 3 3 0 0 0

Z d d 4 3 3 ρ dV = − πv t = −4πρ0 v03 t2 . − dt V dt 3 0

2.28. (a) ρP sa = −r̂ · r̂

31.87 × 10−12 = −7.97 × 10−8 (2 × 10−2 )2

ρP sb = r̂ · r̂

31.87 × 10−12 = 1.99 × 10−8 (4 × 10−2 )2

C/m2 , C/m2 .

(b) QP s = 4πa2 ρP sa + 4πb2 ρP sb = 0. (c) 1 ∂ 1 ∂ ρP = −∇ · P = − 2 (r2 Pr ) = − 2 (31.87 × 10−12 ) = 0. r ∂r r ∂r 24

(d) Z ρP dV = 0.

QP = V

(e) Since P = 0 ( r − 1)E, we have 31.87 × 10−12 = 0 ( r − 1)0.45

or

r = 9.

2.29. Original problem: (E1 , D1 , B1 , H1 ).

D1 = E1 + ξH1 , B1 = ξE1 + µH1 .

Dual problem: E2 = η0 H1 ,

Substitute:

B2 = −η0 D1 ,

η0 D2 = B1 ,

D2 = 2 E2 + ξ2 H2 ,

(*)

B2 = ξ2 E2 + µ2 H2 .

(**)

B2 E2 − = (−η0 H2 ) + ξ η0 η0

E2 η0 D2 = ξ(−η0 H2 ) + µ η0

Compare to (*) and (**): 2 =

η0 H2 = −E1 .

µ , η02

=⇒

B2 = η02 H2 − ξE2 ,

=⇒

D2 = −ξH2 +

ξ2 = −ξ,

µ E2 . η02

µ2 = η02 .

2.30.

D = E + ξH, B = ξE + µH. ∂ ∂E ∂H ∂D = E · ( E + ξH) = E · + ξE · , ∂t ∂t ∂t ∂t ∂ ∂E ∂H ∂B H· = H · (ξE + µH) = ξH · + µH · , ∂t ∂t ∂t ∂t E·

E·

∂D ∂B ∂E ∂H ∂E ∂H +H· = E · + ξE · + ξH · + µH · ∂t ∂t ∂t ∂t ∂t ∂t 1 ∂ ∂ 1 ∂ = E·E+ξ E·H+ µ H·H 2 ∂t ∂t 2 ∂t

∂ µ E · E + ξE · H + H · H = ∂t 2 2 ∂ = Uem . ∂t 25

Thus ∇ · Sem −

∂ Uem = −J · E ∂t

where Sem = E × H and

µ Uem = E · E + ξE · H + H · H 2 2 1 1 = E · ( E + ξH) + H · (ξE + µH) 2 2 1 = (D · E + B · H).

2 2.31.

∂ψ 11 0 z 1 1 0 z 1 z v 1 z v =− f t− + f t+ + g t+ − − g t− , ∂z 2v v 2v v v v 2 v v 2 z 1 1 00 z z v z v ∂2ψ 1 1 00 1 0 1 0 t − t + t + t − = f + f + g − g , ∂z 2 2 v2 v 2 v2 v v2 v 2 v2 v 2 z 1 0 z v ∂ψ 1 z v z = f0 t − + f t+ + g t+ − g t− , ∂t 2 v 2 v 2 v 2 v ∂2ψ 1 00 z 1 00 z v 0 z v 0 z = f t − + f t + + g t + − g t − . ∂t2 2 v 2 v 2 v 2 v The problem is finished by direct substitution.

2.32. (a) φ̂ ∂ E0 ∂B = (rEθ ) = φ̂ β sin(ωt − βr), ∇×E=− ∂t r ∂r r Z E0 β E0 B = φ̂ β sin(ωt − βr) dt = φ̂ cos(ωt − βr), r ω r H(r, t) =

B(r, t) β E0 = φ̂ cos(ωt − βr). µ0 ωµ0 r

(b) I [E(r, t) × H(r, t)] · n̂ dS

Z 2π Z π E0 β E0 = θ̂ cos(ωt − βr) × φ̂ cos(ωt − βr) · r̂ r2 sin θ dθ dφ r ωµ0 r 0 0 β = 4πE02 cos2 (ωt − βr). ωµ0

Psphere (t) =

2.33. (a) ∂E ∂Hz = −ŷ = −ŷβH0 sin(ωt − βx), ∂t ∂x β E(r, t) = ŷ H0 cos(ωt − βx). ω 0

∇ × H = 0

26

(b) ∇ × E = −µ0 so H = ẑ

∂Ey ∂H β2 sin(ωt − βx), = ẑ = ẑ ∂t ∂x ω 0

β2 H0 cos(ωt − βx) = ẑH0 cos(ωt − βx) ω 2 µ0 0

and we must have

β2 = 1. ω 2 µ0 0

So

√ β = ω µ0 0 .

2.34. See Figure 2.

Figure 2 (a) I

Z v0 t 1 1 x̂ cos ωt · x̂ dx. x̂ cos(ωt − βb) · x̂ dx + x v0 t+a x v0 t Z Z v0 t+a Z b ∂B µ0 ω − · dS = − −φ̂ sin(ωt − βz) · φ̂ dz dx. ∂t ηx x=v0 t z=0

v0 t + a both sides = [cos(ωt − βb) − cos ωt] ln . v0 t Z v0 t+a

E · dl =

(b) v × B = ẑv0 I

µ0 cos(ωt − βz). ηρ

Z b

Z b µ0 µ0 (v × B) · dl = v0 cos(ωt − βz) dz − v0 cos(ωt − βz) dz ηv0 t η(v0 t + a) 0 0

µ0 1 1 − [sin(ωt − βb) − sin ωt]. = −v0 ηβ v0 t v0 t + a

I

v0 t + a v0 t

(E + v × B) · dl = [cos(ωt − βb) − cos ωt] ln

µ0 1 1 + v0 − [sin(ωt − βb) − sin ωt]. ηβ v0 t + a v0 t 27

Z v0 t+a Z b

Z

µ0 cos(ωt − βz) · φ̂ dz dx x=v0 t z=0 ηx

µ0 v0 t + a =− [sin(ωt − βb) − sin ωt]. ln ηβ v0 t

B · dS =

d − dt

Z

φ̂

µ0 v0 t + a B · dS = ω [cos(ωt − βb) − cos ωt] ln ηβ v0 t

v0 µ0 v0 [sin(ωt − βb) − sin ωt]. + − ηβ v0 t + a v0 t

2.35.

E(r, θ, t) = θ̂

1 A(t − r/v), r sin θ

H(r, θ, t) = φ̂

1 1 A(t − r/v). η r sin θ

(a) Z π−θ0 V (r, t) = θ0

θ0 1 θ̂ · θ̂A(t − r/v)r dθ = 2A(t − r/v) ln cot . r sin θ 2

(b)

1 1 1 1 A(t − r/v) = r̂ A(t − r/v), Js = n̂ × H = θ̂ × φ̂ η r sin θ η r sin θ Z 2π 1 1 2π I(r, t) = r̂ A(t − r/v) · r̂r sin θ dφ = A(t − r/v). η r sin θ η 0

(c)

η θ0 Rc = ln cot . π 2

28

Chapter 3 3.1. The line charge is shown in Figure 3 along with source and field point position vectors r0 and r.

Figure 3 We have r = x̂x + ŷy = ρ̂ρ and r0 = ẑz 0 , hence the displacement vector is R = r − r0 = ρ̂ρ − ẑz 0 with |R| = (ρ2 + z 0 2 )1/2 . The electric field is given by

Z ∞ Z ∞ Z 1 ρl (z 0 )z 0 dz 0 R 1 ρl (z 0 ) dz 0 0 E(r) = − ẑ ρl (z 0 ) dl = ρ̂ρ 2 0 2 3/2 2 0 2 3/2 4π Γ |R|3 4π −∞ (ρ + z ) −∞ (ρ + z ) as desired. The function ρl (z) = ρ0 sgn(z) is odd about z = 0, so the first integral vanishes: Z ∞ 1 ρ0 ρ0 z 0 dz 0 E(r) = −ẑ = −ẑ ·2 .

2 3/2 2 0 4π 2π ρ (ρ + z ) 0 3.2. The ring is shown in Figure 4 along with source and field point position vectors r0 and r.

We have r = ẑz and r0 = ρ̂0 a = x̂a cos φ0 + ŷa sin φ0 . Hence R = r − r0 = ẑz − x̂a cos φ0 − ŷa sin φ0 and |R| = (z 2 + a2 cos2 φ0 + a2 sin2 φ0 )1/2 = (z 2 + a2 )1/2 .

The electric field is given by Z R 1 E= ρl (φ0 ) dl0 4π Γ |R|3 Z 2π 1 ẑz − x̂a cos φ0 − ŷa sin φ0 ρl (φ0 ) a dφ0 = 3/2 2 2 4π 0 (z + a ) Z 2π

Z 2π 2 a 0 0 0 0 0 0 =− x̂ ρ (φ ) cos φ dφ + ŷ ρ (φ ) sin φ dφ l l 4π (z 2 + a2 )3/2 0 0 Z 2π az + ẑ ρl (φ0 ) dφ0 .

4π (z 2 + a2 )3/2 0 29

Figure 4 When ρl (φ) = ρ0 sin φ, we obtain E = −ŷ When ρl (φ) = ρ0 cos2 φ, we get E = ẑ

a2 ρ0 .

4 (z 2 + a2 )3/2

azρ0 .

2 4 (z + a2 )3/2

3.3. The source and field point positions for this problem are indicated in Figure 5.

Figure 5 Writing r = ẑz and r0 = ρ̂0 ρ0 = x̂ρ0 cos φ0 + ŷρ0 sin φ0 , we obtain R = r − r0 = ẑz − x̂ρ0 cos φ0 − ŷρ0 sin φ0 and 2

2

2

|R| = (z 2 + ρ0 cos2 φ0 + ρ0 sin2 φ0 )1/2 = (z 2 + ρ0 )1/2 .

The electric field is Z Z ∞ Z 2π 0 0 0 0 1 1 0 R 0 0 0 ẑz − x̂ρ cos φ − ŷρ sin φ 0 E= dS = ρs (r ) ρ (ρ , φ ) ρ dφ0 dρ0 s 2 3/2 2 0 4π S |R|3 4π 0 (z + ρ ) 0 30

so that 1 Ex = − 4π

Z ∞ Z 2π 0

0

ρs (ρ0 , φ0 )ρ0 2 cos φ0 0 0 dφ dρ , (z 2 + ρ0 2 )3/2

ρs (ρ0 , φ0 )ρ0 2 sin φ0 0 0 1 dφ dρ , 4π 0 (z 2 + ρ0 2 )3/2 0 Z ∞ Z 2π ρs (ρ0 , φ0 )ρ0 z dφ0 dρ0 .

Ez = 4π 0 (z 2 + ρ0 2 )3/2 0 Z ∞ Z 2π

Ey = −

When ρs (ρ, φ) = ρ0 U (ρ − a), we obtain Ex = Ey = 0,

Ez =

ρ0 1 . sgn(z) p 2 1 + (a/z)2

When ρs (ρ, φ) = ρ0 [1 − U (ρ − a)], we get Ex = Ey = 0,

" # 1 ρ0 sgn(z) 1 − p .

Ez = 2 1 + (a/z)2

3.4. The sphere is indicated in Figure 6.

Figure 6 Writing r = ẑz and r0 = r̂0 r0 = x̂a sin θ0 cos φ0 + ŷa sin θ0 sin φ0 + ẑa cos θ0 , we obtain R = r − r0 = ẑ(z − a cos θ0 ) − x̂a sin θ0 cos φ0 − ŷa sin θ0 sin φ0 and |R| = [(z − a cos θ0 )2 + a2 sin2 θ0 cos2 φ0 + a sin2 θ0 sin2 φ0 ]1/2 = (z 2 + a2 − 2az cos θ0 )1/2 .

31

The electric field is given by Z R 1 ρs (θ0 ) E= dS 0 4π S |R|3 Z 2π Z π ẑ(z − a cos θ0 ) − x̂a sin θ0 cos φ0 − ŷa sin θ0 sin φ0 2 1 ρs (θ0 ) a sin θ0 dθ0 dφ0 , = 2 + a2 − 2az cos θ 0 )3/2 4π 0 (z 0 hence

Z 2π Ex = Ey = 0

since 0

and

a2 Ez = 2

Z π 0

cos φ0 dφ0 = 0 sin φ0

ρs (θ0 )(z − a cos θ0 ) sin θ0 dθ0 . (z 2 + a2 − 2az cos θ0 )3/2

When ρs (θ) = ρ0 , we obtain

a2 ρ0 a+z a−z √ Ez = −√ 2 z 2 a2 + z 2 + 2az a2 + z 2 − 2az " # a2 ρ0 a+z a−z p = −p 2 z 2 (a + z)2 (a − z)2

a2 ρ0 a + z a−z = − 2 z 2 |a + z| |a − z| a2 ρ0 = [sgn(a + z) − sgn(a − z)] 2 z 2   0, |z| < a,     a2 ρ 0 z > a, = ẑ 2 , z   2   −ẑ a ρ0 , z < a. z 2 When ρs (θ) = ρ0 sgn(θ − π2 ), we get Z Z a2 ρ0 π/2 (z − a cos θ0 ) sin θ0 dθ0 a2 ρ0 π (z − a cos θ0 ) sin θ0 dθ0 Ez = − + 2 0 2 π/2 (z 2 + a2 − 2az cos θ0 )3/2 (z 2 + a2 − 2az cos θ0 )3/2

a2 ρ0 2a −√ + sgn(a − z) + sgn(a + z) = 2 z 2 a2 + z 2  2

a ρ0 a   √ , |z| < a, ẑ 2 1 − z a2 + z 2 = 3 a ρ0   −ẑ √ , |z| > a. z 2 a2 + z 2 3.5. (a) Consider Figure 2.6 of the textbook with the only source being surface charge. Begin with ∇ × E = 0. Integrate over a volume region V and use the curl theorem (B.30) Z I (∇ × E) dV = (n̂ × E) dS = 0.

V

S

32

Apply this to each of the regions V1 and V2 in the left side of Figure 2.6: Z Z (n̂ × E) dS = 0, (n̂ × E) dS + S10 S1 Z Z (n̂ × E) dS = 0. (n̂ × E) dS + S20

S2

Add to obtain Z

Z

Z

(n̂20 × E2 ) dS = 0.

(n̂10 × E1 ) dS −

(n̂ × E) dS −

S20

S10

S1 +S2

Now let δ → 0 so that n̂10 = −n̂20 = n̂12 , S1 + S2 = S, S10 = S20 . So Z I [n̂12 × (E1 − E2 )] dS. (n̂ × E) dS =

(*)

S10

S

Next, apply to the surface S in the right side of Figure 2.6: I Z (n̂ × E) dS = 0 = 0 dS. S

(**)

S10

To have (**) give the same result as (*), we must have n̂12 × (E1 − E2 ) = 0. (b) Begin with ∇ · D = 0. Integrate over a volume region V and use the divergence theorem (B.24) Z Z (∇ · D) dV = ρ dV. V

So

V

I

Z D · n̂ dS =

S

ρ dV. V

Apply this to each of the regions V1 and V2 in the left half of Figure 2.6. Since there is no charge within either of these regions, Z Z D · n̂ dS + D · n̂ dS = 0, ZS1 ZS10 D · n̂ dS + D · n̂ dS = 0.

S2

Add to obtain

S20

Z

Z D · n̂ dS −

S1 +S2

Z D1 · n̂10 dS −

S10

D2 · n̂20 dS = 0.

S20

Now let δ → 0 so that n̂10 = −n̂20 = n̂12 , S1 + S2 = S, S10 = S20 . So I Z D · n̂ dS = [n̂12 × (D1 − D2 )] dS. S

(*)

S10

Next, apply to the surface S in the right half of Figure 2.6. The charge contained in V is a surface charge, so I Z D · n̂ dS = S

ρs dS. S10

To have (**) give the same result as (*), we must have n̂12 · (D1 − D2 ) = ρs .

33

(**)

Figure 7 3.6. The geometry appears in Figure 7.

We have ρsp = n̂ · P z=0 = n̂ · (D − 0 E) z=0 = ( − 0 )n̂ · E z=0 = −( − 0 )

∂Φ . ∂n z=0

Z ρ(r0 ) Qδ(z 0 − h)δ(x0 )δ(y 0 ) 0 0 0 0 Φ1 (r) = G1 (r|r ) dV = G1 (r|r0 ) dx dy dz 1 1 V V Q = G1 (r|x0 = 0, y 0 = 0, z 0 = h) 1 Z Z ∞ −kρ |z−h| 1 − 2 −kρ (z+h) e + 1 + 2 e Q 1 = ejkρ ·r d2 kρ 2 1 (2π) 2kρ −∞ Z

0

kρ · r = (x̂kx + ŷky ) · (x̂x + ŷy + ẑz) = kx x + ky y  kx = kρ cos ξ    ky = kρ sin ξ =⇒ kρ · r = kρ ρ cos(ξ − φ) x = ρ cos φ    y = ρ sin φ So, for z < h, Q 1 Φ1 = 1 (2π)2

Z 2π Z ∞ 0

− 2 −kρ (h+z) e−kρ (h−z) + 11 + e 2

2kρ

0

ejkρ ρ cos(ξ−φ) kρ dkρ dξ

Let x = ξ − φ and use Z 2π Z 2π−φ Z π jkρ ρ cos(ξ−φ) jkρ ρ cos x e dξ = e dx = ejkρ ρ cos x dx = 2πJ0 (kρ ρ) −φ

0

So Q 1 Φ1 = 1 2π

Z ∞

−π

2 −kρ (h+z) e−kρ (h−z) + 11 − + 2 e

2

0

J0 (kρ ρ) dkρ

Since n̂ = −ẑ, ∂Φ1 ∂Φ1 Q 1 =− =− ∂n ∂z 1 2π

Z ∞

−kρ (h+z) 2 kρ e−kρ (h−z) + 11 − + 2 kρ e

0

2 34

J0 (kρ ρ) dkρ

Z ∞ 1 − 2 Q 1 1− e−kρ h J0 (kρ ρ) dkρ 1 4π 1 + 2 0

Z ∞ 1 − 0 Q 2 2 = e−kρ h J0 (kρ ρ) dkρ 1 4π 1 + 2 0 1 − 0 Q 2 h = 1 2π 1 + 2 (h2 + ρ2 )3/2

ρsp1 = ( 1 − 0 )

(*)

Next, ZZ ∞

2 2 e−kρ (h−z) j(kx x+ky y) e dkx dky 2kρ −∞ 1 + 2 Z 2π Z ∞ −kρ (h−z) Q 1 2 2 e = ejkρ ρ cos(ξ−φ) kρ dkρ dξ 2 2 (2π) 1 + 2 0 2k ρ 0 Z ∞ Q 1 −kρ (h−z) J0 (kρ ρ) dkρ = e 1 + 2 2π 0 Z ∞ ∂Φ2 ∂Φ2 Q 1 n̂ = ẑ =⇒ = = e−kρ (h−z) J0 (kρ ρ)kρ dkρ ∂n ∂z 1 + 2 2π 0 Z ∞ 1 Q ρsp2 = −( 2 − 0 ) e−kρ h J0 (kρ ρ)kρ dkρ 1 + 2 2π 0 2 − 0 Q h =− 2 1 + 2 2π (h + ρ2 )3/2 Q 1 Φ2 = 2 (2π)2

Z Qp1 =

ρsp1 dS S Z 2π Z ∞

1 − 0 Q 2 h ρ dρ dφ 2 1 2π 1 + 2 (h + ρ2 )3/2 0 0 Z 2π Z ∞ 1 − 0 Q 2 h ρ dρ = dφ 2 1 2π 1 + 2 0 (h + ρ2 )3/2 0 2 1 − 0 =Q 1 1 + 2 =

Z Qp2 =

ρsp2 dS Z 2π Z ∞

S

=− 0

0

2 − 0 Q h ρ dρ dφ 2 1 + 2 2π (h + ρ2 )3/2

2 − 0 = −Q 1 + 2 Q 1 [ 2 ( 1 − 0 ) − 1 ( 2 − 0 )] 1 1 + 2 0 1 − 2 =Q .

1 1 + 2

Qp1 + Qp2 =

Note that Qp1 + Qp2 = 0 when 1 = 2 .

35

(**)

Figure 8 3.7. See Figure 8.

Primary potential: Φp (r) =

Z

Gp (r|r0 )

V

where 1 G (r|r ) = (2π)2 p

0

ρ(r0 ) dV 0

0

ZZ ∞

e−kρ |z−z | jkρ ·(r−r0 ) 2 e d kρ .

2kρ −∞

Secondary potential: 1 Φ (r) = (2π)2 s

ZZ ∞ h i A(kρ )ekρ z + B(kρ )e−kρ z ejkρ ·r d2 kρ .

(A)

−∞

Total potential: Φ(r) = Φp (r) + Φs (r) ) ZZ ∞ ( Z −kρ |z−z 0 | 0 1 e kρ z −kρ z −jkρ ·r0 ρ(r ) 0 = A(kρ )e + B(kρ )e + e dV ejkρ ·r d2 kρ .

(2π)2 2k

ρ −∞ V To find A and B, apply the boundary conditions at z = 0 and z = d:. ) Z −kρ |−z 0 | ZZ ∞ ( 0) 1 e ρ(r 0 Φ z=0 = 0 =⇒ A+B+ e−jkρ ·r dV 0 ejkρ ·r d2 kρ = 0 (2π)2 2kρ

−∞ V Z −kρ z 0 0 e 0 ρ(r ) =⇒ A + B + e−jkρ ·r dV 0 = 0.

V 2kρ

(*)

) ZZ ∞ ( Z −kρ |d−z 0 | 0) ρ(r e 0 Aekρ d + Be−kρ d + e−jkρ ·r dV 0 ejkρ ·r d2 kρ = 0 2kρ

−∞ V Z −kρ (d−z 0 ) 0 e 0 ρ(r ) kρ d −kρ d =⇒ Ae + Be + e−jkρ ·r dV 0 = 0. (**) 2k

ρ V

1 Φ z=d = 0 =⇒ (2π)2

36

Solve (*) and (**) simultaneously to get

Z −jkρ ·r0 0 e ρ(r0 ) −kρ d sinh kρ z −e dV 0 , A= 2k

sinh k d ρ ρ V

Z −jkρ ·r0 sinh kρ (d − z 0 ) e ρ(r0 ) B= − dV 0 .

2k

sinh k d ρ ρ V Back-substitute A and B into (A):

Z Z ∞ Z −jkρ ·r0 0 1 e ρ(r0 ) s −kρ d sinh kρ z Φ (r) = −e dV 0 ekρ z (2π)2 2kρ

sinh kρ d −∞ V

Z −jkρ ·r0 sinh kρ (d − z 0 ) e ρ(r0 ) 0 −kρ z ejkρ ·r d2 kρ + − dV e 2kρ

sinh kρ d V

Z ZZ ∞ 0 0 jkρ ·(r−r0 ) ρ(r0 ) 1 −kρ (d−z) sinh kρ z 2 −kρ z sinh kρ (d − z ) e = −e d kρ dV 0 −e

(2π)2 sinh kρ d sinh kρ d 2kρ V −∞ Z ρ(r0 ) s = G (r|r0 ) dV 0

V where 1 G (r|r ) = (2π)2 s

0

ZZ ∞ 0 0 jkρ ·(r−r0 ) −kρ (d−z) sinh kρ z −kρ z sinh kρ (d − z ) e −e −e d2 kρ . sinh kρ d sinh kρ d 2kρ −∞

3.8. Refer to Figure 9.

Figure 9 In the answer to Problem 3.7, put ∞

X 1 =2 e−(2n+1)kρ d sinh kρ d n=0

to get Gs (r|r0 ) =

1 (2π)2

ZZ ∞ X ∞

0

0

− e−kρ (d−z) e−(2n+1)kρ d (ekρ z − e−kρ z )

−∞ n=0

jkρ ·(r−r0 ) e 0 0 − e−kρ z e−(2n+1)kρ d (ekρ (d−z ) − e−kρ (d−z ) ) d2 kρ .

2kρ 37

Compare this to the primary Green’s function Z Z ∞ −kρ |z−z 0 | 1 e 0 p 0 ejkρ ·(r−r ) d2 kρ .

G (r|r ) = 2 (2π) 2kρ −∞ Since z 0 in Gp is the vertical position of the source point, each term in Gs can be viewed as arising from a point source that is an image of the primary charge Q. Consider the figure, and consider each term in Gs .

0

1. −e−kρ [−z−z +2d+2nd] .

When n = 0 this can be written as 0

0

0

−e−kρ [−z−z +2d] = −e−kρ[−(z+z −2d)] = −e−kρ |z−(−z +2d)| with z + z 0 − 2d < 0 implying that z < −z 0 + 2d. This is the potential for a charge −Q located at z = −z 0 + 2d which is the image charge A in the figure. It represents Q imaged into the conducting plate at z = d.

0

2. −e−kρ [+z+z +2nd] .

When n = 0 this can be written as 0

0

−e−kρ [z+z ] = −e−kρ |z+z | with z + z 0 > 0 implying that z > −z 0 . This is the potential for a charge −Q located at z = −z 0 which is the image charge B in the figure. It represents Q imaged into the conducting plate at z = 0.

0

3. +e−kρ [−z+z +2d+2nd] .

When n = 0 this can be written as 0

0

0

+e−kρ [−z+z +2d] = +e−kρ[−(z−z −2d)] = +e−kρ |z−(z +2d)| with z − z 0 − 2d < 0 implying that z < z 0 + 2d. This is the potential for a charge +Q located at z = z 0 + 2d which is the image charge B 0 in the figure. It represents B imaged into the conducting plate at z = d.

0

4. +e−kρ [+z−z +2d+2nd] .

When n = 0 this can be written as 0

0

+e−kρ [+z−z +2d] = +e−kρ |z−(z −2d)| with z − z 0 + 2d > 0 implying that z > z 0 − 2d. This is the potential for a charge +Q located at z = z 0 − 2d which is the image charge A0 in the figure. It represents A imaged into the conducting plate at z = 0.

For n > 0, each of the terms represents a multiple image into the plate at z = 0 or z = d, and can be identified as above.

3.9. Refer to Figure 10.

To compute the Green’s function we will employ the boundary conditions.

38

Figure 10 (a) Φ1 = Φ2 at z = d. (b) 1

∂Φ1 ∂Φ2 = 2 at z = d. ∂z ∂z

(c) Φ2 = 0 at z = 0.

There are two cases to consider: source in region 1, and source in region 2.

Case 1: Source in region 1.

Z ρ(r0 ) Gp (r|r0 ) Φp1 = dV 0

1 V1 # Z Z ∞ −kρ |z−z 0 | Z " e ρ(r0 ) 1 0 jkρ ·(r−r ) 2 e d k dV 0 . = ρ 2 2kρ 1 −∞ V1 (2π) For z < z 0 we have |z − z 0 | = z 0 − z and # Z " Z Z ∞ −kρ (z 0 −z) 1 e ρ(r0 ) ∂Φp1 0 jkρ ·(r−r ) 2 = e k d k dV 0 . ρ ρ 2 ∂z 2kρ 1 V1 (2π) −∞ Also

ZZ ∞ 1 B(kρ )e−kρ z ejkρ ·r d2 kρ , (2π)2 −∞ ZZ ∞ s ∂Φ1 1 = B(kρ )e−kρ z ejkρ ·r (−kρ ) d2 kρ , ∂z (2π)2 −∞ ZZ ∞ h i 1 kρ z −kρ z Φs2 = Ā(k )e + B̄(k )e ejkρ ·r d2 kρ , ρ ρ (2π)2 −∞ ZZ ∞ h i ∂Φs2 1 kρ z −kρ z = Ā(k )e − B̄(k )e ejkρ ·r kρ d2 kρ . ρ ρ ∂z (2π)2 −∞ Φs1 =

Boundary condition (c) gives Ā(kρ ) + B̄(kρ ) = 0,

so B̄(kρ ) = −Ā(kρ ).

39

Therefore Φs2 = ∂Φs2 ∂z

=

1 (2π)2

ZZ ∞

1 (2π)2

ZZ ∞

Ā(kρ )2 sinh kρ z ejkρ ·r d2 kρ ,

−∞

Ā(kρ )2 cosh kρ z ejkρ ·r kρ d2 kρ .

−∞ p s Boundary condition (a) reads Φ1 + Φ1 = Φs2 at z = d.

Z V1

# 0 ρ(r0 ) e−kρ (z −d) jkρ ·(r−r0 ) 2 e d kρ dV 0 2kρ 1 −∞ ZZ ∞ ZZ ∞ 1 1 −kρ d jkρ ·r 2 + B(kρ )e e d kρ = Ā(kρ )2 sinh kρ d ejkρ ·r d2 kρ , 2 (2π)2 (2π) −∞ −∞

"

ZZ ∞

1 (2π)2

hence

0

e−kρ z ekρ d −jkρ ·r0 ρ(r0 ) e dV 0 + B(kρ )e−kρ d = 2Ā(kρ ) sinh kρ d.

2k

ρ 1 V1

Z

Boundary condition (b):

∂Φp1 ∂Φs1 2 ∂Φs2 + = , ∂z ∂z 1 ∂z 0

e−kρ z ekρ d −jkρ ·r0 ρ(r0 ) 2 e dV 0 − B(kρ )e−kρ d = 2Ā(kρ ) cosh kρ d.

2k

1 ρ 1 V1

Z Calling

0

e−kρ z ekρ d −jkρ ·r0 ρ(r0 ) e dV 0 2k

ρ 1 V1

Z F = we have so far

F + B(kρ )e−kρ d = 2Ā(kρ ) sinh kρ d, 2 F − B(kρ )e−kρ d = 2Ā(kρ ) cosh kρ d.

1 Adding (*) and (**), we obtain 2F = 2Ā(kρ ) sinh kρ d + or Ā(kρ ) =

2 2Ā(kρ ) cosh kρ d 1

1 F S

where

−kρ d

2 = 2Ā(kρ ) sinh kρ d − cosh kρ d , 1

S = 1 sinh kρ d + 2 cosh kρ d.

Subtracting instead, we obtain 2B(kρ )e

1 B(kρ )e−kρ d = Ā(kρ )T

where

B̄(kρ ) = Ā(kρ )ekρ d

T = 1 sinh kρ d − 2 cosh kρ d,

T 1 F kρ d T T = e = F ekρ d .

1 S 1 S 40

(*) (**)

So ρ(r0 ) 1 Φ1 = 2 V1 1 (2π) Z

) Z Z ∞ ( −kρ |z−z 0 | 0 e−kρ z ekρ d kρ d −kρ z T e 0 + e e ejkρ ·(r−r ) d2 kρ dV 0 .

2kρ 2kρ S −∞

This implies 1 G1 (r|r ) = (2π)2 0

Also

jkρ ·(r−r0 ) ZZ ∞ −kρ |z−z 0 | −kρ (z+z 0 −2d) T e e +e d2 kρ .

S 2k ρ −∞

ρ(r0 ) 1 2 V1 1 (2π)

ZZ ∞

1 (2π)2

ZZ ∞

Z Φ2 = so that

G2 (r|r0 ) =

0

2

1 e−kρ z ekρ d jkρ ·(r−r0 ) 2 e d kρ dV 0 sinh kρ z S 2kρ

2

1 ejkρ ·(r−r ) 2 0 d kρ . sinh kρ ze−kρ (z −d) S 2kρ

−∞

0

−∞

Case 2: Source in region 2.

ZZ ∞ 1 B(kρ )e−kρ z ejkρ ·r d2 kρ (2π)2 −∞ ZZ ∞ 1 ∂Φs1 = B(kρ )e−kρ z ejkρ ·r (−kρ ) d2 kρ ∂z (2π)2 −∞ ZZ ∞ h i 1 kρ z −kρ z ejkρ ·r d2 kρ Φs2 = Ā(k )e + B̄(k )e ρ ρ (2π)2 −∞ ZZ ∞ h i s 1 ∂Φ2 kρ z −kρ z Ā(k )e − B̄(k )e ejkρ ·r kρ d2 kρ = ρ ρ ∂z (2π)2 −∞ Φs1 =

ρ(r0 ) Gp (r|r0 ) dV 0 2 V2 # Z " Z Z ∞ −kρ |z−z 0 | e 1 ρ(r0 ) jkρ ·(r−r0 ) 2 = e d k dV 0 . ρ 2 (2π) 2k

ρ 2 V2 −∞

Φp2 =

Z

For z > z 0 we have |z − z 0 | = z − z 0 and # Z " Z Z ∞ −kρ (z−z 0 ) ∂Φp2 1 e ρ(r0 ) jkρ ·(r−r0 ) 2 = − e k d k dV 0 . ρ ρ 2 ∂z (2π) 2k

ρ 2 V2 −∞ Boundary condition (c): Φ2 = 0 at z = 0: 1 (2π)2 Let

ZZ ∞

Ā(kρ ) + B̄(kρ ) ejkρ ·r d2 kρ +

−∞

Z

"

V2

1 (2π)2

0

# 0 e−kρ z jkρ ·(r−r0 ) 2 ρ(r0 ) e d kρ dV 0 = 0 2 −∞ 2kρ

ZZ ∞

e−kρ z −jkρ ·r0 ρ(r0 ) FA = e dV 0 .

2 V2 2kρ Z

41

Then

1 (2π)2

ZZ ∞

Ā(kρ ) + B̄(kρ ) + FA ejkρ ·r d2 kρ = 0

−∞

so Ā(kρ ) + B̄(kρ ) = −FA .

Boundary condition (a): Φs1 = Φs2 + Φp2 at z = d.

ZZ ∞ ZZ ∞ h i 1 1 −kρ d jkρ ·r 2 kρ d −kρ d B(k )e e d k = Ā(k )e + B̄(k )e ejkρ ·r d2 kρ ρ ρ ρ ρ 2 (2π)2 (2π) −∞ −∞ # Z " Z Z ∞ −kρ (d−z 0 ) 1 ρ(r0 ) e 0 jkρ ·(r−r ) 2 + e d k dV 0 ρ 2 2kρ 2 V2 (2π) −∞ B(kρ )e−kρ d = Ā(kρ )ekρ d + B̄(kρ )e−kρ d + e−kρ d FB where

0

ekρ z −jkρ ·r0 ρ(r0 ) e dV 0 2k

ρ 2 V2

Z FB = Boundary condition (b):

∂Φ1 ∂Φ2 = 2 at z = d ∂z ∂z s

∂Φs ∂Φ2 ∂Φp2 1 1 = 2 + at z = d ∂z ∂z ∂z 1

or

ZZ ∞ h i B(kρ )e e Ā(kρ )ekρ d − B̄(kρ )e−kρ d ejkρ ·r kρ d2 kρ −∞ −∞ # Z " Z Z ∞ −kρ (d−z 0 ) 1 e ρ(r0 ) jkρ ·(r−r0 ) 2 + 2 − e k d k dV 0 ρ ρ 2 (2π) 2k

ρ 2 V2 −∞

1 1 (2π)2

ZZ ∞

−kρ d jkρ ·r

1 (−kρ ) d kρ = 2 (2π)2 2

i h − 1 B(kρ )e−kρ d kρ = 2 Ā(kρ )ekρ d − B̄(kρ )e−kρ d kρ − 2 e−kρ d FB kρ We have so far: Ā + B̄ = −FA −kρ d

(*)

kρ d

−kρ d

−kρ d

Be = Āe + B̄e +e FB 1 −kρ d kρ d −kρ d −kρ d Be = −Āe + B̄e +e FB 2 Subtracting (***) from (**), we obtain Be

−kρ d

1 1− = 2Āekρ d 2

Adding (***) to (**), we obtain Be

−kρ d

1 1+ = 2B̄e−kρ d + 2FB e−kρ d 2 42

(**) (***)

Divide to get

1 + 21

2B̄e−kρ d + 2FB e−kρ d 2Āekρ d

1 =

1 − 2 or

Āekρ d Q = B̄e−kρ d + FB e−kρ d

where

Q=

2 + 1 .

2 − 1

(**’)

So far then, Ā + B̄ = −FA ĀQe2kρ d − B̄ = FB

(*) (**’)

Adding (*) and (**’), we get Ā(1 + Qe2Kρ d ) = −FA + BB

2 + 1 kρ d −kρ d = [−FA + FB ]e−kρ d Ā e + e 2 − 1 Z 0 0 −e−kρ z e−kρ d + ekρ z e−kρd −jkρ·r0 ρ(r0 ) −kρ d kρ d kρ d −kρ d Ā[ 2 (e + e ) + 1 (e − e )] = ( 2 − 1 ) e dV 0 2kρ 2 V2 Z e−kρ d sinh kρ z 0 −jkρ·r0 ρ(r0 ) 2 − 1 Ā = e dV 0 S 2k

ρ 2 V2 Ā + B̄ = −FA , −Ā + B̄

1 1 −2kρ d e = − FB e−2kρ d .

Q Q

Adding (*) and (**”) we get

1 1 B̄ 1 + e−2kρ d = − FA + FB e−2kρ d , Q Q and substitution leads to 1 B̄ = − 2kρ

1 sinh kρ (d − z 0 ) + 2 cosh kρ (d − z 0 ) −jkρ ·r0 ρ(r0 ) e dV 0 .

1 sinh kρ d + 2 cosh kρ d 2 V2

Z

Now use Be

1 1− = 2Āekρ d 2

−kρ d

to get B= or B= So

2 2 ekρ d S

ρ(r0 ) 1 Φ1 = 2 V2 2 (2π) Z

2 e2kρ d 2Ā 2 − 1

sinh kρ z 0 jkρ ·r0 ρ(r0 ) e dV 0 .

2k

ρ 2 V2

Z

ZZ ∞ −∞

2 2 ekρ d sinh kρ z 0 jkρ ·(r−r0 ) 2 e d kρ dV 0 , S 2kρ 43

(*) (**”)

G1 (r|r0 ) =

1 (2π)2

ZZ ∞

2 2 kρ d sinh kρ z 0 jkρ ·(r−r0 ) 2 e d kρ , e 2kρ −∞ S

Z Z ∞ −kρ |z−z 0 | ρ(r0 ) 1 2 − 1 kρ z e−kρ d sinh kρ z 0 e Φ2 = + e (2π)2 2kρ S 2kρ V2 2 −∞

0 0 1 sinh kρ (d − z ) + 2 cosh kρ (d − z ) jkρ ·(r−r0 ) 2 − e−kρ z ekρ d e d kρ , 2kρ S Z

G2 (r|r0 ) =

ZZ ∞

e−kρ d sinh kρ z 0 S −∞

0 1 sinh kρ (d − z 0 ) + 2 cosh kρ (d − z 0 ) ejkρ ·(r−r ) 2 − e−kρ (z−d) d kρ .

S 2kρ

1 (2π)2

0

e−kρ |z−z | + ( 2 − 1 )ekρ z

3.10. The point charge and sphere are shown in Figure 11.

Figure 11 We know that the potential due to a point charge q located at r0 is given by

q 1 a/r0 a Φ(r) = − where r0i = 0 2 r0 .

4π |r − r0 | |r − r0i | r Therefore

q ∇ E = −∇Φ = − 4π

1 |r − r0 |

q a + ∇ 4π r0

1 |r − r0i |

By (B.73) we have r − r0 1 =− , |r − r0 | |r − r0 |3

a r − r0 q r − r0 i E= − 0 .

4π |r − r0 |3 r |r − r0i |3 ∇

so

44

.

Now ρs = n̂ · D = r̂ · E r=a =

q r̂ · (r − r0 ) a r̂ · (r − r0i ) − .

4π |r − r0 |3 r0 |r − r0i |3

Here r0 = r̂0 r0 , so that

r = r̂a,

r0i = r̂0 ri0 ,

r0 r̂ · (r − r0 ) = r̂ · r̂a − a r̂0 = a(1 − k 0 r̂ · r̂0 ), a

k0 =

r0 a

and |r − r0 |3 = a3 |r̂ − k 0 r̂0 |3 .

Let γ be the angle between r̂ and r̂0 . Then 2

|r̂ − k 0 r̂0 |3 = (1 + k 0 − 2k 0 cos γ)3/2 , Hence q ρs (r) = ρs (γ) = 4πa2

"

r̂ · (r − r0 ) = a(1 − k 0 cos γ).

1 − k 0 cos γ 1 − ki0 cos γ 0 − k i (1 + k 0 2 − 2k 0 cos γ)3/2 (1 + ki0 2 − 2ki0 cos γ)3/2

where ki0 =

#

a 1 ri0 = 0 = 0. a r k

The total charge is Z Q= ρs (r) dS S Z π = 2π ρs (γ)a2 sin γ dγ 0 # Z π" 2 1 − k 0 cos γ qa 1 − ki0 cos γ 0 = 2 − ki sin γ dγ.

2a 0 (1 + k 0 2 − 2k 0 cos γ)3/2 (1 + ki0 2 − 2ki0 cos γ)3/2 Use

Z

a − cos x (1 − a cos x) sin x dx =√ 3/2 2 (1 + a − 2a cos x) 1 + a2 − 2a cos x

to calculate Z π 0

π k0 + 1 1 − k 0 cos γ k 0 − cos γ k0 − 1 p p p = − =1−1=0 = (1 + k 0 2 − 2k 0 cos γ)3/2 (k 0 + 1)2 (k 0 − 1)2 1 + k 0 2 − 2k 0 cos γ 0

since k 0 = r0 /a > 1. Similarly, Z π ki0 + 1 1 − ki0 cos γ ki0 − 1 p p = − =1+1=2 02 0 3/2 (ki0 + 1)2 (ki0 − 1)2 0 (1 + ki − 2ki cos γ) since ki0 = a/r0 < 1. Finally, qa2 a q a 0 Q = 2 (−ki )(2) = − 0 (2) = −q 0 .

2a 2 r r 45

3.11. The Green’s function for a source inside a grounded conducting sphere can be found by swapping the charge and its image in the Green’s function found in the text for a source external to the sphere. Thus, let r → r0 and r0 → r and obtain   Z 0 ρ(r ) 1  1 a/r  Φ(r) = − 2 dV 0 .

0 a 4π |r − r | 0 V r−r r2

To check this, evaluate the potential at r = a:  Z 0 ρ(r ) 1  1 − Φ(r = ar̂) = 4π |r − r0 | V

 a/a a2 a2

r − r0

 dV 0 = 0.

3.12. Refer to Figure 12.

Figure 12 The total potential is the sum of primary and secondary potentials: Φ(r) = Φ0 (r) + Φs (r), where Φ0 (r) = −E0 z = −E0 r cos θ and, by (3.82), s

Φ (r) =

∞ X

Bn r−(n+1) Pn (cos θ).

n=0

The boundary condition yields Φ(r = a) = 0 = −E0 a cos θ +

∞ X

Bn a−(n+1) Pn (cos θ).

n=0

or E0 aP1 (cos θ) =

∞ X

Bn a−(n+1) Pn (cos θ).

n=0

46

Multiply both sides by Pm (cos θ) sin θ and integrate: Z π E0 a

P1 (cos θ)Pm (cos θ) sin θ dθ = 0

∞ X

Bn a−(n+1)

Z π Pn (cos θ)Pm (cos θ) sin θ dθ.

0

n=0

Use orthogonality relation (E.125) to get ∞

E0 aδ1m

2 X 2 2 = Bn a−(n+1) δnm = Bm a−(m+1) .

3 2n + 1 2m + 1 n=0

So Bm = 0 for m 6= 1, and B1 = E0 a3 .

Therefore Φs (r) = B1 r−2 P1 (cos θ) = E0 a

a 2 r

cos θ.

The secondary electric field is ∂Φs 1 ∂Φs Es (r) = −∇Φs (r) = −r̂ − θ̂ ∂r r ∂θ

1 ∂ ∂ −2 3 cos θ = −E0 a r̂ r + θ̂ 3 ∂r r ∂θ a 3 = E0 [2r̂ cos θ + θ̂ sin θ]. r The total potential, s

Φ(r) = Φ0 (r) + Φ (r) = −E0 r cos θ + E0 a

a 2 r

r a 2 − cos θ = −E0 a cos θ a r

vanishes at r = a as required. Calculation of surface charge density: ρs = r̂ · D r=a = r̂ · [ẑE0 + E0

a 3

r = E0 [r̂ · ẑ + 2 cos θ]

(2r̂ cos θ + θ̂ sin θ)] r=a

= 3 E0 cos θ.

Total surface charge: Z 2π Z π

3 E0 cos θa2 sin θ dθ dφ 0 0 Z π 2 = 6π E0 a sin θ cos θ dθ

Q=

0

= 0.

3.13. The cavity is depicted in Figure 13.

47

Figure 13 Outside the cavity, the total potential is the sum of impressed and scattered potentials: Φ2 (r) = Φ0 (r) + Φs (r), where Φ0 (r) = −E0 z = −E0 r cos θ and, by (3.82), s

Φ (r) =

∞ X

Bn r−(n+1) Pn (cos θ).

n=0

The potential inside the cavity is, by (3.80), Φ1 (r) =

∞ X

An rn Pn (cos θ).

n=0

To find An and Bn , we apply the boundary conditions. Condition (3.36) requires −E0 a cos θ +

∞ X

Bn a−(n+1) Pn (cos θ) =

n=0

∞ X

An an Pn (cos θ).

n=0

Apply (E.125) to get −E0 a + a−2 B1 = A1 a

(1)

and Bn a−(n+1) = An an ,

n 6= 1.

Condition (3.17) requires "∞ # " # ∞ X ∂ X ∂ 0 An rn Pn (cos θ) = 0 r −E0 r cos θ + Bn r−(n+1) Pn (cos θ) ∂r ∂r r=a r=a n=0

or

∞ X

n=0

nAn a

n−1

Pn (cos θ) = − r E0 cos θ − r

n=0

∞ X

(n + 1)Bn a−n−2 Pn (cos θ).

n=0

48

(2)

Apply orthogonality: − r E0 − 2 r B1 a−3 = A1 ,

(3)

− r (n + 1)Bn a−n−2 = nAn an−1 .

(4)

Since (2) and (4) imply An = Bn = 0 for n 6= 1, we write A1 = B1 a−3 − E0 ,

(5)

A1 = −2 r B1 a−3 − r E0 .

(6)

Subtracting (5) and (6), we obtain B1 a−3 (1 + 2 r ) − E0 (1 − r ) = 0 so that B1 = E0 a3 Then

1 − r .

1 + 2 r

1 − r A1 = E0 − E0 = −E0 1 + 2 r

So Φs (r) =

E0 a3 1 − r cos θ, r2 1 + 2 r

3 r 1 + 2 r

Φ1 (r) = −E0 r

.

3 r cos θ.

1 + 2 r

Inside the cavity, E1 = −∇Φ1 = E0

3 r 3 r ∇(r cos θ) = ẑ E0 1 + 2 r 1 + 2 r

since ∇(r cos θ) = ∇z = ẑ. Is the field inside the cavity greater than the impressed field? The inequality 3 r E0 > E0 1 + 2 r holds as long as r > 1, so the answer is yes. Indeed, the impressed field polarizes the material, creating dipoles (Figure 14). The field due to the induced dipoles adds to the impressed field within the cavity.

Figure 14

49

Figure 15 3.14. Refer first to Figure 15.

Use image theory: replace conductor with image charge qi = −Q at z = −d. The field is the sum of the fields due to the charge and its image: E(r) =

Q r − rQ Q r − rQi − .

3 4π 0 |r − rQ | 4π 0 |r − rQi |3

Here p

r − rQ = xx̂ + yŷ + (z − d)ẑ = ρρ̂ + (z − d)ẑ,

|r − rQ | =

r − rQi = xx̂ + yŷ + (z + d)ẑ = ρρ̂ + (z + d)ẑ,

|r − rQi | =

ρ2 + (z − d)2 ,

and so that

p ρ2 + (z + d)2 ,

Q ρρ̂ + (z − d)ẑ ρρ̂ + (z + d)ẑ E(r) = − .

4π 0 [ρ2 + (z − d)2 ]3/2 [ρ2 + (z + d)2 ]3/2

The induced surface charge is given by ρs = 0 ẑ · E z=0 = −

2dQ 1 .

4π (ρ2 + d2 )3/2

For the total charge induced on the ground plane, use Z x dx 1 = −√ 2 (x2 + a2 )3/2 x + a2 to get Z q=

ρs dS = S

Z 2π Z ∞ Z ∞ 2dQ 1 2dQ ρ dρ − ρ dρ dφ = − (2π) = −Q. 2 + d2 )3/2 2 + d2 )3/2 4π 4π (ρ (ρ 0 0 0

50

Figure 16 At points in the z = 0 plane we have E = −ẑ

2d Q .

2 4π 0 (ρ + d2 )3/2

For the force computation, refer to Figure 16.

The stress tensor is given by 1 T̄e = (D · E)Ī − DE. 2 We have

1 1 1 2 1 Q 2 4d2 n̂ · T̄e = (D · E)ẑ · Ī − ẑ · DE = 0 ẑ E · E − Ez Ez = − 0 ẑ Ez = − 0 ẑ .

2 2 2 2 4π 0 (ρ2 + d2 )3 The force can be calculated using Z

x dx (x2 + a2 )3

=−

1 1 2 4 (x + a2 )2

to get Q2 d2 F = − n̂ · T̄e dS = ẑ 2 8π 0 S Z

Z 2π Z ∞ 0

0

ρ dρ Q2 d2 dφ = ẑ (ρ2 + d2 )3 4π 0

Z ∞ 0

ρ dρ Q2 = ẑ . (ρ2 + d2 )3 4π 0 (2d)2

From Coulomb’s law, the force on the image charge is F−Q = ẑ

Q2 4π 0 (2d)2

since the charge separation is 2d.

3.15. See Figure 17.

The volume charge density of these point charges is ρ(r) = 2qδ(r − r0 ) − qδ(r − r0 − d) − qδ(r − r0 + d).

The moments are

51

Figure 17 (0) Z

ρ(r0 ) dV 0 VZ Z Z 0 0 0 0 = 2q δ(r − r0 ) dV − q δ(r − r0 − d) dV − q δ(r0 − r0 + d) dV 0

Q=

V

V

V

= 2q − q − q = 0.

(1) Z

r0 ρ(r0 ) dV 0 VZ Z Z = 2q r0 δ(r0 − r0 ) dV 0 − q r0 δ(r0 − r0 − d) dV 0 − q r0 δ(r0 − r0 + d) dV 0

p=

V

V

V

= 2qr0 − q(r0 + d) − q(r0 − d) = 2qr0 − qr0 − qd − qr0 + qd = 0.

(2) Z Q̄ =

2

(3r0 r0 − r0 Ī)ρ(r0 ) dV 0

V

= 6qr0 r0 − 2qr02 Ī − 3q(r0 + d)(r0 + d) + q(r0 + d) · (r0 + d)Ī − 3q(r0 − d)(r0 − d) + q(r0 − d) · (r0 − d)Ī = q[6r0 r0 − 3r0 r0 − 3r0 d − 3dr0 − 3dd − 3r0 r0 + 3r0 d + 3dr0 − 3dd] + q[−2r02 + r02 + 2d · r0 + d2 + r02 − 2d · r0 + d2 ]Ī = −6qdd + qd2 Ī Note that the first nonzero moment is independent of r0 as expected. To three terms, the potential

52

expansion is Q r̂ · p 1 r̂ · Q̄ · r̂ + + 2 4π 0 r 4π 0 r 2 4π 0 r3 1 r̂ · Q̄ · r̂ =0+0+ 2 4π 0 r3

1 = r̂ · −6qdd + qd2 Ī · r̂ 3 8π 0 r

q = −6(r̂ · d)(d · r̂) + d2 r̂ · Ī · r 3 8π 0 r

q 2 2 −6(r̂ · d) + d = 8π 0 r3 i qd2 h 2 = 1 − 6(r̂ · d̂) .

8π 0 r3

Φ(r) =

Note that this first nonzero term dominates for r d.

3.16.

Z

n

∗ (θ0 , φ0 ) dV 0 ρ(r0 )r0 Ynm

qnm = V

Z 2π Z π Z a = 0

0

0

But

Q n cos 2θ0 r0 πa3

Z 2π

s

2n + 1 (n − m)! m 0 2 Pn (cos θ0 )e−jmφ r0 sin θ0 dr0 dθ0 dφ0 .

4π (n + m)!

0

e−jmφ dφ0 = 2πδm,0 =⇒ qnm = 0 for m 6= 0.

0

For m = 0 we have r

Z a Z π 2n + 1 Q 0 n+2 0 (2π) r dr cos(2θ0 )Pn0 (cos θ0 ) sin θ0 dθ0 4π πa3 0 0 r

Z π 2n + 1 2Q an+3 cos(2θ0 )Pn (cos θ0 ) sin θ0 dθ0 = 4π a3 n + 3 0 r n 2n + 1 a = 2Q In 4π n+3

qn0 =

where

Z π In =

cos(2θ0 )Pn (cos θ0 ) sin θ0 dθ0 .

0

Now

 P0 (cos θ) = 1  P1 (cos θ) = cos θ  P2 (cos θ) = 43 cos 2θ + 14

4 1 cos 2θ = P2 (cos θ) − P0 (cos θ) 3 3

=⇒

so that 4 In = 3

Z π 0

1 P2 (cos θ )Pn (cos θ ) sin θ dθ − 3 0

0

0

0

53

Z π 0

P0 (cos θ0 )Pn (cos θ0 ) sin θ0 dθ0 .

Apply (E.125): Z π

2 2 = δ2n , 2n + 1 5 Z0 π 2 P0 (cos θ0 )Pn (cos θ0 ) sin θ0 dθ0 = δ0n = 2δ0n .

2n + 1 0

So

P2 (cos θ0 )Pn (cos θ0 ) sin θ0 dθ0 = δ2n

2 I0 = − , 3

I2 =

We have

r

8 , 15

and In = 0 for other n.

q00 =

1 2Q 4π 3

2 − 3

r

and q20 =

5 2Q 2 a 4π 5

8 15

4Q =− 9

r

1 4π

r √ 16 5Q 2 1 = a .

75 4π

Therefore ∞

1 X 1 Φ(r) = 0 rn+1 n=0

"

1

n X

2n+1 m=−n

# qnm Ynm (θ, φ)

1 11 1 1 1 q00 Y00 (θ, φ) + q20 Y20 (θ, φ) 0 r 2 0 r3 23 !r !r r r √ 1 1 1 4Q 1 1 1 1 16 5 2 5 0 = P0 (cos θ)e + Qa P2 (cos θ)e0 − 0 2r 4π 9 4π 0 8r3 4π 75 4π

1 1 3 1 2Q 16 · 5 Qa2 =− + cos 2θ + 4π 0 9 r 4π 0 8 · 75 r3 4 4

1 2Q 1 2 3 1 2 = − + Qa cos 2θ + 4π 0 r 9 4π 0 r3 15 4 4 2 2 Q 1 Qa =− + (3 cos 2θ + 1).

9 4π 0 r 30 4π 0 r3 =

3.17. The ring of current is shown in Figure 18.

We have µ0 I B(r) = 4π

I

dl0 ×

R̂ .

R2

Γ 0 0 Beginning with r = xx̂ + yŷ + zẑ and r = x x̂ + y 0 ŷ, we write

R = (x − x0 )x̂ + (y − y 0 )ŷ + zẑ.

Then R2 = (x − x0 )2 + (y − y 0 )2 + z 2 = ρ2 cos2 φ − 2ρa cos φ cos φ0 + a2 cos2 φ0 + ρ2 sin2 φ − 2ρa sin φ sin φ0 + a2 sin2 φ0 + z 2 = ρ2 + a2 + z 2 − 2aρ cos(φ − φ0 ).

54

Figure 18 Furthermore 0

dl0 × R = a dφ0 φ̂ × R = a dφ0 (−x̂ sin φ0 + ŷ cos φ0 ) × [x̂(x − x0 ) + ŷ(y − y 0 )z + ẑz 0 ] = a dφ0 [−ẑ(y − y 0 ) sin φ0 + ŷz sin φ0 − ẑ(x − x0 ) cos φ0 + x̂z cos φ0 ] so that [dl0 × R]z = a dφ0 [−(ρ sin φ − a sin φ0 ) sin φ0 − (ρ cos φ − a cos φ0 ) cos φ0 ] = a dφ0 [−ρ(sin φ sin φ0 + cos φ cos φ0 ) + a(sin2 φ0 + cos2 φ0 )] = a dφ0 [a − ρ cos(φ − φ0 )].

Hence Bz =

µ0 I a 4π

Z 2π 0

a − ρ cos u du. [z 2 + ρ2 + a2 − 2aρ cos u]3/2

Change of variables: u = π − 2x =⇒ cos u = − cos 2x = −[cos2 x − sin2 x] = −[1 − 2 sin2 x].

Z π/2

a + ρ[1 − 2 sin2 x] 2 dx 2 3/2 2 2 2 −π/2 [z + ρ + a + 2aρ(1 − 2 sin x)] Z π/2 µ0 I (a + ρ) − 2ρ sin2 x = a dx π [z 2 + (a + ρ)2 − 4aρ sin2 x]3/2 0 Z π/2 µ0 I 1 (a + ρ) − 2ρ sin2 x a 3/2 = dx, where F = (a + ρ)2 + z 2 , π F [1 − k 2 sin2 x]3/2 0

Bz =

Write

µ0 I a 4π

(a + ρ) − 2ρ sin2 x A B = + [1 − k 2 sin2 x]3/2 [1 − k 2 sin2 x]3/2 [1 − k 2 sin2 x]1/2

and obtain B[1 − k 2 sin2 x] + A = (a + ρ) − 2ρ sin2 x, 55

k2 =

4aρ .

F

A + B = (a + ρ) − Bk 2 sin2 x = −2ρ sin2 x.

B=

2ρ k2

2ρ +A=a+ρ k2 2ρ A=a+ρ− 2 k

A+B =

2ρF 4aρ(a + ρ) − 2ρ[(a + ρ)2 + z 2 ] = 4aρ 4aρ 4a2 ρ + 4aρ2 − 2ρ[a2 + 2aρ + ρ2 + z 2 ] = 4aρ 2 3 2a ρ − 2ρ − 2ρz 2 a2 − ρ2 − z 2 = = 4aρ 2a ) ( Z Z π/2 π/2 µ0 Ia dx dx Bz = +B A πF 3/2 [1 − k 2 sin2 x]3/2 [1 − k 2 sin2 x]1/2 0 0 Z π/2 Z π/2 dx E(k 2 ) dx 2 2 3/2 = 1 − k 2 , 2 1/2 = K(k ).

2 2 [1 − k sin x] [1 − k sin x] 0 0 A=a+ρ−

4aρ F − 4aρ 1 − k2 = 1 − = F F

µ0 Ia B A 2 2 Bz = E(k ) + K(k ) F πF 3/2 F − 4aρ B=

2ρ 2ρF F = = 2 k 4aρ 2a

A=

a − ρ2 − z 2 2a

F − 4aρ = a2 + 2aρ + ρ2 + z 2 − 4aρ = (a − ρ)2 + z 2 Thus µ0 I Bz = 2πF 1/2

a2 − ρ2 − z 2 E(k 2 ) + K(k 2 ) (a − ρ)2 + z 2

Next, [dl0 × R]φ = φ̂ · a dφ0 [−ẑ(y − y 0 ) sin φ0 + ŷz sin φ0 − ẑ(x − x0 ) cos φ0 + x̂z cos φ0 ] = a dφ0 [z sin φ0 cos φ − z cos φ0 sin φ] = a dφ0 z sin(φ − φ0 ).

Therefore µ0 I Bφ = az 4π

Z 2π

sin u du

0

[z 2 + ρ2 + a2 − 2aρ cos u]3/2

56

= 0.

[dl0 × R]ρ = ρ̂ · a dφ0 [−ẑ(y − y 0 ) sin φ0 + ŷz sin φ0 − ẑ(x − x0 ) cos φ0 + x̂z cos φ0 ] = a dφ0 [z cos φ0 cos φ + z sin φ0 sin φ] = a dφ0 z cos(φ − φ0 ).

Therefore µ0 I az Bρ = 4π

Z 2π

cos u du

0

[z 2 + ρ2 + a2 − 2aρ cos u]3/2

.

Change of variables: u = π − 2x =⇒ cos u = −[1 − 2 sin2 x], Z π/2

1 − 2 sin2 x 2 3/2 (2 dx) 2 2 2 −π/2 [z + ρ + a + 2aρ − 4aρ sin x] Z π/2 µ0 I az 1 − 2 sin2 x =− dx. π F 3/2 0 [1 − k 2 sin2 x]3/2

µ0 I Bρ = − az 4π

Write

1 − 2 sin2 x A B .

2 3/2 = 2 3/2 + 2 2 2 [1 − k sin x] [1 − k sin x] [1 − k sin2 x]1/2

Then A + B[1 − k 2 sin2 x] = 1 − 2 sin2 x, A + B = 1, B=

2 , k2

−Bk 2 sin2 x = −2 sin2 x, A=1−

2 k2 − 2 = .

2 k k2

So

µ0 I az E(k 2 ) 2 Bρ = − A + BK(k ) π F 3/2 1 − k2

2 µ0 I az k −2 2 2 E(k ) − 2K(k ) , = − π F 3/2 k 2 1 − k2

k2 =

4aρ .

F

Next use 4aρ 4aρ − 2F k2 − 2 4aρ − 2[a2 + 2aρ + ρ2 + z 2 ] −2[a2 + ρ2 + z 2 ] a2 + ρ2 + z 2 F −2 = = = = = 2 4aρ k2 − 1 4aρ − F 4aρ − [a2 + 2aρ + ρ2 + z 2 ] −a2 + 2aρ − ρ2 − z 2 (a − ρ)2 + z 2 F −1

and the fact that F 3/2 k 2 = 4aρF 1/2 to write 2

µ0 I az a + ρ2 + z 2 2 2 Bρ = 2 E(k ) − 2K(k ) π 4aρF 1/2 (a − ρ)2 + z 2

2

µ0 I z 1 a + ρ2 + z 2 2 2 = E(k ) − K(k ) .

2π ρ F 1/2 (a − ρ)2 + z 2

57

1/2

a 1 2 2 2 1 − k K(k ) − E(k ) , ρ 2

1/2 1 2 µ0 I F 2 2 1 − k K(k ) − E(k ) . = φ̂ 2π ρ 2

µ0 I A(r) = φ̂ πk

B = ∇ × A = ∇ × [φ̂Aφ (ρ, z)] = −ρ̂

k=

2(aρ)1/2 F 1/2

∂Aφ 1 ∂ + ẑ (ρAφ ). ∂z ρ ∂ρ

1 2 µ0 I ∂ 1/2 2 2 1 − k K(k ) − E(k ) F Bz = 2πρ ∂ρ 2

µ0 I 1 ∂ 1/2 1 ∂ = 1 − k 2 K(k 2 ) − E(k 2 ) 1 − k 2 K(k 2 ) − E(k 2 ) .

F + F 1/2 2πρ 2 ∂ρ ∂ρ 2 Let G=

∂ ∂ρ

1 1 ∂ ∂k 1 − k 2 K(k 2 ) − E(k 2 ) = 1 − k 2 K(k) − E(k) .

2 ∂k 2 ∂ρ

Here we have employed the notation K(k), E(k) in order to use derivative formulas from Gradsteyn. k2 =

2k

dk 4aF − 4aρ2(a + ρ) 4aρ =⇒ 2k = .

F dρ F2

dk 4a(a2 + 2aρ + ρ2 + z 2 ) − 8a2 ρ − 8aρ2 = dρ F2 4a = 2 (a2 + 2aρ + ρ2 + z 2 − 2aρ − 2ρ2 ) F 4a = 2 (a2 − ρ2 + z 2 ), F

dk 2a 2 = (a − ρ2 + z 2 ). dρ kF 2

1 2 1 2 dK(k) dE(k) ∂ 1 − k K(k) − E(k) = 1 − k − kK(k) − . ∂k 2 2 dk dk Use Gradsteyn 8.123.2, dK(k) E(k) K(k) = − , dk k[1 − k 2 ] k and Gradsteyn 8.123.4, dE(k) 1 = [E(k) − K(k)]. dk k " # " #

1 − 12 k 2 1 − 21 k 2 ∂ 1 2 1 1 1 − k K(k) − E(k) = E(k) − + K(k) − −k+ ∂k 2 k(1 − k 2 ) k k k " #

1 1 − 12 k 2 − 1 + k 2 1 2 1 2 = E(k) + K(k) −1 + k − k + 1 k 1 − k2 k 2 = E(k)

1 k 1 − K(k) k.

2 21−k 2 58

Hence

2a 2 k E(k) − K(k) (a − ρ2 + z 2 ), G= 2 1 − k2 kF 2

µ0 I 1 2 E(k) ∂ 1/2 2 2 2 1/2 a Bz = 1 − k K(k) − E(k) − K(k) (a − ρ + z ) .

F +F 2πρ 2 ∂ρ F 2 1 − k2 Next use ∂ 1/2 1 ∂F a+ρ F = = 1/2 1/2 ∂ρ ∂ρ 2F F to write

µ0 I 1 1 2 a E(k) 2 2 2 Bz = (a + ρ) 1 − k K(k) − E(k) + − K(k) (a − ρ + z ) 2πρ F 1/2 2 F 1 − k2

1 2 a 2 µ0 I 1 a a2 − ρ2 + z 2 2 2 K(k) (a + ρ) 1 − k − (a − ρ + z ) + E(k) −(a + ρ) + . = 2πρ F 1/2 2 F F 1 − k2 Coefficient of K(k):

1 4aρ a (a + ρ) 1 − − (a2 − ρ2 + z 2 ) 2 F F

1 = (a + ρ)F − 2aρ(a + ρ) − a3 + aρ2 − az 2 F

1 = (a + ρ)(a2 + ρ2 + z 2 + 2aρ) − 2a2 ρ − 2aρ2 − a3 + aρ2 − az 2 F

1 3 = a + ρa2 + aρ2 + ρ3 + az 2 + ρz 2 + 2a2 ρ + 2aρ2 − 2a2 ρ − 2aρ2 − a3 + aρ2 − az 2 F

1 2 ρ(a + 2aρ + ρ2 + z 2 ) = F

ρ = (a + ρ)2 + z 2 F = ρ.

Coefficient of E(k): a a2 − ρ2 + z 2 F 1 − k2 a = −a − ρ + (a2 − ρ2 + z 2 ) F − 4aρ

1 −(a + ρ)[F − 4aρ] + a(a2 − ρ2 + z 2 ) = F − 4aρ

1 = −(a + ρ)(a2 + 2aρ + ρ2 + z 2 ) + 4a2 ρ + 4ρa2 + a3 − ρ2 a + az 2 F − 4aρ ρ (a2 − ρ2 − z 2 ). = F − 4aρ

−a−ρ+

So

µ0 I a2 − ρ 2 − z 2 Bz = ρK(k) + ρ E(k) F − 4aρ 2πρF 1/2 2

a − ρ2 − z 2 µ0 I 2 2 = E(k ) + K(k ) 2πF 1/2 (a − ρ)2 + z 2 59

as before. Lastly,

∂Aφ 1 2 µ0 I ∂ 1/2 2 2 F 1 − k K(k ) − E(k ) Bρ = − =− ∂z 2πρ ∂z 2 (

) 1 2 ∂F 1/2 ∂ 1 µ0 I 1 − k K(k) − E(k) + F 1/2 1 − k 2 K(k) − E(k) . =− 2πρ 2 ∂z ∂z 2 ∂ ∂z

1 ∂ 1 ∂k 1 − k 2 K(k) − E(k) = 1 − k 2 K(k) − E(k) .

2 ∂k 2 ∂z

Use ∂ ∂k

1 2 1 k 1 1 − k K(k) − E(k) = E(k) − K(k) k 2 2 1 − k2 2

k E(k) − K(k) , = 2 1 − k2

k2 =

4aρ dk ∂ 1 ∂F 1 =⇒ 2k = 4aρ F −1 = −4aρ 2 = −8aρz 2 F dz ∂z F ∂z F dk 4aρ 1 k2 z kz =⇒ =− 2 z=− =− , dz F k F k F

and

1 ∂F z ∂F 1/2 = = 1/2 .

1/2 ∂z ∂z 2F F

µ0 I 1 2 z kz k E(k) 1/2 Bρ = − 1 − k K(k) − E(k) − K(k) − F + 2πρ 2 2 1 − k2 F F 1/2

µ0 I z 1 2 1 k2 1 2 =− 1 − k K(k) − E(k) − E(k) + k K(k) 2πρ F 1/2 2 2 1 − k2 2

µ0 I 1 2 1 2 1 k2 z =− K(k) 1 − k + k − E(k) 1 + .

2 2 2 1 − k2 2πF 1/2 ρ

Coefficient of E(k): 1+

1 − k 2 + 12 k 2 1 − 12 k 2 a2 + ρ2 + z 2 1 k2 1 k2 − 2 = = = = .

2 1 − k2 1 − k2 1 − k2 2 k2 − 1 (a − ρ)2 + z 2

So µ0 I Bρ = 2πF 1/2

2

z a + ρ2 + z 2 2 2 E(k ) − K(k ) ρ (a − ρ)2 + z 2

as before.

3.18. Let B(r) be the magnetic field produced by loop 1. As shown in Problem 3.17, this can be written as B(r) = Bz (ρ, z)ẑ + Bρ (ρ, z)ρ̂.

60

The force on loop 2 is given by I I2 dl2 × B

F2 = Γ2

where dl2 × B = a2 dφφ̂ × [Bz (a2 , d)ẑ + Bρ (a2 , d)ρ̂] = a2 dφρ̂Bz (a2 , d) − a2 dφẑBρ (a2 , d).

So Z 2π

Z 2π [x̂ cos φ + ŷ sin φ] dφ − Bρ (a2 , d)a2

F2 = Bz (a2 , d)a2

= −ẑBρ (a2 , d)2πa2 .

The formula for Bρ (ρ, d) is given in the solution to Problem 3.17.

3.19. See Figure 19.

Figure 19

I J = ẑ 2 =⇒ B = πa Maxwell’s stress tensor:

ẑ dφ 0

0

( µ0 Iρ φ̂ 2πa ρ ≤ a, 2, µ0 I φ̂ 2πρ , ρ ≥ a.

1 T̄m = (B · H)Ī − BH. 2

Magnetic force: Z Fm = −

T̄m · dS. S

Thus, the force on the half of the wire occupying y > 0 is Z Z Fm = T̄m · ŷ dx dz. x

z

61

We have



2 2  µ0 Iρ 2 ŷ − φ̂µ0 Iρ 2 cos φ, ρ ≤ a, 2 2πa 2πa 2

T̄m · ŷ =  µ0 I 2 ŷ − φ̂µ I cos φ ρ ≥ a.

0 2πρ 2 2πρ

Note: in the plane y = 0 the relation φ̂ = ŷ cos φ holds. So (

Ix 2 − µ20 2πa ŷ, x ≤ a, 2 T̄m · ŷ =

µ0 I 2 − 2 2πx ŷ, x ≥ a.

Therefore "Z

Z l Fm = −ŷ

dy 0

a

µ0 −a 2

I 2πa2

2

Z −a

µ0 x dx + −∞ 2 2

I 2π

2

dx + x2

"

x3 a 1 −a 1 ∞ − − 3a4 −a x −∞ x a

µ0 I 2 2 1 1 = −ŷ + + 8π 2 3a a a

µ0 I 2 8 µ0 I 2 = −ŷ = −ŷ 2 .

2 8π 3a 3π a

Fm = −ŷ l

µ0 I 2 8π 2

Z ∞ a

µ0 2

I 2π

2

#

This matches equation (3.136).

3.20. (a) Neglecting fringing, we have Φ(r) = Φ(z). Laplace’s equation d2 Φ(z) =0 dz 2 can be integrated twice to yield the general solution Φ(z) = C1 z + C2 .

Then

dΦ(z) = −C1 z. ∂z Boundary condition at z = d is n̂ · D = ρs . Using n̂ = −ẑ and ρs = Q/A, we obtain E = −∇Φ = −ẑ

−ẑ · [− C1 ẑ] = or C1 = Therefore E=−

Q ẑ, A

Q .

A

Φ(z) =

62

Q A

Q z + C2 .

A

# dx , x2

Boundary condition at z = 0 is Φ(0) = 0. This yields C2 = 0, so Φ(z) =

Q z.

A

Next, 1 W = 2

N

Z ρΦ dV + V

1X Qi Vi 2 i=1

where ρ = 0 and N = 2. On the bottom plate, V = 0. On the top plate, Z d Z d Qd Q dz = .

E · dl = V =− A 0 A 0 So 1 W = Q 2

Qd A

=

Q2 d .

2 A

(b) 1 W = 2

Z

1 D · E dV = 2 V

Z dZ Z 0

y

x

Q Q Q2 d − ẑ · − ẑ dx dy dz = .

A A 2 A

(c) dW d F = −ẑ = −ẑ dz dz

Q2 z 2 A

= −ẑ

Q2 .

2 A

(d) See Figure 20.

Figure 20

1 (D · E)Ī − DE · (−ẑ) 2 1 = − D · Eẑ + DEz 2 2

Q 2 Q =− ẑ + ẑ 2 A A 2 Q = ẑ, 2 A2 Z Z I Q2 Q2 Fe = − T̄e · dS = − ẑ dx dy = − ẑ.

2 2 A S x y 2 A T̄e · n̂ =

63

3.21. (a) Neglecting fringing, we have Φ(r) = Φ(z). Laplace’s equation d2 Φ(z) =0 dz 2 can be integrated twice to yield the general solution Φ(z) = C1 z + C2 .

Applying the boundary conditions on Φ at z = 0 and z = d, we find that C2 = 0 and C1 = V0 /d.

Hence V0 Φ(z) = z. d Also, dΦ(z) V0 E = −∇Φ = −ẑ = −ẑ . dz d Next, Z N 1 1X W = ρΦ dV + Qi Vi 2 V 2 i=1

where ρ = 0 and N = 2. On the bottom plate, V = 0. On the top plate, V = V0 . Therefore 1 W = QV0 .

2 Note that on the top plate we have n̂ = −ẑ so that

V0 V0 ρs = n̂ · D = −ẑ · −ẑ = , d d Z Z V0 V0 Q= dx dy = A , d y x d and

1 W = 2

V0 V0 A A V0 = . d 2d

(b) Z 1 W == D · E dV 2 V

Z Z Z 1 d V0 V0 = − ẑ · − ẑ dx dy dz 2 0 y x d d =

AV02 .

2d

(c) dW d F = −ẑ = −ẑ dz dz

AV02 2z

64

= −ẑ z=d

AV02 .

2d2

(d) Refer to Figure 20 again.

1 T̄e · n̂ = (D · E)Ī − DE · (−ẑ) 2 1 = − D · Eẑ + DEz 2

2 V0 2 V0 =− ẑ + ẑ 2 d d 2 V = 02 ẑ, 2d I Z Z AV02 V02 ẑ dx dy = − ẑ.

Fe = − T̄e · dS = − 2 2d2 S x y 2d 3.22. Use Qm =

N X

cmn Vn .

n=1

We can find cmn by setting Vn = 0 for n 6= m and computing the ratio cmn =

Qm V0

with Vm = V0 .

Similarly, we can find cnm by setting Vm = 0 for m 6= n and computing the ratio Qn V0

cnm =

with Vn = V0 .

But Green’s reciprocation theorem states that N X

qk0 Φk =

N X

qk Φ0k .

k=1

k=1

Let the primed case be the first situation considered above, and let the unprimed case be the second situation. Then Qn Vn = Qm Vm .

But Vn = Vm = V0 , so Qn = Qm . Hence cnm = 3.23.

Qm =

N X n=1

Write

cmn Vn =

N X

Qn Qm = cmn = .

V0 V0 "

cmn Vn + Vm

n=1 N X

N X k=1

cmk = Cmm .

k=1

65

cmk − Vm

N X k=1

# cmk .

Then Qm = Cmm Vm +

= Cmm Vm +

= Cmm Vm +

N X

N X

cmn Vn −

n=1

k=1

N X

N X

k=1 N X

cmk Vk −

cmk Vm

cmk Vm

k=1

[−cmk ][Vm − Vk ].

k=1

Since Vm − Vk = 0 when k = m, we can exclude the k = m term from the sum. Let Cmn = −cmn for m 6= n. Then N X Qm = Cmm Vm + Cmk (Vm − Vk ). k=1 k6=m

3.24.

1 W = 2 =

N N N 1X 1X X ρ(r)Φ(r) dV + cin Vn Qi Vi = 2 2 V

Z

i=1

N N 1 XX

2

cin Vn Vi =

i=1 n=1

i=1

N N 1 XX

2

!

Vi

n=1

cmn Vn Vm .

m=1 n=1

3.25. The flux ψm through loop m is Z

Z B(r) · n̂ dS =

ψm =

N X

Bn (r) · n̂ dS,

Sm n=1

Sm

where Bn is the field produced by loop n. Write Bn (r) = ∇ × An (r), where An is the vector potential produced by loop n. Then Z ψm =

N X

[∇ × An (r)] · n̂ dS

Sm n=1

=

=

N Z X

[∇ × An (r)] · n̂ dS

n=1 Sm N I X

An (r) · dl

n=1 Γm

by Stokes’ theorem. Next, substitute An (r) =

µ0 4π

In dl0 0 Γn |r − r |

I

66

to get N I X

I In dl0 µ0 · dl ψm = 4π Γn |r − r0 | n=1 Γm

I I N X µ0 dl · dl0 = In 4π Γm Γn |r − r0 | =

n=1 N X

Lmn In .

n=1

We have Lmn = Lnm by inspection.

3.26. By (3.148) we have N

1X Im ψm .

2

W =

m=1

Using ψm =

N X

Lmn In ,

n=1

we obtain

N

N X

m=1

n=1

1X Im W = 2

!

Lmn In

N

N

1 XX Lmn In Im . = 2 m=1 n=1

3.27. Let J be the actual distribution of current in a conducting body. In this case the dissipated power is Z Z 1 1 1 J · E dV = J · J dV. P = 2 V 2 V σ Now let J0 = J + δJ be any other steady current distribution in the body. The dissipated power becomes Z 1 1 P 0 = P + δP = (J + δJ) · (J + δJ) dV 2 V σ so that Z Z 1 1 1 δP = J · δJ dV + |δJ|2 dV. σ 2 σ V V But Z Z Z 1 J · δJ dV = E · δJ dV = − ∇Φ · δJ dV V σ VZ V = − [∇ · (Φ δJ) − Φ∇ · δJ] dV IV Z = − Φ δJ · dS + Φ∇ · δJ dV = 0.

S

Hence

1 δP = 2

V

Z

1 |δJ|2 dV ≥ 0, σ V

as required.

67

3.28. Use the standard series expansion ln(1 − x) = −

∞ X 1 n=1

n

xn

as follows: R2 = ρ2> − 2ρ< ρ> cos φ + ρ2< = (ρ> − ρ< ejφ )(ρ> − ρ< e−jφ )

ρ< −jφ ρ< jφ 2 e 1− e , = ρ> 1 − ρ> ρ>

ρ< −jφ 1/2 ρ< jφ 1− ln R = ln 1− e e ρ> ρ>

1 ρ< jφ 1 ρ< −jφ = ln ρ> + ln 1 − + ln 1 − e e 2 ρ> 2 ρ>

∞ ∞ 1 X 1 ρ< jφ n 1 X 1 ρ< −jφ n = ln ρ> − e − e 2 n ρ> 2 n ρ> n=1 n=1

∞

1 X 1 ρ< n jnφ e + e−jnφ = ln ρ> − 2 n ρ> n=1

∞ X 1 ρ< n = ln ρ> − cos nφ. n ρ>

∴

ρ2>

n=1

3.29. (a) Φ(r, θ, φ) =

∞ X l h X

i Alm rl + Blm r−(l+1) Ylm (θ, φ).

l=0 m=−l

(b) Φ(r, θ) =

∞ h i X Al rl + Bl r−(l+1) Pl (cos θ). l=0

(c) Since Pl (cos θ) θ=0 = 1, Φ(r = z) =

∞ h X

i Al rl + Bl r−(l+1) .

l=0

(d) If Φ(r = z) can be found as an expansion in integer powers of z for an azimuthally-symmetric problem, then Al and Bl can be determined. (e) We have Q Φ(z) = .

2 2 4π (z + c − 2zc cos α)1/2 If z > c then

∞ 1 1 X c l = Pl (cos α) z z (z 2 + c2 − 2zc cos α)1/2 l=0

68

and

∞

Φ(z) =

Q X c l+1 Pl (cos α).

4π c z l=0

Hence

∞

Φ(r, θ) =

Q X c l+1 Pl (cos α)Pl (cos θ), 4π c r

r > c.

l=0

For r < c,

∞

Q X r l Φ(r, θ) = Pl (cos α)Pl (cos θ), 4π c c

r > c.

l=0

3.30. (a) D = ¯ · E = θ̂

8 0 2 0 + r̂ 2 C/m2 .

2 r r

(b) P = D − 0 E = θ̂

8 0 C/m2 . r2

(c) ρP = −∇ · P = 3.31.

1 E(r) = 4π 0

8 0 cos θ C/m3 . r3 sin θ

1 ρ(r ) −∇ dV 0 R V

Z

0

Z Z 1 1 1 1 0 0 0 dV = − ρ(r )∇ · ∇ dV 0 ∇ · E(r) = − ∇· ρ(r )∇ 4π 0 R 4π R 0

Z V Z V 1 1 1 0 2 0 =− ρ(r )∇ dV = − ρ(r0 )[−4πδ(r − r0 )] dV 0 4π 0 V R 4π 0 V 1 = ρ(r).

0 3.32. See Figure 21.

Figure 21

1 T̄ = (D · E)Ī − DE = 2 69

V02 0 d2

1 Ī − ẑẑ , 2

V02 0 1 V02 0 T̄ · n̂ = Ī − ẑẑ · ẑ = −ẑ , d2 2 2d2 Z V 2 0 −F = T̄ · n̂ dS = −ẑ 0 2 A, 2d S

V 2 0 A F = ẑ 0 2 .

2d 3.33. See Figure 22.

Figure 22

1 T̄e = (D · E) Ī − DE, 2 1 T̄e z=0 = ẑE0 · 0 ẑE0 Ī − 0 E0 ẑE0 ẑ, 2

1 1 T̄e z=0 · n̂ = ẑE0 · 0 ẑE0 Ī − 0 E0 ẑE0 ẑ · ẑ = − 0 E02 ẑ, 2 2 I Fe = − T̄e z=0 · n̂ dS = 5 0 E02 ẑ N. S

3.34. We have ∇2t Φ(x, y) = 0 in the regions y > 0 and y < 0. Write Z ∞ 1 Φ̃(k, y)ejkx dk Φ(x, y) = 2π −∞ and obtain ∇2t Φ = so that

1 2π

Z ∞ −∞

Write the solutions as

∂2 ∂2 + ∂x2 ∂y 2

Φ̃(k, y)ejkx dk = 0

∂2 2 − k Φ̃(k, y) = 0. ∂y 2

( A(k)e−|k|y , y > 0, Φ̃(k, y) = B(k)e|k|y , y < 0.

Boundary condition on continuity of potential: Φ(x, 0+ ) = Φ(x, 0− ) =⇒ Φ̃(k, 0+ ) = Φ̃(k, 0− ) =⇒ A(k) = B(k).

70

Other boundary condition: ŷ · [ 1 E(x, 0+ ) − 2 E(x, 0− )] = ρ0 δ(x) =⇒ =⇒

∂Φ(x, y) ∂Φ(x, y) = 1 + ρ0 δ(x) ∂y ∂y y=0− y=0+ Z ∞ ∂ Φ̃ ∂ Φ̃ 2 = 1 + ρδ(x)e−jkx dx. ∂y y=0− ∂y y=0+ −∞

2

So 2 |k|A(k) = 1 (−|k|)A(k) + ρ0 which gives A(k) = Therefore Φ(x, y) =

1 2π

ρ0 . |k|( 1 + 2 )

Z ∞

ρ0 e∓|k|y ejkx dk |k|( +

) 1 2 −∞

for y ≷ 0.

3.35. ρ/ = ∇ · E # " # " e−r/a e−r/a =∇· r̂ + ∇ · r̂ r2 ra " #

−r/a 1 ∂ 1 2e −r/a + 2 r . = ∇ · −e ∇ r r ∂r ra Use the identity ∇ · (φA) = A · ∇φ + φ∇ · A with φ = −e−r/a and A = ∇(1/r) to get

1 1 1 1 r −r/a −r/a ∇·E=∇ −e ∇· ∇ · ∇ −e + 2 − e−r/a r r r a a2 !

r̂ e−r/a e−r/a e−r/a = − 2 · r̂ + e−r/a 4πδ(r) + 2 − r a r a ra2 = 4πδ(r) −

e−r/a . ra2

So ρ = 4π δ(r) − 3.36. −ρ/ = ∇2 Φ = ∇2

e−r/a . ra2

−ar a

e + ∇2 e−ar . r 2

71

Use ∇2 (φψ) = ∇ · (∇[φψ]) = ∇ · (φ∇ψ + ψ∇φ) = ∇ψ · ∇φ + φ∇ · (∇ψ) + ∇φ · ∇ψ + ψ∇ · (∇φ) = 2∇ψ · ∇φ + φ∇2 ψ + ψ∇2 φ with φ = 1/r and ψ = e−ar to get

a

1 2 −ar 1 −ar −ar 2 1 · ∇(e ) + ∇ (e ) + e ∇ + ∇2 e−ar . ∇ Φ = 2∇ r r r 2 2

Now use

1 r̂ ∇ = − 2, r r

and ∇2 (e−ar ) =

∇(e

−ar

) = −r̂ae

−ar

,

1 ∇ = −4πδ(r), r 2

1 ∂ 2 1 a −ar −ar 2 2 −ar 2 (r (−a)e ) = [−2ar + a r ]e = a − 2 e r2 ∂r r2 r

to find that ∇2 Φ =

a3 −ar e − 4πδ(r).

2

So ρ = 4π δ(r) −

a3 −ar e .

2

3.37.

Z 1 r − r0 E= ρ(r0 ) dV 0 4π 0 V |r − r0 |3 Z ∞Z ∞Z a 1 (x − x0 )x̂ + (y − y 0 )ŷ + (z − z 0 )ẑ = dx0 dy 0 dz 0 . ρ(z 0 ) 4π 0 −∞ −∞ −a [(x − x0 )2 + (y − y 0 )2 + (z − z 0 )2 ]3/2 We have 1 Ex = 4π 0

Z ∞Z a

0

ρ(z ) −∞

−a

Z ∞

u du 2 0 2 0 2 3/2 −∞ [u + (y − y ) + (z − z ) ]

dy 0 dz 0 = 0

by the odd symmetry of the integrand in the u integral. Similarly Ey = 0. The z component is given by

Z a Z ∞ Z ∞ 1 u du 0 0 Ez = ρ(z )(z − z ) dx0 dz 0 2 + (y − y 0 )2 + (z − z 0 )2 ]3/2 4π 0 −a [u −∞ Z a Z−∞ ∞ 2 1 = ρ(z 0 )(z − z 0 ) du dz 0 2 0 2 4π 0 −a −∞ u + (z − z ) Z a 1 z − z0 = π ρ(z 0 ) dz 0 .

2π 0 −a |z − z 0 |

72

In other words,

Z a  1   ρ(z 0 ) dz 0 , z > a,   2  0 −a    1 Z a ρ(z 0 ) dz 0 , z < −a, Ez = − 2  0 −a   Z z Z a   1 1  0 0  ρ(z ) dz − ρ(z 0 ) dz 0 , −a < z < a.  2 0 −a 2 0 z

3.38. Write

1 Φ(r) = 4π 0

Z

η(r0 )

S

1 dS 0 |r − r0 |

where r0 = ar̂0 = ax̂ sin θ0 cos φ0 + aŷ sin θ0 sin φ0 + aẑ cos θ0 ,

r = zẑ,

dS 0 = a2 sin θ0 dθ0 dφ0 .

So r − r0 = −ax̂ sin θ0 cos φ0 − aŷ sin θ0 sin φ0 + (z − a cos θ0 )ẑ, |r − r0 | = [a2 sin2 θ0 cos2 φ0 + a2 sin2 θ0 sin2 φ0 + z 2 − 2az cos θ0 + a2 cos2 θ0 ]1/2 = [a2 sin2 θ0 + a2 cos2 θ0 + z 2 − 2az cos θ0 ]1/2 = [a2 + z 2 − 2az cos θ0 ]1/2 , and we have Z π Z 2π η(θ0 ) 1 0 Φ(r) = dφ a2 sin θ0 dθ0 dφ0 1/2 2 2 0 4π 0 0 0 [a + z − 2az cos θ ] Z π a2 η(θ0 ) sin θ0 dθ0 √ = .

2 0 0 a2 + z 2 − 2az cos θ0 3.39.

J=∇×H

a −ρ/a = ∇ × H0 e φ̂ ρ = ∇ × (ΦA)

where

Φ = H0 ae−ρ/a and A =

1 φ̂. ρ

Use ∇ × (ΦA) = ∇Φ × A + Φ∇ × A and the fact that

∇×

1 φ̂ ρ

=

δ(ρ) ẑ. ρ

(The latter identity can be verified by writing

Z 2π Z b Z 2π 1 1 ∇× φ̂ · ẑρ dρ dφ = φ̂ · φ̂ρ dφ = 2π ρ ρ 0 0 0 73

and

Z 2π Z b 0

0

δ(ρ) ẑ · ẑρ dρ dφ = 2π, ρ

where we have used Stokes’ theorem to obtain the first equality.) We have

1 −ρ/a δ(ρ) −ρ/a ẑ J = ∇ H0 ae × φ̂ + H0 ae ρ ρ 1 δ(ρ) = −H0 e−ρ/a (ρ̂ × φ̂) + H0 ae−ρ/a ẑ ρ ρ

a −ρ/a 1 = H0 e ẑ δ(ρ) − . ρ a The presence of δ(ρ) indicates a line current along the z-axis.

3.40. Write Hp = H0 ẑ = −∇Φpm ,

H = −∇Φm where ∇2 Φm = 0,

and Φm = Φpm + Φsm .

We have Φpm = −H0 r cos θ = −H0 rP1 (cos θ) and can expand Φsm =

∞ X    Gn rn Pn (cos θ),   n=0 ∞ X

    

0 ≤ r < a,

hn r−(n+1) Pn (cos θ),

r ≥ a.

n=0

Boundary condition (1): continuity of Φm at r = a gives ∞ X

Gn an Pn (cos θ) − H0 aP1 (cos θ) =

n=0

∞ X

hn a−(n+1) Pn (cos θ) − H0 aP1 (cos θ).

n=0

Application of orthogonality gives Gn an = hn a−(n+1) so that hn = Gn a2n+1 .

Boundary condition (2): continuity of normal B, r̂ · (B1 − B2 ) = 0 at r = a.

Since B = µH = −µ∇Φm , we get "∞ # X µ Gn nrn−1 Pn (cos θ) − H0 P1 (cos θ) n=0

" = µ0 −

∞ X

# (n + 1)hn r−n−2 Pn (cos θ) − H0 P1 (cos θ)

n=0

r=a

74

r=a

or

∞ X

Gn nan−1 µ + (n + 1)hn a−n−2 µ0 Pn (cos θ) = H0 (µ − µ0 )P1 (cos θ).

n=0

Application of orthogonality gives G1 µ + 2h1 a−3 µ0 = H0 (µ − µ0 ),

h1 = G1 a3 ,

or G1 (µ + 2µ0 ) = H0 (µ − µ0 ), So G1 = H0

hn = Gn = 0 for n 6= 1.

µ − µ0 , µ + 2µ0

h1 = H0 a3

µ − µ0 . µ + 2µ0

Substitution yields

µ − µ0 3µ0 Φm (r ≤ a) = H0 r cos θ − H0 r cos θ = H0 r − cos θ µ + 2µ0 µ + 2µ0 and Φm (r ≥ a) = H0 a

a 2 µ − µ r

0

µ + 2µ0

So H(r ≤ a) =

cos θ − H0 r cos θ.

3µ0 H0 ẑ µ + 2µ0

and a 3 µ − µ a 3 µ − µ 0 0 H(r ≥ a) = −2H0 cos θr̂ + H0 sin θθ̂ − H0 ẑ r µ + 2µ0 r µ + 2µ0 a 3 µ − µ 0 (2 cos θr̂ − sin θθ̂) − H0 ẑ. = −H0 r µ + 2µ0 3.41. We have

µ0 A= 4π

Js (r0 ) dS 0 0 S |r − r |

Z

where r0 = ρ0 ρ̂0 ,

r = zẑ,

|r − r0 | =

q z 2 + ρ0 2 ,

and Js = η0 v = η0 ωρφ̂.

So 0

η ωρ0 φ̂ p0 ρ0 dρ0 dφ0 2 2 0 0 0 z +ρ Z a Z 2π µ0 η0 ωρ0 0 0 0 p = ρ dρ φ̂ dφ0 2 2 0 4π 0 0 z +ρ = 0.

A(0, 0, z) =

µ0 4π

Z 2π Z a

75

Next, µ0 B= 4π

Js (r0 ) × (r − r0 ) 0 dS . |r − r0 |3 S

Z

Here

0

Js (r0 ) × (r − r0 ) = η0 ωρ0 φ̂ × (zẑ − ρ0 ρ̂0 ) = η0 ωρ0 (zρ̂0 + ρ0 ẑ) so we have µ0 B(0, 0, z) = 4π

Z 2π Z a 0

η0 ωρ0

2

0

zρ̂0 + ρ0 ẑ dρ0 dφ0 (z 2 + ρ0 2 )3/2

which simplifies to

µ0 η0 ω 2z 2 + a2 √ B(0, 0, z) = ẑ − 2|z| .

2 z 2 + a2 3.42. We have

Q0 r̂ Q0 r̂ − ∇ · 4π 0 r2 4π 0 ra

Q0 r̂ Q0 r̂ = ∇· − ∇· 2 4π 0 r 4π 0 a r

Q0 Q0 1 ∂ = 4πδ(r) − (r) 4π 0 4π 0 a r2 ∂r ρ = , 0

∇·E=∇·

so

ρ = Q0 δ(r) −

1 .

4πar2

Then Z Q=

ρ dV V

Z

Q0 = Q0 δ(r) dV − 4πa V Z a Q0 = Q0 − 4π dr 4πa 0 = Q0 − Q0

Z 2π Z π Z a 0

0

0

1 2 r sin θ dr dθ dφ r2

= 0.

3.43. (a)

z 2 −1 = 0, n̂ · P = ẑ · ẑP0 a z=±a z=±a

∂ z 2 2z −∇ · P = −P0 − 1 = −P0 2 . ∂z a a

76

(b) By the result of an earlier problem we have Z z Z a 1 1 0 0 ρ(z ) dz − ρ(z 0 ) dz 0 Ez = 2 0 −a 2 0 z Z z Z a 2z 0 2z 0 1 1 P0 2 dz 0 + P0 2 dz 0 =− 2 0 −a a 2 0 z a P0 P0 =− (z 2 − a2 ) + (a2 − z 2 ) 2 2 0 a 2 0 a2

1 z 2 = − P0 −1 .

0 a 3.44. For ρ > a we have Φ(ρ, φ) =

∞ X

ρ−n [An cos nφ + Bn sin nφ].

n=1

The boundary condition gives ∞ X

( V0 , 0 ≤ φ ≤ π, a−n [An cos nφ + Bn sin nφ] = 0, π < φ < 2π. n=1 By orthogonality we find that An = 0 for all n, and that Bn = Therefore Φ(ρ, φ) =

Bn = 0 for all even n,

2V0 n a for odd n. nπ 2V0 X 1 ρ n sin nφ π n a n=1 odd

for ρ > a.

3.45. The general solution takes the form Φ(r, θ) =

∞ X

(Gn rn + Hn r−(n+1) )Pn (cos θ).

n=0

Case 1: r < a. Write Φ(r, θ) =

∞ X

Gn rn Pn (cos θ).

n=0

Impose the boundary condition: ∞ X

( V0 , 0 ≤ θ ≤ π/2, Φ(a, θ) = Gn an Pn (cos θ) = 0, π/2 < θ ≤ π. n=0 77

Apply orthogonality: ∞ X

Gn a

n

Z π/2

Z π

Pm (cos θ) sin θ dθ

Pn (cos θ)Pm (cos θ) sin θ dθ = V0 0

0

n=0

which gives V0 , 2

G0 =

G2n = 0 for n = 1, 2, . . . ,

and G2n+1 = Therefore

∞

Φ(r, θ) =

a2n+1

4n + 3 4n + 4

V0 X r 2n+1 + 2 a n=0

Case 2: r > a. Write Φ(r, θ) =

V0

4n + 3 4n + 4

∞ X

P2n (0).

P2n (0)P2n+1 (cos θ) for r < a.

Hn r−(n+1) Pn (cos θ).

n=0

Impose the boundary condition: ∞ X

( V0 , 0 ≤ θ ≤ π/2, Φ(a, θ) = Hn a−(n+1) Pn (cos θ) = 0, π/2 < θ ≤ π. n=0 Apply orthogonality: ∞ X n=0

−(n+1)

Z π

Z π/2 Pn (cos θ)Pm (cos θ) sin θ dθ = V0

Hn a

0

Pm (cos θ) sin θ dθ 0

which gives H0 =

V0 a, 2

H2n = 0 for n = 1, 2, . . . ,

and H2n+1 = V0 a

2n+2

4n + 3 4n + 4

P2n (0).

Therefore ∞

V0 a X a 2n+2 Φ(r, θ) = + 2 r r n=0

4n + 3 4n + 4

P2n (0)P2n+1 (cos θ) for r > a.

3.46. The primary potential of the dipole is Φp =

p0 cos θ.

4π 0 r2

The secondary potentials are Φs1 =

∞ X

An rn Pn (cos θ) for r < a,

n=0

78

Φs2 =

∞ X

(Bn rn + Cn r−(n+1) )Pn (cos θ) for a < r < b,

n=0

and Φs3 =

∞ X

Dn r−(n+1) Pn (cos θ) for r > b.

n=0

Boundary condition 1 at r = a: Φ1 (r = a) = Φ2 (r = a) ∞ ∞ X X =⇒ An an Pn (cos θ) = (Bn an + Cn a−n−1 )Pn (cos θ) n=0

=⇒ An a

n=0 2n+1

= Bn a

2n+1

+ Cn .

Boundary condition 2 at r = a: D1n (r = a) = D2n (r = a) ∂Φ2 ∂Φ1 = − =⇒ − 0 ∂r ∂r ! ∞ X p0 n−1 =⇒ − 0 − cos θ + An na Pn (cos θ) 2π 0 a2 n=0 ∞ X

! p0 = − − cos θ + [Bn nan−1 − (n + 1)Cn a−n−2 ]Pn (cos θ) 2π 0 a2 n=0 p0 p0 − A1 0 = − B1 + 2C1 a−3 =⇒ 0 2π 0 a3 2π 0 a3 p0 =⇒ ( − 0 ) = B1 a3 − 2C1 − A1 0 a3 .

2π 0 Boundary condition 1 at r = b Φ2 (r = b) = Φ3 (r = b) ∞ ∞ X X n −(n+1) =⇒ (Bn b + Cn b )Pn (cos θ) = Dn b−(n+1) Pn (cos θ) n=0

n=0

=⇒ B1 b + C1 b−2 = D1 b−2 =⇒ B1 b3 + C1 = D1 .

79

Boundary condition 2 at r = b: D2n (r = b) = D3n (r = b) ∂Φ2 ∂Φ3 =⇒ − = − 0 ∂r ∂r ! ∞ X p0 cos θ + =⇒ − − [Bn nbn−1 − (n + 1)Cn b−n−2 ]Pn (cos θ) 2π 0 b3 n=0 ! ∞ X p0 = − 0 − cos θ − Dn (n + 1)b−n−2 Pn (cos θ) 2π 0 b3 n=0 p0 3 = B1 b − 2 C1 + 2 0 D1 . =⇒ ( − 0 ) 2π 0 Hence we have four equations in four unknowns: A1 a3 = B1 a3 + C1 , F = B1 a3 − 2C1 − A1 0 a3 ,

F = ( − 0 )

p0 , 2π 0

B1 b3 = D1 − C1 , F = B1 b3 − 2 C1 + 2 0 D1 .

Solve these for A1 , B1 , C1 , D1 and substitute into the equations

p0 + A1 r cos θ, Φ1 = 4π 0 r2

p0 −2 Φ2 = + B 1 r + C1 r cos θ, 4π 0 r2 and

Φ3 =

p0 −2 + D1 r cos θ.

4π 0 r2

3.47. First show that the potential exterior to the sphere is

a 2 Φ(r, θ) = E0 a − r cos θ. r So 1 ∂Φ ∂Φ E = −∇Φ = −r̂ − θ̂ ∂r r ∂θ

a 3 a 3 = r̂E0 2 + 1 cos θ + θ̂E0 − 1 sin θ. r r Now

Z 2π Z π/2 Fz = ẑ · F = − 0

0

80

ẑ · T̄ · n̂a2 sin θ dθ dφ

where

|E|2 ẑ · T̄ · n̂ = 0 (ẑ · E)(E · n̂) − (ẑ · n̂) 2

|E|2 (ẑ · r̂) . = 0 (ẑ · E)(E · r̂) − 2

So

Z 2π Z π/2 |E|2 Er (Er (ẑ · r̂) + Eθ (ẑ · θ̂)) − Fz = 0 a (ẑ · r̂) sin θ dθ dφ 2 0 0 r=a

Z 2π Z π/2 E 2 + Eθ2 + (ẑ · θ̂)Er Eθ (ẑ · r̂) Er2 − r sin θ dθ dφ = 0 a2 2 0 0 r=a

Z 2π Z π/2 2 Er − Eθ2 2 = 0 a cos θ − Er Eθ sin θ sin θ dθ dφ.

2 0 0 r=a 2

But Eθ (a, θ) = 0 by the boundary condition on tangential electric field, and Er (a, θ) = 3E0 cos θ, so Z Z 9 0 a2 E02 2π π/2 cos3 θ sin θ dθ dφ Fz = 2 0 0

π/2 9 0 a2 E02 1 4 = 2π − cos θ 2 4 0 =

9π 0 a2 E02 .

4

3.48. The shielding effectiveness on the axis of the hole is given by (3.232):   Ezs (0, z) 1  −1 a 1  (z < 0). =− tan − |z| E0 π |z| + a a

|z|

We need to compute the limit of this expression as z → 0− . Use lim tan−1

z→0−

a π = |z| 2

and lim

1

z→0− |z| + a a |z|

=

" lim |z| lim

z→0−

z→0− |z| a

1 = [0] a =0 to give

Ezs (0, z) 1 =− .

E0 2 z→0− lim

81

1 2

+a

#

3.49. The shielding effectveness on the axis of the hole is given by (3.317):   1  −1 a Ezs (0, z) 1  =− (z < 0). tan − |z| E0 π |z| + a |z|

a

When |z| a we can use the small argument approximation for the arc tangent 1 tan−1 (x) ≈ x − x3 3 to give   1a Ezs (0, z) 1 a3 1  = −1 ≈− − − |z| 3 a E0 π |z| 3 |z| π + |z|

a

" #! a 1 1 a3 a . − − |z| 3 |z|3 |z| 1 + a22 |z|

Using the first two terms of the binomial series we have for |a/z| 1 the approximation 1 1+

a2 |z|2

≈1−

a2 . |z|2

This gives Ezs (0, z) 1 ≈− E0 π Thus

SE ≈ 20 log10

a a a3 1 a3 − + 3− |z| |z| |z| 3 |z|3

3π |z|3 2 a3

= 20 log10

3π 2

=−

2 a3 .

3π |z|3

+ 20 log10

z 3 a

or SE ≈ 13.46 dB + 60 log10 |z/a|

(|z/a| 1).

3.50. That the dielectric shell provides no shielding when the field is z-directed can be argued on purely physical grounds. If E is z-directed, then all the dipoles induced in the dielectric will be z-directed, and we can write the polarization as P(r) = ẑPz (ρ, φ).

Thus, the equivalent polarization volume charge density in the dielectric shell is ρP = −∇ · P =

∂ Pz (ρ, φ) = 0. ∂z

Similarly, the equivalent polarization surface charge densities on the inner and outer surfaces of the dielectric shell are ρP s |ρ=a = n̂ · P(a, φ) = −ρ̂ · ẑPz (a, φ) = 0 and ρP s |ρ=b = n̂ · P(b, φ) = ρ̂ · ẑPz (b, φ) = 0.

Since there is no equivalent polarization charge, by (3.74) there is no scattered potential. Thus, the field everywhere is identical to the incident field, and the shell provides no shielding whatsoever.

82

3.51. We wish to solve the following four equations to determine D: Da = BI1 (γa) + CK1 (γa) A + µ0 H̃0 b = BI1 (γb) + CK1 (γb) b

γa 0 Da = BI1 (γa) + CK10 (γa) µr

A γb 0 − + µ0 H̃0 b = BI1 (γb) + CK10 (γb) . b µr

(*) (**) (***) (****)

Subtracting (***) from (*) and isolating C gives C = −f B where f=

0 I1 (γa) − γa N µr I1 (γa) . = 0 M K1 (γa) − γa µr K1 (γa)

Substituting C = −f B into (**) and (****) and subtracting gives

γb γb 0 0 2µ0 H̃0 b = B I1 (γb) − f K1 (γb) + I1 (γb) − f K1 (γb) . µr µr

(A)

Substituting C = −f B into (*) gives Da = B [I1 (γa) − f K1 (γa)] .

Dividing (B) by (A) and substituting f = M/N gives Da M I1 (γa) − N K1 (γa) = 2µ0 H̃0 b M I1 (γb) − N K1 (γb) − µγbr M I10 (γb) µγbr N K10 (γb) Substituting M and N and factoring the denominator we have Da U = V 2µ0 H̃0 b where U=

γa K1 (γa)I10 (γa) − I1 (γa)K10 (γa) µr

γa 0 γb 0 V = K1 (γa) − K (γa) I1 (γb) + I1 (γb) − µr 1 µr

γb 0 γa 0 I1 (γa) − I (γa) K1 (γb) + K1 (γb) . µr 1 µr Finally, using the Wronskian relation (E.93) I10 (z)K1 (z) − I1 (z)K10 (z) = and simplifying, gives (3.247).

83

1 z

(B)

Chapter 4 4.1. By (4.35), Z ∞

Im ˜c (r, Ω) dΩ Ω−ω −∞ Z 0 Z ∞ 1 Im ˜c (r, Ω) 1 Im ˜c (r, Ω) = − P.V. dΩ − P.V. dΩ. π Ω−ω π Ω−ω −∞ 0

1 Re ˜ (r, ω) − 0 = − P.V. π c

In the first integral let x = −Ω: 1 Re ˜ (r, ω) − 0 = − P.V. π c

Z 0

Im ˜c (r, −x) 1 (−dx) − P.V. −x − ω π ∞

Z ∞ 0

Im ˜c (r, Ω) dΩ. Ω−ω

Use symmetry condition (4.27), Im ˜c (r, −ω) = − Im ˜c (r, ω), to get Z ∞ Z ∞ 1 − Im ˜c (r, Ω) 1 Im ˜c (r, Ω) Re ˜ (r, ω) − 0 = − P.V. dΩ − P.V. dΩ π −Ω − ω π Ω−ω 0

Z0 ∞ 1 1 1 = − P.V. Im ˜c (r, Ω) + dΩ π Ω+ω Ω−ω 0

Z ∞ Ω−ω+Ω+ω 1 c Im ˜ (r, Ω) dΩ = − P.V. π (Ω + ω)(Ω − ω) Z0 ∞ Ω Im ˜c (r, Ω) 2 = − P.V. dΩ. π Ω2 − ω 2 0 c

This is (4.37). Next examine (4.36): Z ∞ 1 Re ˜c (r, Ω) − 0 σ0 (r) c Im ˜ (r, ω) = P.V. dΩ − π Ω−ω ω −∞ Z 0 Z ∞ c 1 1 σ0 (r) Re ˜ (r, Ω) − 0 Re ˜c (r, Ω) − 0 = P.V. dΩ + P.V. dΩ − . π Ω−ω π Ω−ω ω −∞ 0 Let x = −Ω in the first integral: Z 0 Z ∞ Re ˜c (r, −x) − 0 Re ˜c (r, Ω) − 0 1 1 σ0 (r) c Im ˜ (r, ω) = P.V. (−dx) + P.V. dΩ − . π −x − ω π Ω−ω ω ∞ 0 Use the symmetry condition Re ˜c (r, −ω) = Re ˜c (r, ω):

Z ∞ 1 1 σ0 (r) 1 c c [Re ˜ (r, Ω) − 0 ] − Im ˜ (r, ω) = P.V. + dΩ − π Ω+ω Ω−ω ω −∞

Z ∞ 1 −Ω + ω + Ω + ω σ0 (r) = P.V. [Re ˜c (r, Ω) − 0 ] dΩ − π (Ω + ω)(Ω − ω) ω −∞ Z ∞ c 2ω Re ˜ (r, Ω) − 0 σ0 (r) = P.V. dΩ − .

2 − ω2 π Ω ω −∞ This is (4.38).

84

4.2. We have ˜2xx = ˜2yy = 0

ωp2 (ω − jν) 1− ω[(ω − jν)2 − ωc2 ]

!

!

= 0

ωp2 (ω − jν) 1− ω[ω 2 − 2jνω − ν 2 − ωc2 ]

= 0

ωp2 (ω − jν) 1− ω[(ω 2 − ν 2 − ωc2 ) − 2jνω]

= 0

ωp2 (ω − jν)[(ω 2 − ν 2 − ωc2 ) + 2jνω] 1− ω[(ω 2 − ν 2 − ωc2 )2 + 4ν 2 ω 2 ]

= 0 = 0

!

!

! ωp2 [ω(ω 2 − ν 2 − ωc2 ) + 2ν 2 ω] + jωp2 [2νω 2 − 2νω(ω 2 − ν 2 − ωc2 )] 1− ω[(ω 2 − ν 2 − ωc2 )2 + 4ν 2 ω 2 ] ! ωp2 [ω(ω 2 − ν 2 − ωc2 ) + 2ν 2 ω] ωp2 [2νω 2 − 2νω(ω 2 − ν 2 − ωc2 )] 1− +

.

0 ω[(ω 2 − ν 2 − ωc2 )2 + 4ν 2 ω 2 ] jω[(ω 2 − ν 2 − ωc2 )2 + 4ν 2 ω 2 ]

Next, ωp2 ωc ω[ω 2 − 2jνω − ν 2 − ωc2 ] ωp2 ωc = −j 0 ω[(ω 2 − ν 2 − ωc2 ) − 2jνω] ωp2 ωc [(ω 2 − ν 2 − ωc2 ) + 2jνω] = −j 0 ω[(ω 2 − ν 2 − ωc2 )2 + 4ν 2 ω 2 ] 2 ωp2 ωc2 2νω ωp ωc [(ω 2 − ν 2 − ωc2 ) + 2jνω] + 0 . = 0 jω[(ω 2 − ν 2 − ωc2 )2 + 4ν 2 ω 2 ] ω[(ω 2 − ν 2 − ωc2 )2 + 4ν 2 ω 2 ]

˜cxy = −˜ cyx = −j 0

Next, !

˜czz = 0

ωp2 (ω + jν) 1− ω(ω 2 + ν 2 )

!

= 0

ωp2 ω 1− ω(ω 2 + ν 2 )

+ 0

ωp2 ν . jω[ω(ω 2 + ν 2 )]

Finally, ˜czx = ˜cxz = ˜czy = ˜cyz = 0.

Hence we can write [˜¯ c ] = [˜¯ ] + where ˜xx = ˜yy = 0

˜] [σ̄ jω

ωp2 [ω(ω 2 − ν 2 − ωc2 ) + 2ν 2 ω] 1− ω[(ω 2 − ν 2 − ωc2 )2 + 4ν 2 ω 2 ] 85

! ,

σ̃xx = σ̃yy = 0

ωp2 [2νω 2 − 2νω(ω 2 − ν 2 − ωc2 )] , [(ω 2 − ν 2 − ωc2 )2 + 4ν 2 ω 2 ]

˜xy = −˜ yx = 0

ωp2 ωc2 2νω , ω[(ω 2 − ν 2 − ωc2 )2 + 4ν 2 ω 2 ]

ωp2 ωc [(ω 2 − ν 2 − ωc2 ) + 2jνω] , σ̃xy = −σ̃yx = 0 ω[(ω 2 − ν 2 − ωc2 )2 + 4ν 2 ω 2 ] ! ωp2 ω ˜zz = 0 1 − , ω(ω 2 + ν 2 ) σ̃zz = 0

ωp2 ν , [ω(ω 2 + ν 2 )]

and ˜zx = ˜xz = ˜zy = ˜yz = σ̃zx = σ̃xz = σ̃zy = σ̃yz = 0. ˜ ] are real matrices, and since [˜¯ ] = −[˜¯ ]T and [σ̄ ˜ ] = −[σ̄ ˜ ]T , we see that [˜¯ ] and [σ̄ ˜] Since [˜¯ ] and [σ̄ are hermitian.

4.3. Debye formulas: Re ˜(ω) − ∞ =

s − ∞ , 1 + ω2τ 2

Im ˜(ω) = −

ωτ ( s − ∞ ) .

1 + ω2τ 2

Note that we must modify the Kramers–Kronig relations slightly because lim ˜c (ω) = ∞

ω→∞

(not 0 ).

So Z ∞

Re ˜(ω) − ∞ dΩ Ω−ω −∞ Z ∞ s − ∞ 1 1+Ω2 τ 2 = P.V. dΩ π −∞ Ω − ω Z ω−δ s − ∞ Z ∞ s − ∞ 1 1 1+Ω2 τ 2 1+Ω2 τ 2 = lim dΩ + lim dΩ. π δ→0 −∞ Ω − ω π δ→0 ω+δ Ω − ω

Im ˜(ω) =

1 P.V. π

To compute the integral, consider the contour integral I s − ∞ 1 dΩ

I= π τ 2 C 1 + j τ1 1 − j τ1 (Ω − ω) where C is shown in Figure 23. We have (informally) I Z Z Z Z = + + + = πj res(Ω = ω) + 2πj res(Ω = j τ1 ).

C C1 C2 Cω C∞ |{z} |{z} =0

=0

86

Figure 23 Since

Z

Z ∞

Z +

C1

,

= P.V. −∞

C2

we have

s − ∞ 1 1 πj res(Ω = ω) + 2πj res(Ω = j ) τ π τ2 " # s − ∞ 1 1 2 = (πj) + π τ2 (ω + j τ1 )(ω − j τ1 ) (2j τ1 )(j τ1 − ω) " # 1 s − ∞ 1 2 = j − 1 τ2 (ω − j τ1 ) ω + j τ1 2j τ

Im ˜(ω) =

=

s − ∞ 1 −ωτ · 1 · τ2 ω − j τ ω + j τ1

=−

ωτ ( s − ∞ ) .

1 + ω2τ 2

Next, Z ∞ 1 Im ˜c (Ω) dΩ Re ˜ − ∞ = − P.V. π −∞ Ω − ω Z ∞ 1 −Ωτ ( s − ∞ ) = − P.V. dΩ π (1 + Ω2 τ 2 )(Ω − ω) −∞ c

By the same procedure as above we obtain Re ˜c − ∞ =

1 τ ( s − ∞ ) · πj res(Ω = ω) + 2πj res(Ω = j τ1 ) 2 π τ

where the residues are found for the function f (Ω) =

Ω . (Ω + j τ1 )(Ω − j τ1 )(Ω − ω) 87

So " # 1 2j 1 τ ( −

) ω s ∞ τ Re ˜c − ∞ = · (πj) + π τ2 (ω + j τ1 )(ω − j τ1 ) (2j τ1 )(j τ1 − ω) # " 2j τ1 j ω τ ( s − ∞ ) · − 1 = τ2 (ω − j τ1 ) ω + j τ1 2j τ # " ω − ω − j τ1 τ ( s − ∞ ) j = · τ2 (ω − j τ1 ) ω + j τ1 s − ∞ = 2 2 . ω τ +1 4.4. Extend (2.12) and (2.13) to the frequency domain. Then, for the original problem we have D̃1 = ˜¯ 1 · Ẽ1 + ˜ξ̄ 1 · H̃1 , ˜ · Ẽ + µ̄ ˜ · H̃ .

B̃ = ζ̄ 1

1

1

(1) (2)

1

1

For the dual problem we have D̃2 = ˜¯ 2 · Ẽ2 + ˜ξ̄ 2 · H̃2 , ˜ · Ẽ + µ̄ ˜ · H̃ .

B̃ = ζ̄ 2

2

2

(3) (4)

2

2

Now substitute the solution to the dual problem (4.181)–(4.184) into (1): !

B̃2 Ẽ 2 = ˜¯ · (−η0 H̃2 ) + ˜ξ̄ 1 · −η0 η0 or

B̃2 = (η02˜¯ 1 ) · H̃2 + (−˜ξ̄ 1 ) · Ẽ2 .

Compare this to (4). We find that ˜ = −˜ξ̄ , ζ̄ 2 1

˜ 2 = η02˜¯ 1 . µ̄

Next substitute (4.181)–(4.184) into (2): Ẽ2 η0

˜ · (−η H̃ ) + µ̄ ˜1 · η0 D̃2 = ζ̄ 0 2 1 or ˜ ) · H̃ + D̃2 = (−ζ̄ 2 1

˜1 µ̄ η02

· Ẽ2 .

Comparison to (3) shows that ˜ ¯ 2 =

˜1 µ̄ , η02

˜ξ̄ = −ζ̄ ˜ .

2 1

88

!

4.5. ∇(e

−jk·r

∂ ∂ ∂ e−jkx x e−jky y e−jkz z ) = x̂ + ŷ + ẑ ∂x ∂y ∂z ∂ ∂ ∂ = x̂e−jky y e−jkz z e−jkx x + ŷe−jkx x e−jkz z e−jky y + ẑe−jkx x e−jky y e−jkz z ∂x ∂y ∂z = (−jkx x̂ − jky ŷ − jkz ẑ)e−jkx x e−jky y e−jkz z

= −j(x̂kx + ŷky + ẑkz )e−j(kx x+ky y+kz z) = −jke−jk·r . ∇ · (Ee−jk·r ) = e−jk·r ∇ · E + E · ∇(e−jk·r ) = E · (−jke−jk·r ) = −jk · Ee−jk·r . ∇ × (Ee−jk·r ) = e−jk·r ∇ × E − E × ∇(e−jk·r ) = −E × (−jke−jk·r ) = −jk × Ee−jk·r . ∇2 (Ee−jk·r ) = e−jk·r ∇2 E + E∇2 (e−jk·r ) + 2(∇e−jk·r · ∇)E = E∇2 (e−jk·r ) = E∇ · (∇e−jk·r ) = E∇ · (−jke−jk·r )

∂ −jk·r ∂ −jk·r ∂ −jk·r = E −jkx e − jky e − jkz e ∂x ∂y ∂z = E[(−jkx )(−jkx ) + (−jky )(−jky ) + (−jkz )(−jkz )]e−jk·r = −E(kx2 + ky2 + kz2 )e−jk·r = −k 2 Ee−jk·r .

4.6. First,

σ σ ˜c = 0 r − j = Re ˜c + j Im ˜c =⇒ Re ˜c = r 0 , Im ˜c = − . ω 0 ω r p σ √ c k = ω µ˜ = ω µ0 0 r − j = β − jα. ω 0 By (4.215), v v "s s u # u

2 u u1 c 2 p [Im ˜ ] ω√ 1 t σ t c β = ω µ0 Re ˜ 1+ +1 = r √ 1+ 1+ .

2 [Re ˜c ]2 c ω 0 r 2 89

We have vp =

ω . β

vg =

dω , dβ

To find

write √ r β = √ ωf (ω), c 2

2 #1/2 A g(ω) = 1 + , ω "

1/2

f (ω) = [1 + g(ω)]

,

A=

σ .

0 r

√

r 0 dβ = √ ωf (ω) + f (ω) . dω c 2 1 1 g 0 (ω) f 0 (ω) = [1 + g(ω)]−1/2 g 0 (ω) = .

2 2 f (ω) " 2 #−1/2 A A2 1 .

1+ (−2A2 ω −3 ) = − 3 g 0 (ω) = 2 ω ω g(ω) The ω-β diagram is shown in Figure 24. The velocities vp and vg are shown in Figure 25. It appears

3

f (GHz)

Light Line: ε=εr ε0 2

1

0

0

100

200

β (r/m)

300

400

500

Figure 24 that vg > vp at all frequencies, but we should check for low frequencies. For small ω we have r r r r √ 1 ω σ ω 1 σ ω 1 σ σ −j k≈ = √ −j √ =⇒ β = ω . c ω 0 c 2 ω 0 c 2 ω 0 c 2 0 r ω √ 2 0 . vp = = ωc β σ 90

0.15

vg

v/c

0.10

vp

0.05

0.00

6

7

8

9

10

11

log 10 (f)

Figure 25 r dβ 1 −1/2 1 σ = ω , dω 2 c 2 0 r √ 2 0 dω = 2 ωc = 2vp . vg = dβ σ Thus, since vg > vp at all frequencies, this model of water exhibits anomalous dispersion!

From Figure 25 we see that the relaxation effect shifts the region over which vg differs the most from vp up in frequency significantly. From Figure 24 we see that relaxation model produces a nonzero velocity at low frequencies.

4.7. For perpendicular polarization we have ki = (x̂ sin θi + ẑ cos θi )k1 ,

kr = (x̂ sin θi − ẑ cos θi )k1 ,

kt = (x̂ sin θt + ẑ cos θt )k2 .

Use (4.261): i

Ẽi⊥ = ŷẼ0 e−jk ·r = ŷẼ0 e−jk1 (x sin θi +z cos θi ) , r

Ẽr⊥ = ŷẼ0 Γ̃⊥ e−jk ·r = ŷẼ0 Γ̃⊥ e−jk1 (x sin θi −z cos θi ) , t

Ẽt⊥ = ŷẼ0 T⊥ e−jk ·r = ŷẼ0 T̃⊥ e−jk2 (x sin θt +z cos θt ) .

Use the equations following (4.261): Ẽ0 k1 i k̂ × Ẽi⊥ = (ẑ sin θi − x̂ cos θi )e−jk1 (x sin θi +z cos θi ) , k1 η1 η1 k1 r Ẽ0 Γ̃⊥ H̃r⊥ = k̂ × Ẽr⊥ = (ẑ sin θi + x̂ cos θi )e−jk1 (x sin θi −z cos θi ) , k1 η1 η1 k2 t Ẽ0 T̃⊥ H̃t⊥ = k̂ × Ẽt⊥ = (ẑ sin θt − x̂ cos θt )e−jk2 (x sin θt +z cos θt ) . k2 η2 η2

H̃i⊥ =

91

Since regions 1 and 2 are lossless, and θi < θc , we know that Γ̃⊥ and T̃⊥ are real numbers.

In region 1 we have Ẽtot = Ẽi⊥ + Ẽr⊥ h i = ŷẼ0 e−jk1 x sin θi e−jk1 z cos θi + Γ̃⊥ ejk1 z cos θi h i = ŷẼ0 e−jk1 x sin θi (1 + Γ⊥ )e−jk1 z cos θi + 2jΓ⊥ sin(k1 z cos θi ) and, similarly, H̃tot = ẑ

i Ẽ0 −jk1 x sin θi h e (1 + Γ̃⊥ )e−jk1 z cos θi + 2j Γ̃⊥ sin(k1 z cos θi ) η1 h i Ẽ0 + x̂ e−jk1 x sin θi (−1 + Γ̃⊥ )e−jk1 z cos θi + 2j Γ̃⊥ sin(k1 z cos θi ) . η1

Note the traveling and standing wave terms in each final expression. In region 2 we have Ẽtot = Ẽt⊥ = ŷẼ0 T̃⊥ e−jk2 x sin θt e−jk2 z cos θt and H̃tot = H̃t⊥ =

Ẽ0 T̃⊥ (ẑ sin θt − x̂ cos θt )e−jk2 x sin θt e−jk2 z cos θt . η2

These represent pure traveling waves. The case of parallel polarization proceeds similarly.

4.8. Referring to the results of Problem 4.7, we have 1 Re[Ětot × Ȟ∗tot ] 2 1 |Ě0 |2 = x̂ Re[e−jk1 z cos θi + Γ⊥ ejk1 z cos θi ][ejk1 z cos θi + Γ⊥ e−jk1 z cos θi ] sin θi 2 η1 1 |Ě0 |2 − ẑ Re[e−jk1 z cos θi + Γ⊥ ejk1 z cos θi ][−ejk1 z cos θi + Γ⊥ e−jk1 z cos θi ] cos θi .

2 η1

Sav,1 =

Then ẑ · Sav,1 = −

1 |Ě0 |2 Re[−1 + Γ2⊥ − Γ⊥ e2jk1 z cos θi + Γ⊥ e−2jk1 z cos θi ] cos θi .

2 η1

At z = 0, ẑ · Sav,1 =

1 |Ě0 |2 Re(1 − Γ2⊥ ) cos θi .

2 η1

In region 2, ẑ · Sav,2 =

1 |Ě0 |2 T⊥2 cos θt .

2 η2

Continuity of ẑ · Sav at z = 0 would require that 1 |Ě0 |2 1 |Ě0 |2 T⊥2 Re(1 − Γ2⊥ ) cos θi = cos θt .

2 η1 2 η2 92

In view of the fact that T⊥ = 1 + Γ⊥ , this reduces to 1 1 (1 − Γ⊥ ) cos θi = (1 + Γ⊥ ) cos θi . η1 η2 Next use Z1⊥ =

η1 , cos θi

Z2⊥ =

η2 , cos θt

to get 1 − Γ⊥ 1 + Γ⊥ = Z1⊥ Z2⊥ which yields Γ⊥ =

Z2⊥ − Z1⊥ .

Z2⊥ + Z1⊥

We see that ẑ · Sav is continuous at z = 0. The case of parallel polarization is similar.

4.9. For perpendicular polarization we have ki = (x̂ sin θi + ẑ cos θi )k1 ,

kr = (x̂ sin θi − ẑ cos θi )k1 ,

where αc =

kt = x̂ sin θi − jẑαc ,

q k12 sin2 θi − k22 .

Use (4.261): i

Ẽi⊥ = ŷẼ0 e−jk ·r = ŷẼ0 e−jk1 (x sin θi +z cos θi ) , r

Ẽr⊥ = ŷẼ0 Γ̃⊥ e−jk ·r = ŷẼ0 Γ̃⊥ e−jk1 (x sin θi −z cos θi ) , t

Ẽt⊥ = ŷẼ0 T̃⊥ e−jk ·r = ŷẼ0 T̃⊥ e−jk2 x sin θt e−αc z .

Use the equations following (4.261): k1 i Ẽ0 k̂ × Ẽi⊥ = (ẑ sin θi − x̂ cos θi )e−jk1 (x sin θi +z cos θi ) , k1 η1 η1 k1 r Ẽ0 Γ̃⊥ k̂ × Ẽr⊥ = (ẑ sin θi + x̂ cos θi )e−jk1 (x sin θi −z cos θi ) , H̃r⊥ = k1 η1 η1 1 t Ẽ0 T̃⊥ H̃t⊥ = k × Ẽt⊥ = (ẑk1 sin θi + j x̂αc )e−jk2 x sin θt e−αc z . k2 η2 k2 η2 H̃i⊥ =

Since regions 1 and 2 are lossless, and θi < θc , we know that Γ̃⊥ and T̃⊥ are real numbers.

In region 1 we have Ẽtot = Ẽi⊥ + Ẽr⊥ h i = ŷẼ0 e−jk1 x sin θi e−jk1 z cos θi + Γ̃⊥ ejk1 z cos θi h i = ŷẼ0 e−jk1 x sin θi ejφ⊥ /2 e−j(k1 z cos θi +φ⊥ /2) + ej(k1 z cos θi +φ⊥ /2) = ŷẼ0 e−jk1 x sin θi ejφ⊥ /2 2 cos(k1 z cos θi + φ⊥ /2).

93

Similarly, H̃tot = ẑ

Ẽ0 −jk1 x sin θi jφ⊥ /2 e e 2 cos(k1 z cos θi + φ⊥ /2) sin θi η1 Ẽ0 + x̂ e−jk1 x sin θi ejφ⊥ /2 2j sin(k1 z cos θi + φ⊥ /2) cos θi . η1

These are pure standing waves. In region 2 we have the evanescent wave expressions Ẽtot = Ẽt⊥ = ŷẼ0 T̃⊥ e−jk2 x sin θi e−αc z and H̃tot = H̃t⊥ =

Ẽ0 T̃⊥ (ẑk1 sin θi + j x̂αc )e−jk2 x sin θt e−αc z . k2 η2

The case of parallel polarization proceeds similarly.

4.10. We have ẑ · Sav,1 = ẑ ·

1 Re[Ětot × Ȟ∗tot ] 2

1 ∗ Re[Ětot,y Ȟtot,x ] 2 1 = Re{|Ě0 |2 2 cos(k1 z cos θi + φ⊥ /2)[−j2 sin(k1 z cos θi + φ⊥ /2)] cos θi } 2 = 0. =

The rest of the problem is similar.

4.11. In region 1 we have ˜1 = 0 1r , In region 2,

(a)

µ̃1 = µ0 µ1r .

σ ˜c2 = 2r 0 − j , ω

µ̃2 = µ0 µ2r .

p √ √ µ̃1 ˜1 = ω µ0 0 µ1r 1r , r p σ √ √ . k2 = ω µ̃2 ˜c2 = ω µ0 0 µr2 r2 1 − j ω 0 2r k1 = β1 = ω

The definition of a good conductor is through the inequality σ 1. ω 0 2r So

r

σ 1−j ≈ ω 0 2r

r −j

σ 1−j = √ ω 0 2r 2

94

r

σ . ω 0 2r

This implies r

σ √ 1−j µ2r 2r √ ω 0 2r 2 r r p ωσ ω µ̃2 σ = µ̃2 (1 − j) = (1 − j) 2 2 √

k2 = β2 − jα2 ≈ ω µ0 0

or

r β 2 = α2 =

ω µ̃2 σ .

2

Examining the ratio q

ω µ̃2 σ

β2 2 ≈ √ = β1 ω µ̃1 ˜1

s

µ̃2 σ = 2ω µ̃1 ˜1

r

σ ω 0 2r

r

µ2r 2r 1 √ 1 µ1r 1r 2

we find that β2 β1 .

Thus A = β22 − α22 − (β12 − α12 ) sin2 θi ≈ −β12 sin2 θi , B = 2(β2 α2 − β1 α1 sin2 θi ) ≈ 2β22 , √ τ t = (A2 + B 2 )1/4 ≈ 2β2 .

Then

1 1 2β22 −1 B −1 γ = tan ≈ tan 2 A 2 −β12 sin2 θi 1 = tan−1 [−∆] where ∆ 1 2 = −π/4. t

Finally, −1

θt = tan Since

β1 sin θi τ t cos γ t

.

β1 β1 ≈√ 1, τt 2β2

we have θt ≈ tan−1 δ

where δ 1.

Hence θt ≈ 0. (b) For perpendicular polarization, t

i −jk ·r Ẽ2 = ŷT̃⊥ Ẽ⊥ e ,

kt = k02 + jk002 ≈ ẑβ2 (1 − j).

95

So i −jβ2 z −β2 z Ẽ2 ≈ ŷT̃⊥ Ẽ⊥ e e ,

i −jβ2 z −β2 z J̃2 = σ Ẽ2 = ŷσ T̃⊥ Ẽ⊥ e e .

The current per unit width is J˜s =

Z ∞

J̃2 · ŷ dz Z ∞ i e−(1+j)β2 z dz = σ T̃⊥ Ẽ⊥ 0

0

σ i −(1+j)β2 z ∞ = T̃⊥ Ẽ⊥ [e ]0 −(1 + j)β2 1 σ i T̃⊥ Ẽ⊥ . = 1 + j β2 But

1 1−j = , 1+j 2

so i K̃ = ŷσ T̃⊥ Ẽ⊥

1−j .

2β2

Now i k t T̃⊥ Ẽ⊥ e−jβ2 z e−β2 z H̃2 ≈ −x̂ z k2 η2 i T̃⊥ Ẽ⊥ ≈ −x̂ e−jβ2 z e−β2 z . η2

Here s η2 =

µ̃2 = ˜c2

s

µ̃2 = ˜2r 0 − j ωσ

r

1 µ̃2 q 2r 0 1 − j

r σ ω 0 2r

≈

µ2r µ0 1 + j √ 2r 0 2

r

ω 0 2r = (1 + j) σ

So i T̃⊥ Ẽ⊥ H̃2 ≈ −x̂ 1+j

r

2σ −jβ2 z −β2 z e e ω µ̃2 r 2σ i 1−j = −x̂T̃⊥ Ẽ⊥ 2 ω µ̃2 i 1−j σ = −x̂T̃⊥ Ẽ⊥ 2 β2

since

r β2 =

ω µ̃2 σ .

2

Thus, at z = 0, i −ẑ × H̃t ≈ −ẑ × (−x̂)T̃⊥ Ẽ⊥

1−j σ i 1−j ≈ K̃ = ŷσ T̃⊥ Ẽ⊥ .

2 β2 2β2 96

r

ω µ̃2 .

2σ

(c) Zs =

i T̃⊥ Ẽ⊥ Ẽy 2β2 = (1 + j)β2 . = = 1−j i (1 − j)σ H̃x z=0 T̃⊥ Ẽ⊥ 2β2 σ

Let δ = 1/β2 be the skin depth. Then Zs = (1 + j)

1 = Rs + jXs , σδ

Rs = Xs =

1 . σδ

(d) At z = 0, Sav =

1 Re[Ě × Ȟ∗ ]|z=0 2

i 1−j σ −x̂T̃⊥ Ě⊥ 2 β2

i = ŷT̃⊥ Ě⊥ × i 2 = ẑ|T̃⊥ Ě⊥ |

σ .

2β2

But i Ǩ = ŷσ T̃⊥ Ě⊥

So i 2 | Ǩ · Ǩ∗ = σ 2 |T̃⊥ Ě⊥

Hence

R̃1 =

1−j .

2β2

(1 − j)(1 + j) i 2 1 | 2. = σ 2 |T̃⊥ Ě⊥ 2 2β2 β2

1 1 ẑ (Ǩ · Ǩ∗ )Rs = ẑ (Ǩ · Ǩ∗ ) 2 2

4.12. From the text we have

∗

β2 σ

1 i 2 σ | = Sav . = ẑ|T̃⊥ Ě⊥ 2 2β2 β2

b1 Γ1 + Γ2 P̃12 . = a1 1 + Γ1 Γ2 P̃12

With Γ2 = −Γ1 , R̃1 = Γ1

1 − P̃12 .

1 − Γ21 P̃12

Use of the expansion (1 − x)−1 = 1 + x + x2 + x3 + · · · gives R̃1 (ω) = Γ1 (1 − P̃12 )(1 + Γ21 P̃12 + Γ41 P̃14 + Γ61 P̃16 + · · · ) = Γ1 [1 + Γ21 P̃12 − P̃12 − Γ21 P̃14 + Γ41 P̃14 − Γ41 P̃16 + Γ61 P̃16 + · · · ] = Γ1 [1 + (Γ21 − 1)P̃12 + Γ21 (Γ21 − 1)P̃14 + Γ41 (Γ21 − 1)P̃16 + · · · ].

Now use P̃12n (ω) ↔ δ(t − nτ ),

τ=

∆ cos θt . v1

Also (Γ21 − 1) = −(1 − Γ1 )(1 + Γ1 ) = −T1 T2 .

97

So i i i i E r (t) = Γ1 E⊥ (t) − Γ1 T1 T2 E⊥ (t − 2τ ) − Γ31 T1 T2 E⊥ (t − 3τ ) − Γ51 T1 T2 E⊥ (t − 5τ ) + · · · .

Each of these terms may be interpreted in terms of multiple reflections within the slab, as shown in Figure 26.

Figure 26 4.13. See Figure 27.

Figure 27 We have ki = kxi x̂ + kzi ẑ,

kxi = k sin θi ,

kzi = k cos θi ,

(kxi )2 + (kzi )2 = k 2 .

Region 0: i

i

i

i

i −j(kx x+kz z) Ẽi⊥ = ŷẼ⊥ e , r −j(kx x−kz z) Ẽr⊥ = ŷẼ⊥ e ,

i −x̂kzi + ẑkxi Ẽ⊥ i i e−j(kx x+kz z) , k η r i i x̂kz + ẑkx Ẽ⊥ −j(kxi x−kzi z) H̃r⊥ = e . k η

H̃i⊥ =

98

Region 1: kx2 + kz2 = k02 , + −x̂kz + ẑkx Ẽ⊥ e−j(kx x+kz z) , k0 η0 − x̂kz + ẑkx Ẽ⊥ H̃− = e−j(kx x−kz z) . ⊥ k0 η0

+ −j(kx x+kz z) Ẽ+ , ⊥ = ŷẼ⊥ e

H̃+ ⊥ =

− −j(kx x−kz z) Ẽ− , ⊥ = ŷẼ⊥ e

Region 2: (kxt )2 + (kzt )2 = k 2 , t

t

t −j(kx x+kz z) Ẽt⊥ = ŷẼ⊥ e ,

H̃t⊥ =

t −x̂kzt + ẑkxt Ẽ⊥ t t e−j(kx x+kz z) . k η

Note: Applying the boundary conditions, we find that to match the phase of the field at the interface we must have kx = kxt = kxi , so kzt = kzi ,

kz =

q q k02 − (kxi )2 = k02 − k 2 sin2 θi .

If k02 − k 2 sin2 θi < 0, then kz = −j k̄z where k̄z = k 2 sin2 θi > k02 , or

k02 1 = k2 r

sin2 θi >

so that

q k 2 sin2 θi − k02 . This condition holds if

1 θi > sin−1 √ = θc . r

r , Ẽ t , Ẽ i , and Ẽ + , we apply the boundary conditions.

To find Ẽ⊥ ⊥ ⊥ ⊥

(1) Tangential E continuous at z = z1 : i

i

− jkz z1 + −jkz z1 r jkz z1 i −jkz z1 e , e + Ẽ⊥ e = Ẽ⊥ e + Ẽ⊥ Ẽ⊥

or i P1 Ẽ⊥ +

1 r 1 − + Ẽ = Q1 Ẽ⊥ + Ẽ P1 ⊥ Q1 ⊥

where

i

P1 = e−jkz z1 ,

Q1 = e−jkz z1 .

(2) Tangential H continuous at z = z1 : + − k i Ẽ i k i Ẽ r 1 kz Ẽ⊥ kz Ẽ⊥ 1 − z ⊥ P1 + z ⊥ =− Q1 + k η k η P1 k0 η0 k0 η0 Q1

or

1 Z1⊥

1 − 1 + r 1 i Ẽ⊥ − P1 Ẽ⊥ = Ẽ⊥ − Q1 Ẽ⊥ P1 Z2⊥ Q1

where Z1⊥ =

kη , kzi

Z2⊥ =

99

k0 η0 . kz

(3) Tangential E continuous at z = z2 : i

+ −jkz z2 − jkz z2 t −jkz z2 , Ẽ⊥ e + Ẽ⊥ e = Ẽ⊥ e

or + + Q2 Ẽ⊥

where

1 − t Ẽ = P2 Ẽ⊥ Q2 ⊥

i

P2 = e−jkz z2 ,

Q2 = e−jkz z2 .

(4) Tangential H continuous at z = z2 : − or −

+ − kz Ẽ⊥ kz Ẽ⊥ 1 k i Ẽ t Q2 + = − z ⊥ P2 k0 η0 k0 η0 Q2 k η

1 1 1 + − 1 Ẽ⊥ Q2 + Ẽ⊥ =− Ẽ t P2 .

Z2⊥ Z2⊥ Q2 Z1⊥ ⊥

So we have

1 r − 1 + Ẽ = Ẽ⊥ + Ẽ⊥ Q1 , P1 ⊥ Q1

Z1⊥ 1 r − 1 + i Ẽ = Ẽ⊥ − Ẽ⊥ Q1 , − P1 Ẽ⊥ + P1 ⊥ Z2⊥ Q1

− 1 + t P2 = Ẽ⊥ Q2 , Ẽ⊥ + Ẽ⊥ Q2

Z1⊥ − 1 + t − Ẽ⊥ P2 = Ẽ⊥ − Ẽ⊥ Q2 .

Z2⊥ Q2

i + P1 Ẽ⊥

(1) (2) (3) (4)

Subtracting (2) from (1) we get − i 2P1 Ẽ⊥ = Ẽ⊥

1 Q1

Z1⊥ 1− Z2⊥

+ + Ẽ⊥ Q1

Z1⊥ .

1+ Z2⊥

Adding (4) to (3) we get

Z1⊥ − + − 1 + − Ẽ⊥ − Ẽ⊥ Q2 = Ẽ⊥ + E⊥ Q2 , Z2⊥ Q2 hence − + 2 Ẽ⊥ = −Ẽ⊥ Q2

Z2⊥ − Z1⊥ + 2 = −Ẽ⊥ Q2 Γ.

Z2⊥ + Z1⊥

Substitution into (1) yields + Ẽ⊥ =

2Z2⊥ P1 Q1 1 i i Ẽ⊥ = τ Ẽ⊥ .

2 2 Q Z1⊥ + Z2⊥ Q2 1 2 −Γ Q22

Adding (2) to (1), we find that r Ẽ⊥ = P12

Q2 Γ 1 − Q12 2 2

Q Γ2 − Q12 2

100

i Ẽ⊥ .

(*)

From (3) we obtain t Ẽ⊥ =

P1 Q1 1 − Γ2 i Ẽ .

P2 Q2 Q21 − Γ2 ⊥ Q22

Let

Q2 r Γ 1 − Q12 Ẽ⊥ 2 R = i = P12 Q21 2 Ẽ⊥ Γ − Q2

,

T =

t Ẽ⊥ i Ẽ⊥

=

P1 Q1 1 − Γ2 .

P2 Q2 Q21 − Γ2 Q22

2

Now use

1 Re{Ě × Ȟ∗ }.

2

Sav = (a) In region 0 at z = z1 we have i Ě = Ěi⊥ + Ěr⊥ = ŷĚ⊥

Ȟ = Ȟi⊥ + Ȟr⊥ =

e

−jkzi z1

jkzi z1

+ Re

−jkxi x

e

i = ŷĚ⊥

R i e−jkx x , P1 + P1

ii 1 i h i i Ě⊥ −x̂kzi + ẑkxi e−jkz z1 + x̂kzi + ẑkxi ejkz e−jkx x . kη

So

1 R R∗ ki i i i i∗ P1 + −P1∗ + ∗ ejkx x · ẑ Re ŷĚ⊥ e−jkx x × z x̂Ě⊥ 2 P1 kη P

∗ 1 R 1 1 R i 2 | −1 + 2 − = Re − |Ě⊥ + |R|2 2 Z1⊥ P1 P12 1 1 |Ě i |2 (1 − |R|2 ). = 2 Z1⊥ ⊥

ẑ · Sav =

(b) In region 1 at z = z2 we obtain, similarly, ẑ · Sav = − (c) In region 2 at z = z2 ,

k̄z i 2 | Im{Γ}. |τ |2 |Ě⊥ η0 k0

1 1 ẑ · Sav = |T |2 |Ě i |2 .

2 Z1⊥ ⊥

Power is conserved if 1 − |R|2 = |T |2 .

4.14. Begin with the wave equation (4.319): 1 ˜ · H̃ = 0. ∇ × (∇ × H̃) − ω 2 µ̄

Assuming H̃(r) = H̃0 e−jkz z where kz = β − jα, we have

∇ × H̃ = ∇ × H̃0 e−jkz z = −jkz ẑ × H̃0 e−jkz z

101

and

∇ × (∇ × H̃) = −jkz ∇ × ẑ × H̃0 e−jkz z h i = −jkz e−jkz z ∇ × (ẑ × H̃0 ) − (ẑ × H̃0 )e−jkz z = (−jkz )2 ẑ × (ẑ × H̃0 )e−jkz z = kz2 e−jkz z H̃0 .

Substitute into (4.319) to get kz2 ˜ · H̃0 e−jkz z H̃0 e−jkz z = ω 2 µ̄ ˜ or ˜ · H̃0 . kz2 H̃0 = ω 2 ˜µ̄ Now let

(1)



 µ̃xx µ̃xy 0 ˜ = µ̃yx µ̃yy 0  . µ̄ 0 0 µ̃0

Write (1) in component form: kz2 H̃0x = ω 2 ˜µ̃xx H̃0x + ω 2 ˜µ̃xy H̃0y , kz2 H̃0y = ω 2 ˜µ̃yx H̃0x + ω 2 ˜µ̃yy H̃0y , or

2

−ω 2 ˜µ̃xy H̃0x 0 kz − ω 2 ˜µ̃xx = . −ω 2 ˜µ̃yx kz2 − ω 2 ˜µ̃yy 0 H̃0y

A nontrivial solution requires the determinant of the coefficient matrix to vanish: (kz2 − ω 2 ˜µ̃xx )(kz2 − ω 2 ˜µ̃yy ) − ω 4 ˜2 µ̃yx µ̃xy = 0, kz4 + [−ω 2 ˜(µ̃xx + µ̃yy )]kz2 + ω 4 ˜2 (µ̃xx µ̃yy − µ̃xy µ̃yx ) = 0.

But µ̃xx = µ̃yy and µ̃yx = −µ̃xy , so kz4 + [−2ω 2 ˜µ̃xx ]kz2 + ω 4 ˜2 (µ̃2xx + µ̃2xy ) = 0.

We get kz2 =

2ω 2 ˜µ̃xx ±

q 4ω 4 ˜2 µ̃2xx − 4ω 4 ˜2 (µ̃2xx + µ̃2xy ) 2

q = ω 2 ˜µ̃xx ± ω 2 ˜ −µ̃2xy = ω 2 ˜(µ̃xx ∓ j µ̃xy ).

Now examine (4.118) and (4.119). Assume α 1. Near resonance when ω ≈ ω0 , the denominator is [ω 2 (1 + α2 ) − ω02 ]2 + 4α2 ω 2 ω02 ≈ [ω02 (1 + α2 ) − ω02 ]2 + 4α2 ω02 ω02 ≈ ω04 α4 + 4α2 ω04 ≈ 4α2 ω04 .

102

So

ωM ωM −α2 ω03 + j2ω03 α − jµ0 , µ̃xx ≈ µ0 − µ0 ωM = µ0 1 − 4 2 4ω 2αω 4α ω0 0 0 µ̃xy ≈

2µ0 αω03 ωM − jµ0 ωM ω03 α2 ωM ωM α ≈ µ0 . − jµ0 4 2 2ω0 α 4ω0 4α ω0

Thus

ωM ωM ωM ωM α µ̃xx + j µ̃xy ≈ µ0 1 − − jµ0 + jµ0 + µ0 4ω0 2αω0 2ω0 α 4ω0

ωM ωM α ≈ µ0 1 − + 4ω0 4ω0

ωM ≈ µ0 1 − , a real number.

4ω0 Therefore kz2 = (β− − jα− )2 is approximately real, hence α− ≈ 0. Also,

ωM ωM α ωM ωM µ̃xx − j µ̃xy ≈ µ0 1 − − jµ0 − µ0 − jµ0 4ω0 2αω0 2ω0 α 4ω0

ωM ωM ≈ µ0 1 − − jµ0 4ω0 αω0 ωM . ≈ −jµ0 αω0 So ωM 2 kz2 = (β+ − jα+ )2 ≈ −jµ0 ω ˜, αω0 0 r ωM 1−j √ β+ − jα+ ≈ √ ω0 µ0 0 , αω0 2 r ωM α+ ≈ k0 α− .

2αω0 4.15. For TE polarization, j H̃z0 (2) (2) H̃z = − H̃z0 H0 (kρ), Ẽφ = −ZT E H1 (kρ), 4 4 where k = β − jα and ZT E = k/ω˜ c . For kρ 1, use r r 2 −j(kρ−π/4) 2 −j(kρ−3π/4) (2) (2) e , H1 (kρ) ∼ e .

H0 (kρ) ∼ πkρ πkρ Then n o 1 Re φ̂Ěφ × ẑȞz∗ 2 r r

1 Ȟz0 2 −j(βρ−3π/4) j 2 j(βρ−π/4) −αρ ∗ = ρ̂ Re −ZT E e Ȟz0 e e 2 4 πkρ 4 πk ∗ ρ

|Ȟz0 |2 2 −2αρ jπ/2 1 = ρ̂ Re −ZT E e je 2 16|k| πρ e−2αρ = ρ̂ Re{ZT E } |Ȟz0 |2 .

16π|k|ρ

Sav =

103

The power passing through a cylinder of radius ρ and length l is Z Sav · ρ̂ dS Pav = S Z l Z 2π e−2αρ Re{ZT E } = |Ȟz0 |2 ρ dφ dz 16π|k|ρ 0 0 e−2αρ = 2πl Re{ZT E } |Ȟz0 |2 .

16π|k| Hence

e−2αρ Pav = Re{ZT E } |Ȟz0 |2 . l 8|k|

For a lossless region we have k = β − jα = β, r √ ω µ k µ ZT E = = = = η (real), ω ω

and

η|Ȟz0 |2 Pav = . l 8k

4.16. We have Ẽzs =

∞ X

Dn Hn(2) (k0 ρ) cos nφ

n=0

where Dn = Ẽ0 n j −n

Jn (k0 a) (2)

.

Hn (k0 a)

For k0 ρ 1 we have Hn(2) (k0 ρ) ≈

r

2 −j(k0 ρ− π −nπ) 4 e . πk0 ρ

Therefore ∞ X

Jn (k0 a)

r

P∞

2 −jk0 ρ jπ/4 jnπ/2 e e e cos nφ (2) Hn (k0 a) πk0 ρ n=0 r ∞ X 2j e−jk0 ρ Jn (k0 a) = n (2) cos nφ. √ Ẽ0 πk0 ρ Hn (k0 a) n=0

Ẽzs ≈

Ẽ0 n j

−n

Then

2 πk0

2

Ẽ02

Jn (k0 a) cos nφ n=0 n (2) Hn (k0 a)

σ2D = 2πρ

ρẼ02 2

∞

4 X Jn (k0 a) = n (2) cos nφ . k0 Hn (k0 a) n=0

104

For small ka, use the approximations (E.50), (E.52), and (E.53).

Since N0 (x) is logarithmic for x 1 while Nn (x) ∼ x−n for n > 0, the n = 0 term from the sum dominates.

J0 (k0 a) ≈ 1,

2 2 2 e0.57722 (2) ≈ j ln(0.891k0 a), H0 (k0 a) ≈ jN0 (k0 a) ≈ j [ln k0 a − ln 2 − 0.57722] ≈ j ln k0 a π π 2 π 4 1 σ2D ≈ 2 k0 j π ln(0.891k0 a)

2

=

1 π2a .

2 k0 a ln (0.891k0 a)

4.17. Start with Ẽi = ŷẼ0 e−jk0 x = (ρ̂ sin φ + φ̂ cos φ)Ẽ0 e−jk0 ρ cos φ and Ẽ0 −jk0 x Ẽ0 e = ẑ e−jk0 ρ cos φ . η0 η0 Internal to the cylinder (ρ < a) we have H̃i = ẑ

H̃z =

∞ X

Bn Jn (kρ) cos nφ,

n=0 ∞

X jZT E ∂Hz Ẽφ = = Bn jZT E Jn0 (kρ) cos nφ, k ∂ρ n=0

∞

X ZT E n ZT E 1 ∂Hz Ẽρ = −j = j Bn Jn (kρ) sin nφ. k ρ ∂φ k ρ n=0

External to the cylinder (ρ > a) we have H̃zs = Ẽφs = Ẽρs =

∞ X n=0 ∞ X n=0 ∞ X n=0

Dn Hn(2) (k0 ρ) cos nφ, jη0 Dn Hn(2)0 (k0 ρ) cos nφ, j

η0 n Dn Jn (k0 ρ) sin nφ. k0 ρ

Boundary condition 1: H̃z is continuous at ρ = a.

H̃zi + H̃zs = H̃z =⇒

∞

∞

n=0

n=0

X Ẽ0 −jk0 a cos φ X Dn Hn(2) (k0 a) cos nφ = Bn Jn (ka) cos nφ. e + η0

Apply orthogonality: Z π Z π ∞ X Ẽ0 −jk0 a cos φ e cos mφ dφ + Dn Hn(2) (k0 a) cos mφ cos nφ dφ −π η0 −π n=0 Z π ∞ X = Bn Jn (ka) cos mφ cos nφ dφ, n=0

−π

105

which gives Ẽ0 (2) 2π m j −m Jm (k0 a) + 2πDm Hm (k0 a) = 2πBm Jm (ka). η0

(1)

Boundary condition 2: Ẽφ is continuous at ρ = a. cos φẼ0 e−jk0 a cos φ +

∞ X

Dn jZT E Hn(2)0 (k0 a) cos nφ =

n=0

∞ X

Bn jZT E Jn0 (ka) cos nφ.

n=0

Apply orthogonality: Z π Z π ∞ X −jk0 a cos φ (2)0 cos mφ cos nφ dφ cos φ cos mφe dφ + Dn jη0 Hn (k0 a) Ẽ0 −π

=

−π

n=0 ∞ X

Z π

Bn jZT E Jn0 (ka)

cos mφ cos nφ dφ, −π

n=0

which gives 0 (2)0 0 m Ẽ0 j2πj −m Jm (k0 a) + Dm jη0 2πHm (k0 a) = Bm jZT E 2πJm (ka).

Solving (1) and (2) simultaneously, we get Dm = −

0 (ka) − η0 J (ka)J 0 (k a) Jm (k0 a)Jm Ẽ0 m 0 ZT E m m j −m (2) (2)0 η0 Hm (k0 a)J 0 (ka) − η0 Hm (k0 a)Jm (ka) m

and

ZT E

(2)0

(2)

0 (k a)H Ẽ0 Jm (k0 a)Hm (k0 a) − Jm m (k0 a) 0 Bm = m j −m .

(2)0 (2) ZT E 0 η0 Jm (ka)Hm (k0 a) − J (ka)Hm (k0 a) η0

m

For a PEC cylinder, use ZT E →0 η0 Then H̃zs = −

∞ X Ẽ0 n=0

Also,

=⇒

η0

Dm = −

n j −n

0 (k a) Ẽ0 Jm 0 m j −m (2)0 . η0 Hm (k0 a)

Jn0 (k0 a) (2)0 Hn (k0 a)

Hn(2) (k0 ρ) cos nφ.

∞ Ẽ0 −jk0 ρ cos φ X Ẽ0 i H̃z = e = n j −n Jn (k0 ρ) cos nφ. η0 η0 n=0

Thus, the total field external to the cylinder is " # ∞ 0 (k a) X J Ẽ 0 0 n H̃ztot = n j −n Jn (k0 ρ) − (2)0 Hn(2) (k0 ρ) cos nφ η0 Hn (k0 a) =

Ẽ0 η0

n=0 ∞ X

n j −n

(2)0 n=0 Hn (k0 a)

h

i Jn (k0 ρ)Hn(2)0 (k0 a) − Jn0 (k0 a)Hn(2) (k0 ρ) cos nφ.

106

(2)

4.18. The incident field is given by Ẽi = ẑẼ0 e−jk0 ρ cos φ ,

H̃i = −(ρ̂ sin φ + φ̂ cos φ)

Ẽ0 −jk0 ρ cos φ e . η0

In region 1 (a ≤ ρ ≤ b), Ẽz1 (ρ, φ) = H̃φ1 (ρ, φ) =

∞ X

[An Jn (kρ) + Bn Hn(2) (kρ)] cos nφ,

n=0 ∞ X n=0

j − η

[An Jn0 (kρ) + Bn Hn(2)0 (kρ)] cos nφ.

In region 2 (ρ ≥ b), Ẽz2 (ρ, φ) = H̃φ2 (ρ, φ) =

∞ X

Cn Hn(2) (k0 ρ) cos nφ,

n=0 ∞ X n=0

j − η0

Cn Hn(2)0 (k0 ρ) cos nφ.

Boundary condition 1: Ẽz1 = 0 at ρ = a: An Jn (ka) + Bn Hn(2) (ka) = 0 =⇒ Bn = −An

Jn (ka) (2)

.

Hn (ka)

So Ẽz1 = H̃z1 =

∞ X

" An

n=0 ∞ X n=0

Jn (kρ) − Hn(2) (kρ)

j − η

#

Jn (ka)

cos nφ,

(2)

Hn (ka)

" An Jn0 (kρ) − Hn(2)0 (kρ)

Jn (ka)

#

(2)

cos nφ.

Hn (ka)

Boundary condition 2: Etan continuous at ρ = b: −jk0 b cos φ

Ẽ0 e

+

∞ X

Cn Hn(2) (k0 b) cos nφ =

n=0

∞ X

" An

Jn (kb) − Hn(2) (kb)

n=0

Apply orthogonality to obtain Ẽ0 2π n j −n Jn (k0 b) + 2πCn Hn(2) (k0 b) = 2πAn Fn or An Fn − Cn Hn(2) (k0 b) = n j −n Jn (k0 b)Ẽ0 where

" Fn = Jn (kb) − Hn(2) (kb) 107

Jn (ka) (2)

Hn (ka)

# cos nφ.

Jn (ka) (2)

Hn (ka)

# cos nφ.

Boundary condition 3: Htan continuous at ρ = b: " #

∞ ∞ X Jn (ka) j j Ẽ0 −jk0 b cos φ X 0 (2)0 (2)0 + − Cn Hn (k0 b) cos nφ = − An Jn (kb) − Hn (kb) (2) cos nφ. − cos φ e η0 η0 η Hn (ka) n=0

n=0

Apply orthogonality to get

j Ẽ0 j −n 0 (2)0 −j2π n j Jn (k0 b) + 2πCn − Hn (k0 b) = 2π − An G n η0 η0 η or An

η0 Gn − Cn Hn(2)0 (k0 b) = Ẽ0 n j −n Jn0 (k0 b) η

where

" Jn0 (kb) − Hn(2)0 (kb)

Gn =

Jn (ka)

#

(2)

cos nφ.

Hn (ka)

Solving the equations An Fn − Cn Hn(2) (k0 b) = Ẽ0 n j −n Jn (k0 b), η0 An Gn − Cn Hn(2)0 (k0 b) = Ẽ0 n j −n Jn0 (k0 b), η simultaneously, we obtain (2)0

Cn = − n j and

−n

Ẽ0

(2)

Fn Hn (k0 b) − ηη0 Gn Hn (k0 b) Fn Jn0 (k0 b) − ηη0 Gn Jn (k0 b)

2j −n πk0 b Ẽ0 n j An = − .

(2)0 (2) Fn Hn (k0 b) − ηη0 Gn Hn (k0 b)

4.19. See Figure 28. By symmetry the fields should be even about the line φ = φ0 . Hence we expand them in terms of cos n(φ − φ0 ).

For ρ > ρ0 we have Ẽz2 = H̃φ2 =

∞ X

Cn cos n(φ − φ0 )Hn(2) (k0 ρ),

n=0 ∞ X n=0

−j η0

Cn cos n(φ − φ0 )Hn(2)0 (k0 ρ).

For ρ < ρ0 we have Ẽz1 = H̃φ1 =

∞ h i X An Jn (k0 ρ) + Bn Hn(2) (k0 ρ) cos n(φ − φ0 ), n=0 ∞ X n=0

−j η0

h i 0 (2)0 An Jn (k0 ρ) + Bn Hn (k0 ρ) cos n(φ − φ0 ).

108

Figure 28 Boundary condition 1: Ẽz = 0 at ρ = a. By orthogonality we get Bn = −An

Jn (k0 a) (2)

,

Hn (k0 a)

so Ẽz1 = =

∞ X n=0 ∞ X

" An Jn (k0 ρ) −

Jn (k0 a) (2)

Hn (k0 a)

# Hn(2) (k0 ρ)

cos n(φ − φ0 )

h i A0n Jn (k0 ρ)Hn(2) (k0 a) − Jn (k0 a)Hn(2) (k0 ρ) cos n(φ − φ0 )

n=0

where A0n = Also, H̃φ1 =

∞ X −j n=0

η0

An .

(2) Hn (k0 a)

h i A0n Jn0 (k0 ρ)Hn(2) (k0 a) − Jn (k0 a)Hn(2)0 (k0 ρ) cos n(φ − φ0 ).

Boundary condition 2: Ẽz continuous at ρ = ρ0 . By orthogonality h i Cn Hn(2) (k0 ρ0 ) = A0n Jn (k0 ρ0 )Hn(2) (k0 a) − Jn (k0 a)Hn(2) (k0 ρ0 ) = A0n Fn , say.

Boundary condition 3: H̃φ is discontinuous at the surface current δ(φ − φ0 ) .

J̃s = ẑI˜ ρ0 We have

− ˜δ(φ − φ0 ) , H̃φ (ρ+ 0 , φ) − H̃φ (ρ0 , φ) = I ρ0

109

∞ X −j η0

n=0

−

Cn cos n(φ − φ0 )Hn(2)0 (k0 ρ0 )

∞ X −j n=0

η0

h i δ(φ − φ0 ) A0n Jn0 (k0 ρ)Hn(2) (k0 a) − Jn (k0 a)Hn(2)0 (k0 ρ) cos n(φ − φ0 ) = I˜ . ρ0

Multiply by cos m(φ − φ0 ) and apply the orthogonality relation Z π 2π cos n(φ − φ0 ) cos m(φ − φ0 ) dφ = δmn n −π to get Cn Hn(2)0 (k0 ρ0 )

−j η0

2π − A0n Gn n

−j η0

2π = n

δ(φ − φ0 ) I˜ I˜ cos m(φ − φ0 ) dφ = ρ0 ρ0 −π

Z π

where Gn = Jn0 (k0 ρ)Hn(2) (k0 a) − Jn (k0 a)Hn(2)0 (k0 ρ).

Solving the equations Cn Hn(2) (k0 ρ0 ) − A0n Fn = 0, Cn Hn(2)0 (k0 ρ0 ) − A0n Gn = n jη0

I˜ , 2πρ0

simultaneously and back-substituting, we obtain An = n jη0

(2) (2) I˜ Hn (k0 ρ0 )Hn (k0 a) 2πρ0 Fn Hn(2)0 (k0 ρ0 ) − Gn Hn(2) (k0 ρ)

Bn = − n jη0 and Cn = n jη0 4.20. Let

k ψ̃(x, y) = 2π

(2) I˜ Hn (k0 ρ0 )Jn (k0 a) 2πρ0 Fn Hn(2)0 (k0 ρ0 ) − Gn Hn(2) (k0 ρ)

(2) I˜ Fn Hn (k0 a) .

2πρ0 Fn Hn(2)0 (k0 ρ0 ) − Gn Hn(2) (k0 ρ)

Z

A(k cos ξ)e−jkx cos ξ e±jky sin ξ sin ξ dξ.

An important question is: What is the path of integration? Compare with 1 u(x, y) = 2π

Z ∞+j∆

A(kx )e−jkx x e−j

√ 2

k −kx2 y

dkx .

−∞+j∆

The integration variable must go through the same values during integration in each case.

Simple but important case: k real =⇒ ∆ → 0.

110

Use cos z = cos(u + jv) = cos u cosh v − j sin u sinh v, sin z = sin(u + jv) = sin u cosh v + j cos u sinh v.

Case 1: y > 0. Choose ky = −k sin ξ. (Note: minus sign is chosen by convention. Others (e.g., Born and Wolf) choose a plus sign.) p −jky = −j k 2 − kx2 p − kx2 − k 2

ξ

kx = k cos ξ

−jky = jk sin ξ

ξr + jξi

−k cosh ξi

jk(−j sinh ξi ) = k sinh ξi

[−∞, 0] p −j k 2 − kx2

[π − j∞, −π]

[∞, −k]

[−∞, 0]

ξr

k cos ξr

jk sin ξr

[−π, 0]

[−k, k]

[0, −jk, −jk, 0]

kx

[0, −jk, −jk, 0] p − kx2 − k 2

ξr + jξi

k cosh ξi

jk(j sinh ξi ) = −k sinh ξi

[k, ∞]

[0, −∞]

[0, j∞]

[k, ∞]

[0, −∞]

kx kx [−∞, −k] kx [−k, k]

Figure 29 The case for y < 0 is treated similarly. The resulting contour C< is identical to C> .

4.21. For a magnetic line source we have TEz fields so that Ẽz = 0. Let  Z ∞+j∆ 1   Ã+ (kx )e−jkx x e−jky y dkx , y > 0,   2π −∞+j∆ H̃z (x, y) = Z ∞+j∆  1    Ã− (kx )e−jkx x e+jky y dkx , y < 0.

2π −∞+j∆ From (4.225) we have Ẽx =

1 ∂ H̃z , jω˜ c ∂y 111

so

 Z ∞+j∆ −ky 1   Ã+ (kx )e−jkx x e−jky y dkx ,   2π ω c −∞+j∆ Ẽx =

Z ∞+j∆  +ky 1    Ã− (kx )e−jkx x e+jky y dkx , 2π −∞+j∆ ω˜ c

y > 0, y < 0.

BC 1) Tangential H̃ is continuous at y = 0. This implies Ã+ (kx ) − Ã− (kx ) = 0

(A)

BC 2) n̂12 × (Ẽ1 − Ẽ2 ) = −J̃ms = −I˜m δ(x). This implies −Ã+ (kx ) − Ã− (kx ) = I˜m Adding (A) and (B) we obtain

ω˜ c ky

(B)

c

ω˜

−2Ã− (kx ) = I˜m ky so

c

ω˜

Ã+ (kx ) = Ã− (kx ) = −I˜m .

2ky

Finally, ω˜ c I˜m H̃z (x, y) = − 2π

Z ∞+j∆

e−jky |y| −jkx x e dkx .

2ky −∞+j∆

4.22. See Figure 30. In region 1 we need both upward and downward traveling waves:

Figure 30 + Ẽz1 = − Ẽz1 =

1 2π

Z ∞+j∆

1 2π

Z ∞+j∆

Ã+ (kx )e−jky y e−jkx x dkx ,

−∞+j∆

ky =

p k 2 − kx2

Ã− (kx )ejky y e−jkx x dkx ,

−∞+j∆

+ H̃x1 =

1 2π

− H̃x1 =

1 2π

Z ∞+j∆ −∞+j∆ Z ∞+j∆ −∞+j∆

ky ω µ̃

ky − ω µ̃

Ã+ (kx )e−jky y e−jkx x dkx ,

112

Ã− (kx )ejky y e−jkx x dkx .

In region 2 we need only upward traveling waves: 1 Ẽz2 = 2π

Z ∞+j∆

1 H̃x2 = 2π

Z ∞+j∆

B̃(kx )e−jky y e−jkx x dkx ,

−∞+j∆

−∞+j∆

ω µ̃ ky

B̃(kx )e−jky y e−jkx x dkx .

Boundary condition 1: Ẽz = 0 at y = 0.

Ã− (kx ) = −Ã+ (kx ),

+ − Ẽz1 = Ẽz1 + Ẽz1 =

1 H̃x1 = 2π

Z ∞+j∆

1 2π +

Z ∞+j∆ −∞+j∆

à (kx ) −∞+j∆

h i Ã+ (kx ) e−jky y − ejky y e−jkx x dkx ,

kx ω µ̃

h

i e−jky y + ejky y e−jkx x dkx .

Boundary condition 2: Ẽz continuous at y = h. h i + −jky h jky h à e −e = B̃e−jky h , i h B̃ = Ã+ 1 − e2jky h .

Boundary condition 3: ˜ H̃x1 − H̃x2 = J˜zs = Iδ(x) = I˜

1 2π

Z ∞+j∆

e−jkx x dkx ,

−∞+j∆

i h i h ω µ̃ Ã+ e−jky h + ejky h − Ã+ 1 − e2jky h e−jky h = I˜ , ky ω µ̃ −jky h e .

Ã+ = I˜ 2ky Thus 1 2π

Z ∞+j∆

1 Ẽz2 = 2π

Z ∞+j∆

Ẽz1 =

i ω µ̃ h −jky (y+h) I˜ e − ejky (y−h) e−jkx x dkx , −∞+j∆ 2ky

h i ω µ̃ I˜ e−jky y e−jky h − ejky h e−jkx x dkx −∞+j∆ ky Z ∞+j∆ i 1 ω µ̃ h −jky (y+h) = I˜ e − e−jky (y−h) e−jkx x dkx 2π −∞+j∆ ky

113

We can combine these two formulas using the absolute value function since ( y − h, y > h, |y − h| = h − y, y < h.

The result, Ẽz (x, y) = −

ω µ̃I˜ 2π

Z ∞+j∆

e−jky |y−h| − e−jky (y+h) −jkx x e dkx , 2ky −∞+j∆

matches (4.403) and (4.406) when split into two terms.

4.23. The impressed magnetic field is, from (4.402), Z ω˜ c1 I˜m ∞+j∆ e−jky1 |y−h| −jkx x i e dkx , H̃z (x, y) = − 2π 2ky1 −∞+j∆

ky1 =

q

k12 − kx2 .

For 0 ≤ y < h the impressed field can be written as Z ω˜ c1 I˜m ∞+j∆ ejky1 (y−h) −jkx x i H̃z (x, y) = − e dkx .

2π 2ky1 −∞+j∆ Use Ampere’s law in the form 1 ∂ H̃z jω˜ ∂y

Ẽx = to obtain

˜ Z ∞+j∆ ejky1 (y−h)

Im Ẽxi (x, y) = − 2π

2

−∞+j∆

e−jkx x dkx ,

0 ≤ y < h.

The scattered field in region 1 is s H̃z1 (x, y) =

1 2π

s Ẽx1 (x, y) =

1 2π

Z ∞+j∆

Ã1 (kx )e−jky1 y e−jkx x dkx ,

−∞+j∆ Z ∞+j∆ −∞+j∆

ky1 − c ω˜ 1

Ã1 (kx )e−jky1 y e−jkx x dkx .

The scattered field in region 2 is s H̃z2 (x, y) =

1 2π

s Ẽx2 (x, y) =

1 2π

Z ∞+j∆

Ã2 (kx )ejky2 y e−jkx x dkx ,

−∞+j∆ Z ∞+j∆ −∞+j∆

ky2 ω c2

Ã2 (kx )ejky2 y e−jkx x dkx .

Boundary condition 1: Ẽtan continuous at y = 0: Z Z ∞+j∆ Z ∞+j∆ ky1 ky2 I˜m ∞+j∆ e−jky1 h −jkx x 1 1 −jkx x − e dkx − Ã1 e dkx = Ã2 e−jkx x dkx , 2π −∞+j∆ 2 2π −∞+j∆ ω˜ c1 2π −∞+j∆ ω˜ c2 or −I˜m

ky1 ky2 e−jky1 h − c Ã1 = c Ã2 .

2 ω˜ 1 ω˜ 2 114

Boundary condition 2: H̃tan continuous at y = 0: −ω˜ c1 I˜m

e−jky1 h + Ã1 = Ã2 .

2ky1

Solving simultaneously, we obtain Ã1 =

˜c ky1 Ã2 = 2F c 2 c ˜2 ky1 + ˜1 ky2

˜c1 ky2 − ˜c2 ky1 F, ˜c1 ky2 + ˜c2 ky1

where

c

−jky1 h

ω˜ e F = −I˜m 1 ky1 2

.

Substitution yields Z I˜m ω˜ c1 ∞+j∆ e−jky1 (y+h) s H̃z1 (x, y) = − RT E (kx )e−jkx x dkx , 2π 2k y1 −∞+j∆ Z ∞+j∆ −jky2 (y−hky1 /ky2 ) c ˜ e Im ω˜ 2 s TT E (kx )e−jkx x dkx , H̃z2 (x, y) = − 2π 2ky2 −∞+j∆

where RT E =

˜c1 ky2 − ˜c2 ky1 , ˜c1 ky2 + ˜c2 ky1

TT E =

2˜ c1 ky2 = 1 + RT E . ˜c1 ky2 + ˜c2 ky1

4.24. See Figure 31.

Figure 31 Write i

i

H̃i = ẑH̃0 e−jkx x e−jky y ,

kxi = −k cos φ0 ,

kyi = −k sin φ0 , i

i

Ẽi = η H̃0 (x̂ sin φ0 − ŷ cos φ0 )e−jkx x e−jky y =

η H̃0 i i (−kyi x̂ + kxi ŷ)e−jkx x e−jky y . k 115

q kxi = − k 2 − (kyi )2 ,

Represent the scattered magnetic field in terms of an inverse transform: H̃zs =

1 2π

Z ∞+j∆

Ã(ky )e−jkx x e−jky y dky ,

kx =

−∞+j∆

q k 2 − ky2 .

To find Ẽs use ∇ × H̃ = jω˜ Ẽ = x̂ to obtain Ẽzs =

1 2π

Z ∞+j∆

Ã(ky )

−∞+j∆

∂ H̃z ∂ H̃z − ŷ ∂y ∂x

kx e−jkx x e−jky y dky . ω˜

Next we apply the boundary condition where Ẽtot = Ẽi + Ẽs .

ŷ · Ẽtot = 0 at x = 0, We obtain

ki 1 i η H̃0 x e−jky y + k 2π

Z ∞+j∆

1 2π

Z ∞+j∆

Next use i

e−jky y =

Ã(ky )

−∞+j∆

kx e−jky y dky = 0. ω˜

1 e−jky y dky i) j(k − k y −∞+j∆ y

to get ki Ã(ky ) = −η H̃0 x k

Z ∞+j∆

ω˜

kx

1 . j(ky − kyi )

Substitution into (1) yields H̃zs = −

1 2π

ki η H̃0 x k −∞+j∆

ω˜

kx

1 e−jkx x e−jky y dky j(ky − kyi )

Let us evaluate this integral using the residue theorem.

1 kxi ω˜

1 s i −jkx x −jky y H̃z = 2πj lim (ky − ky ) −η H̃0 e e .

2π k kx j(ky − kyi ) ky →kyi Use lim kx =

ky →kyi

q

k 2 − (kyi )2 = −kxi

to get ki H̃zi = −η H̃0 x k

ω˜

−kxi

i

i

e−jky y ejkx x =

Finally then,

i

i

ω˜ η i i H̃0 e−jky y ejkx x . k

H̃s = ẑH̃0 e−jky y ejkx x .

116

(1)

Figure 32 4.25. See Figure 32.

We have ∆n = 2∆ and ∆n−1 = ∆, θ0 = 0 =⇒ kx0 = 0 =⇒ kzn = kn ,

Zn = ηn .

So √ kn = ω µ0 0 = k0 ,

kn−1 = ω

p µ0 4 0 = 2k0 ,

Zn =

p µ0 / 0 = η0 ,

Zn−1 =

p µ0 /4 0 = η0 /2.

Use (4.494): 1 (η0 /2 + η0 )2 1 (η0 /2 − η0 )2 cos(k0 2∆ + 2k0 ∆) − cos(k0 2∆ − 2k0 ∆) 4 η0 η0 /2 4 η0 η0 /2 2 2 1 3 1 1 = 2 cos(4k0 ∆) − 2 4 2 4 2 1 9 = cos(4k0 ∆) − .

8 8

cos κL =

So

8 1 cos(4k0 ∆) = cos κL + , 9 8

1 −1 8 4k0 ∆ = ± cos cos κL + + 2nπ (n = 0, ±1, ±2, . . .).

9 8

With L = 2∆ + ∆ = 3∆ we get 1 k0 ∆ = ± cos−1 4

1 π [8 cos(3κ∆) + 1] + n .

9 2

A plot of k0 ∆ vs. κ∆ is shown below. The periodicity of the propagation constant is seen, as are the pass bands and stop bands. The stop bands indicate ranges of frequency for which real values of the propagation constant are not allowed. Thus, the wave is evanescent within the stop bands.

117

4 n=2, plus sign

3 n=2, minus sign

k0∆

STOP BAND 2

n=1, plus sign

n=1, minus sign

1

STOP BAND n=0, plus sign

0

0

1

2

3

4

5

κ∆

6

7

8

Figure 33 4.26. See Figure 34.

Using T-matrices we can write ! ! !

(n) (n) (n+1) (n+1) (n+2) (n+2) T11 T12 T11 T12 an T11 T12 an+3 . = (n) (n) (n+1) (n+1) (n+2) (n+2) bn b n+3 T21 T22 T21 T22 T21 T22 But, by periodicity of the medium,

T11 T12 an+3 an jκL an+3 = =e T21 T22 bn+3 bn bn+3

(1)

where L = ∆n + ∆n+1 + ∆n+2 and

T11 T12 T21 T22

(n)

=

(n)

T11 T12 (n) (n) T21 T22

!

(n+1)

(n+1)

T11 T12 (n+1) (n+1) T21 T22

!

(n+2)

(n+2)

T11 T12 (n+2) (n+2) T21 T22

! .

Rewriting (1) we have the eigenvalue equation

T11 − ejκL T12 an+3 0 = T21 T22 − ejκL bn+3 0 for the allowed values of κ. Setting the determinant to zero we have T11 T22 − T12 T21 − ejκL (T11 + T22 ) + e2jκL = 0.

Use

(n)

T11 T22 − T12 T21 =

(n)

(n+1)

(n+1)

(n+2)

(n+2)

T11 T12 T T12 T11 T12 T11 T12 = 11 (n) (n) (n+1) (n+1) (n+2) (n+2) .

T21 T22 T21 T22 T21 T22 T21 T22 118

(2)

Figure 34 Since

(n)

(n)

Zn−1 T11 T12 , (n) (n) = Zn T21 T22 we have T11 T22 − T12 T21 =

Zn−1 Zn Zn+1 Zn−1 Zn Zn+1 = = 1.

Zn Zn+1 Zn+2 Zn Zn+1 Zn−1

So 1 − ejκL (T11 + T22 ) + e2jκL = 0 or e−jκL + ejκL = T11 + T22 or

T11 + T22 .

2 This is the desired eigenvalue equation for κ. Specific formulas for T11 and T22 may be found by multiplying out (2) and simplifying using expressions from Section 4.14.2 of the text. cos κL =

4.27. ωp2 =

2 × 1011 (1.602 × 10−19 )2 N qe2 = , 0 me (8.854 × 10−12 )(9.109 × 10−31 )

∴ ωp = 25.2 × 106 s−1 .

(a) ( 0 ωp2 ν ωp2 5.71 × 10−10 S/m at 5 MHz, σ̃ = 2 ≈

ν = 0 ω + ν2 ω2 1.43 × 10−12 S/m at 100 MHz, # " # ( " ωp2 ωp2 0.355 0 at 5 MHz, ≈ 0 1 − 2 = ˜ = 0 1 − 2 2 ω +ν ω 0.998 0 at 100 MHz. (b) ωc =

qe 1.602 × 10−19 B0 = · 0.5 × 10−4 = 8.79 × 106 s−1 , me 9.109 × 10−31 119

"

# ( ωp2 0.300 0 at 5 MHz, ˜ = 0 1 − 2 = 2 ω − ωc 0.998 0 at 100 MHz, " # ( ωp2 0.355 0 at 5 MHz, ˜z = 0 1 − 2 = ω 0.988 0 at 100 MHz, ( ωc ωp2 0.196 0 at 5 MHz, δ̃ = 0 = 2 2 ω(ω − ωc ) 2.26 × 10−5 0 at 100 MHz. 4.28. (a) Re ˜ − 0 = 0 ωp2

ω02 − ω 2 .

2 (ω0 − ω 2 )2 + 4ω 2 Γ2

(*)

Setting ∂ Re ˜ =0 ∂ω and using the quotient rule for differentiation, we obtain (ω02 − ω 2 )2 = 4ω02 Γ2 .

Solving for ω, we get ω = ω0

p

1 ± 2Γ/ω0

and choose the minus sign for a minimum. Substitution back into (*) gives the desired result for Re ˜. (b) ! ωp2 ω→0 =⇒ Re ˜ = 5 0 = 0 1 + 2 , ω0 and

1 ωp2 ω0 [Im ˜]max = −16 0 = − 0 2 .

2 ω0 Γ

So 5=1+

ωp2 , ω02

32 =

ωp2 . Γω0

Combining these, we obtain Γ 1 = . ω0 8 Then W = 2Γ

Γ 1 W =2 = . ω0 ω0 4

=⇒

4.29. (a) Ě(r, ω̌) = θ̂

E0 −jβr jξE e e . r sin θ

(b) Ȟ(r, ω̌) = r̂ × θ̂

E0 /η0 −jβr jξE E0 /η0 −jβr jξE e e = φ̂ e e . r sin θ r sin θ 120

(c)

E0 /η0 −jβr jξE E0 −jβr jξE lim r η0 r̂ × Ȟ + Ě = lim r η0 r̂ × φ̂ e e + θ̂ e e r→∞ r→∞ r sin θ r sin θ

E0 −jβr jξE E0 −jβr jξE = lim −θ̂ e e + θ̂ e e r→∞ sin θ sin θ = 0.

E0 −jβr jξE E0 /η0 −jβr jξE lim r r̂ × Ě − η0 Ȟ = lim r r̂ × θ̂ e e − η0 φ̂ e e r→∞ r→∞ r sin θ r sin θ

E0 −jβr jξE E0 −jβr jξE = lim φ̂ e e − φ̂ e e r→∞ sin θ sin θ = 0.

4.30.

E(r, t) = E0 cos(ωt + ξ E ),

D(r, t) = D0 cos(ωt + ξ D ).

1 1 we (r, t) = E · D = E0 · D0 cos(ωt + ξ E ) cos(ωt + ξ D ).

2 2 Z 2π/ω

1 E0 · D0 cos(ωt + ξ E ) cos(ωt + ξ D ) dt 2 0 Z 2π/ω h i ih D 1 jξE jωt ω 1 D E = E0 · D 0 e e + e−jξ e−jωt ejξ ejωt + e−jξ e−jωt dt 2π 2 4 0

Z Z 2π/ω ω 1 1 jξE jξD 2π/ω j2ωt −jξ E −jξ D = E0 · D 0 e e e dt + e e e−j2ωt dt 2π 2 4 0 0 Z 2π/ω Z 2π/ω jξ E −jξ D −jξ E jξ D +e e dt + e e dt

ω hwe (r, t)i = 2π

0

i 1 1h E D E D = E0 · D0 ejξ e−jξ + e−jξ ejξ 2 4 i 11h D E D jξ E E0 e · D0 e−jξ + (E0 ejξ · D0 e−jξ )∗ = 24 h i 11 E D 2 Re E0 ejξ · D0 e−jξ = 24

1 = Re Ě(r) · Ď∗ (r) .

4 4.31. Since k 2 = ω 2 µ̃(Re ˜ + j Im ˜) = (β − jα)2 = β 2 − 2jαβ − α2 , we have β 2 − α2 = ω 2 µ̃ Re ˜, Therefore α=

−2αβ = ω 2 µ Im ˜.

ω 2 µ̃ Im ˜ 2β 121

0

and we can back-substitute: 2

β −

ω 2 µ̃ Re ˜ 2β

2

= ω 2 µ̃ Re ˜.

Rearrange to obtain 1 β 4 − β 2 ω 2 µ̃ Re ˜ − ω 4 µ̃(Im ˜)2 = 0.

4 2 Use the quadratic formula to solve for β , picking the positive sign so that β 2 > 0: s " # 2 µ̃ Re (Im ˜)2 ω ˜ 2 1+ 1+ . β = 2 (Re ˜)2 So

v "s # u u1 p (Im ˜)2 t β = ω µ̃ Re ˜ 1+ +1 .

2 (Re ˜)2

Then s s " # " # 2 µ̃ Re 2 µ̃ Re 2 2 ω ˜ (Im

˜ ) ω ˜ (Im

˜ ) α2 = β 2 − ω 2 µ̃ Re ˜ = 1+ 1+ − ω 2 µ̃ Re ˜ = −1 + 1 + 2 (Re ˜)2 2 (Re ˜)2 so

v "s # u u1 p (Im ˜)2 t α = ω µ̃ Re ˜ 1+ −1 .

2 (Re ˜)2

4.32. Phasor fields: Ě(r) = x̂E0 e−jβz ,

Ȟ(r) = ŷ

E0 −jβz e . η0

Complex Poynting theorem:

Z I Z 1 1 1 1 ∗ ∗ ∗ ∗ Ě · J̌ dV = (Ě × Ȟ ) · dS − 2j ω̌ Ě · Ď − B̌ · Ȟ dV − 2 V 2 S 4 V 4 The first term is zero because J̌ = 0. Next, I Z Z E02 1 1 E02 1 E02 ∗ ∗ Ě × Ȟ = ẑ =⇒ (Ě × Ȟ ) · dS = dS − dS = 0. η0 2 S 2 top η0 2 bottom η0 Finally, 1 1 E2 1 E2 1 1 1 Ě · Ď∗ − B̌ · Ȟ∗ = 0 E02 − µ0 20 = 0 E02 − µ0 0 = 0.

4 4 4 4 η0 4 4 µ0 / 0 4.33. p k = β − jα = ω µ̃˜ =ω

r 1 c 2 ω c 2ω µ0 2 + 10 = 2 + 10 = + 10 = β, µ0 c2 ω c ω c

dβ 2 dω c = , ∴ vg = = . dω c dβ 2 ω ωc ω vp = = 2ω = . β 2ω + 10c + 10 c 122

4.34.

E(r, t) = Re Ě(r)ejωt = x̂E0 cos(ωt − βz) − ŷE0 sin(ωt − βz),

Ex2 + Ey2 = E02 cos2 (ωt − βz) + sin2 (ωt − βz) = E02 .

This is a case of left-hand circular polarization.

4.35.

Z 2π Z π Z Ě ∗ 1 2 1 ∗ ∗ Pav = Ěθ Ȟφ dS = Ě0 e−jβr 0 ejβr Ě × Ȟ · dS = sin2 θr2 sin θ dθ dφ 2 2 2 η (βr) 0 S 0 0 S Z π Z 2π 2 16π | Ě | |Ě0 |2 2 0 = . sin3 θ dθ = dφ 2 η0 β 2 0 η 0 3β 0 Z

4.36. (a) Z 2π

I

t

Ȟ · dl =

ǐ = C

0

Ě0 a Ě0 a φ̂ · φ̂ρ dφ = 2π . ηρ η

(b) Z L Z 2π Z b σ |Ě0 |2 a2 −2αz σ Ě · Ě∗ dV = e ρ dρ dφ dz ρ2 0 0 a 2 V 2 σ|Ě0 |2 a2 1 − e−2αL = 2π ln(b/a).

2 2α Z

Pd =

(c) Z

Z 2π Z b

∗

n̂ · (Ě × Ȟ ) dS =

Pcs (z) = S

0 2 2 |Ě0 | a = 2πe−2αz ln(b/a). η∗

|Ě0 |2 a2 −2αz e ρ dρ dφ η ∗ ρ2 a

(d) Ps = 0 = Pd + j2ω̌(Wm − We )Pf =⇒ Pd = − Re[Pf ].

Pf = Pcs z=L − Pcs z=0 = −

|Ě0 |2 a2 2π(1 − e−2αL ) ln(b/a), η∗

σ|Ě0 |2 a2 1 − e−2αL 2π ln(b/a) = |Ě0 |2 a2 Re[1/η ∗ ]2π(1 − e−2αL ) ln(b/a), 2 2α σ α= .

4 Re[1/η ∗ ] 4.37. From (4.454) we have the structure reflection coefficients for each case as R̃A =

Γ̃ + Γ̃2 P̃A2 − Γ̃P̃ 2 − P̃ 2 P̃A2 , 1 + Γ̃P̃A2 − Γ̃2 P̃ 2 − Γ̃P̃ 2 P̃B2

R̃B =

123

Γ̃ + Γ̃2 P̃B2 − Γ̃P̃ 2 − P̃ 2 P̃B2 .

1 + Γ̃P̃B2 − Γ̃2 P̃ 2 − Γ̃P̃ 2 P̃B2

Solving each equation for P̃ 2 gives, respectively, P̃ 2 =

Γ̃ + Γ̃2 P̃A2 − R̃A − Γ̃P̃A2 R̃A

P̃ 2 =

Γ̃ + Γ̃2 P̃B2 − R̃B − Γ̃P̃B2 R̃B

(*)

Γ̃ + P̃A2 − Γ̃2 R̃A − Γ̃P̃A2 R̃A Γ̃ + P̃B2 − Γ̃2 R̃B − Γ̃P̃B2 R̃B

.

(**)

Equating and clearing the denominators gives

Γ̃ + Γ̃2 P̃A2 − R̃A − Γ̃P̃A2 R̃A Γ̃ + P̃B2 − Γ̃2 R̃B − Γ̃P̃B2 R̃B

= Γ̃ + Γ̃2 P̃B2 − R̃B − Γ̃P̃B2 R̃B Γ̃ + P̃A2 − Γ̃2 R̃A − Γ̃P̃A2 R̃A .

Multiplying an collecting terms we find Γ̃4 − C Γ̃3 + C Γ̃ − 1 = 0 where

C=

R̃A − R̃B

1 + R̃A R̃B 1 + P̃A2 P̃B2 + P̃A2 − P̃B2 R̃B P̃A2 − R̃A P̃B2

.

This takes the form of (4.461) with the solution given by s C C 2 Γ̃ = ± − 1.

2 2 Once Γ̃ is found, P̃ may be computed using either (*) or (**).

4.38. As in the two thickness method, let ∆B = κ∆A . Then P̃A2 = e−2jkz ∆ = Q and P̃B2 = e−2jkz (κ∆) = Qκ . The transmission coefficients are then T̃A = T̃B =

(1 − Γ̃2 )P̃A (1 − Γ̃2 )Q = , 1 − Γ̃2 P̃A2 1 − Γ̃2 Q2

(1 − Γ̃2 )P̃B (1 − Γ̃2 )Qκ = .

1 − Γ̃2 P̃B2 1 − Γ̃2 Q2κ

Solving each equation for Γ̃2 gives Q − T̃A , Q − T̃A Q2 Qκ − T̃B Γ̃2 = .

Q − T̃B Q2κ

Γ̃2 =

Equating the expressions and clearing the denominators we find Q2κ − T̃A Q2κ−1 −

T̃A κ+1 T̃A κ−1 Q + Q + T̃A Q − 1 = 0.

T̃B T̃B 124

(*) (**)

For the special case of κ = 2 this reduces to Q4 − CQ3 + CQ − 1 = 0 where C = T̃A +

T̃A .

T̃B

This takes the form of (4.461) with the solution given by s C C 2 − 1.

Q= ± 2 2 Once Q and thus P̃A is found, Γ̃ may be computed using either (*) or (**).

125

Chapter 5 5.1. We assume that µ, , and σ are even about z = 0. For the case of even symmetry we have the following: source/field components

symmetry about z = 0

i Jxi , Jyi , Jmz i , Ji Jzi , Jmx my

even odd even odd

Ex , Ey , Hz Ez , Hx , Hy All the terms

∂Ez , ∂y

∂Ey , ∂z

µ

∂Hx , ∂t

i Jmx ,

in equation (5.1) are odd in z, so symmetry is obeyed. The rest of the problem is similar.

5.2. Recall that the field due to a Hertzian dipole on the z axis is given by

I˜ dz e−jkr η0 2 I˜ dz e−jkr η0 1 Ẽ = r̂ 2 + cos θ + θ̂ jωµ0 + + sin θ.

4π r r jω 0 r2 4π r r jω 0 r2 Write Ẽ = f1 (r) cos θr̂ + f2 (r) sin θθ̂.

Case 1: horizontal dipole. See Figure 35.

Figure 35 ˜ and assume Ĩi and Ĩ are directed oppositely. Then ri = r, θi = θ, and Choose di = d, |I˜i | = |I|, (Ẽri )tan = −(Ẽr )tan ,

(Ẽθi )tan = −(Ẽθ )tan .

So (Ẽ + Ẽi )tan = 0 126

Figure 36 and the boundary condition is satisfied.

Case 2: vertical dipole. See Figure 36. ˜ and assume Ĩi and Ĩ are co-directed. Then ri = r and θi = π − θ so that Choose di = d, |I˜i | = |I|, cos θi = − cos θ and sin θi = sin θ. We have (Ẽri )tan = −(Ẽr )tan ,

(Ẽθi )tan = −(Ẽθ )tan .

So (Ẽ + Ẽi )tan = 0 and the boundary condition is satisfied.

Summary: horizontal currents image in opposite directions, while vertical currents image in the same direction.

5.3. The impressed field is given by ˜ Z ∞+j∆ e−jky |y−h|

ω µ̃I Ẽzi (x, y) = − 2π

−∞+j∆

2ky

e−jkx x dkx .

The scattered field consists of both upward and downward traveling waves: Z ∞+j∆ h i 1 s Ẽz (x, y) = Ã+ (kx )e−jky y + Ã− (kx )ejky y e−jkx x dkx .

2π −∞+j∆ To find Ã+ and Ã− , apply the boundary condition Ẽzi + Ẽzs = 0 at y = ±d to get the equations ω µ̃I˜ −jky (d−h) e = Ã+ e−jky d + Ã− ejky d , 2ky ω µ̃I˜ −jky (d+h) e = Ã+ ejky d + Ã− e−jky d .

2ky Solving simultaneously, we obtain Ã− =

ω µ̃I˜ −jky d sin ky (d + h) e , 2ky sin 2ky d

Ã+ = 127

ω µ̃I˜ −jky d sin ky (d − h) e .

2ky sin 2ky d

Therefore ˜ Z ∞+j∆ e−jky d sin ky (d − h)

ω µ̃I Ẽzs = 2π

2ky

−∞+j∆

sin 2ky d

e

−jky y

sin ky (d + h) jky y −jkx x + e e dkx . sin 2ky d

The term in brackets can be rewritten algebraically to put this into the form of (5.8).

5.4. Assume J̃i = 0, J̃im 6= 0. From (5.71) we have ˜c i 0 e−jkR Ãh (r) = J̃m (r ) dV 0 = ˜c R V 4π Z

Z

J̃im (r0 )G(r|r0 ) dV 0 .

V

Examine (5.53): ω ∇(∇ · Ãh ) − jω Ãh k2 Z Z ω˜ c i 0 0 0 c J̃m (r )G(r|r ) dV − jω˜ J̃im (r0 )G(r|r0 ) dV 0

= −j 2 ∇∇ · k V V Z h i 1 i 0 0 i 0 0 c J̃m (r )G(r|r ) + 2 ∇∇ · J̃m (r )G(r|r ) dV 0 . = −jω˜

k V

H̃ = −j

Use

h i ∇ · J̃im (r0 )G(r|r0 ) = J̃im (r0 ) · ∇G(r|r0 ) + G(r|r0 ) ∇ · J̃im (r0 ) | {z } =0

to write c

Z

H̃ = −jω˜

c

ZV

= −jω˜

J̃im (r0 )G(r|r0 ) +

1 ∇∇G(r|r0 ) · J̃im (r0 ) k2

dV 0

Ḡe (r|r0 ) · J̃im (r0 ) dV 0

V

where

∇∇ Ḡe (r|r ) = Ī + 2 G(r|r0 ). k 0

Next, examine (5.63): 1 ∇ × Ãh ˜c Z = −∇ × J̃im (r0 )G(r|r0 ) dV 0 V Z =− ∇ × [J̃im (r0 )G(r|r0 )] dV 0 .

Ẽ(r) = −

V

Use h i ∇ × J̃im (r0 )G(r|r0 ) = ∇G(r|r0 ) × J̃im (r0 ) + G(r|r0 ) ∇ × J̃im (r0 ) = −J̃im (r0 ) × ∇G(r|r0 ) {z } | =0

128

and (B.15): h i J̃im (r0 ) × ∇G(r|r0 ) = J̃im (r0 ) × ∇G(r|r0 ) · Ī

= ∇G(r|r0 ) × Ī · J̃im (r0 ) = Ḡm (r|r0 ) · J̃im (r0 ).

So

Z

Ḡm (r|r0 ) · J̃im (r0 ) dV 0 .

Ẽ(r) = V

5.5. Start with

J̃ · n̂0 dS 0 .

4πR S

I V = lim ∇ ∆V →0

Assume J̃ = ẑJ˜z with J˜z constant. Then

I I R 1 0 0 dS = − lim J̃ · n̂0 dS 0 .

V = lim J̃ · n̂ ∇ ∆V →0 S ∆V →0 S 4πR 4πR3 We have

 ˜  Jz , 0 J̃ · n̂ = −J˜z ,   0,

on the top surface of the cube, on the bottom surface, on the other surfaces.

Assume r = 0 (center of cube) so that ( aẑ + x0 x̂ + y 0 ŷ, −R = −(r − r ) = r = −aẑ + x0 x̂ + y 0 ŷ, 0

0

on top, on bottom.

Then Z

R V = − lim 2 J˜z dS 0 ∆V →0 4πR3 top Z aZ a aẑ + x0 x̂ + y 0 ŷ dx0 dy 0 = lim 2 J˜z a→0 4π(a2 + x0 2 + y 0 2 )3/2 −a −a Z Z 2a ˜ a a dx0 dy 0 = ẑ lim Jz 2 02 0 2 3/2 a→0 4π −a −a (a + x + y ) Z a 2a ˜ dy 0 p Jz · 2a = ẑ lim 2 2 a→0 4π −a (a2 + y 0 ) 2a2 + y 0 2a ˜ 4π = ẑ lim Jz · a→0 4π 6a 1˜ = ẑ Jz .

3 Setting this equal to L̄ · J̃ = L̄ · (ẑJ˜z ), we find that Lzz = 1/3. Since (J˜z ẑ) · n̂ = 0 on the side surfaces, we have Lxz = Lyz = 0. By letting J̃ = x̂J˜x and J̃ = ŷJ˜y , we see immediately that Lxx = Lyy = Lzz = 1/3 and Lyx = Lyz = Lzx = Lzz = 0. Therefore   1/3 0 0 1 L̄ =  0 1/3 0  = Ī.

3 0 0 1/3 129

5.6. Start with

J̃ · n̂0 dS 0 .

S 4πR

I V = lim ∇ ∆V →0

Assume J̃ = ẑJ˜z with J˜z constant. Then

I I R 1 0 0 dS = − lim J̃ · n̂0 dS 0 .

V = lim J̃ · n̂ ∇ ∆V →0 S ∆V →0 S 4πR 4πR3 We have

 ˜  on the top surface of the cylinder, Jz , 0 ˜ J̃ · n̂ = −Jz , on the bottom surface,   0, on the curved surface.

Assume r = 0 (center of cylinder) so that ( aẑ + x0 x̂ + y 0 ŷ = aẑ + ρ0 cos φ0 x̂ + ρ0 sin φ0 ŷ, on top, 0 0 −R = −(r − r ) = r = −aẑ + x0 x̂ + y 0 ŷ = −aẑ + ρ0 cos φ0 x̂ + ρ0 sin φ0 ŷ, on bottom.

Then Z

R dS 0 J˜z 3 ∆V →0 4πR top Z 2π Z a aẑ + ρ0 cos φ0 x̂ + ρ0 sin φ0 ŷ 0 0 0 = lim 2 J˜z ρ dρ dφ a→0 4π(a2 + ρ0 2 )3/2 0 0 " # a 1 ˜ 1 = ẑ lim 2(2π) Jz a − p a→0 4π ρ0 2 + a2 0

1 1 = −ẑ lim J˜z a √ − a→0 2a2 a

1 = ẑJ˜z 1 − √ 2 ˜ = 0.293Jz ẑ.

V = − lim 2

Setting this equal to L̄ · (ẑJ˜z ), we find that Lzz = 0.293.

5.7. (a) 1 e−jkr = lim = 0. r→∞ 4πr r→∞ 4πr

lim ψ̃(r, ω) = lim

r→∞

(b)

∂ ψ̃ ∂ e−jkr 1 e−jkr = = (−jkr − 1) 2 , ∂r ∂r 4πr 4π r !

e−jkr e−jkr ∂ ψ̃ 1 e−jkr lim r jk ψ̃ + = lim r jk = lim − = 0. + (−jkr − 1) 2 r→∞ r→∞ r→∞ ∂r 4πr 4π r 4πr

130

5.8. Begin with ∇2 A = ∇(∇ · A) − ∇ × (∇ × A) Use ∇·A=

and ∇ = ∇t + û

∂ . ∂u

∂Au ∂ · (At + ûAu ) = ∇t · At + ∇t + û ∂u ∂u

to write

∂Au ∂ ∇t · A t + ∇(∇ · A) = ∇t + û ∂u ∂u ∂Au ∂ 2 Au ∂At + û + ∇t . = ∇t (∇t · At ) + û∇t · ∂u ∂u2 ∂u With the help of (B.104) we obtain ∂Au ∂ 2 At ∂Au − ∇ t × ∇ t × At + − ∇t 2 ∂u ∂u ∂u ∂ 2 At = ∇t (∇t · At ) + − ∇t × ∇t × At . ∂u2

(∇2 A)t = ∇t (∇t · At ) + ∇t

Finally, ∂At ∂At ∂ 2 Au + − ∇t · + ∇2t Au (∇2 A)u = ∇t · 2 ∂u ∂u ∂u

∂2 = ∇2t + 2 Au ∂u = ∇ 2 Au .

5.9. (B.88): û · At = û · [A − ûAu ] = û · A − Au = Au − Au = 0.

(B.89):

(û · ∇t )φ =

∂ û · ∇ − û ∂u

φ = û · ∇φ −

∂φ = 0. ∂u

(B.90):

∂ ∂φ φ = ∇φ − û . ∇t φ = ∇ − û ∂u ∂u (B.91):

∂φ ∂φ ∂φ û · (∇φ) = û · ∇t φ + û = û · (∇t φ) + = , ∂u ∂u ∂u

∂ ∂φ ∂φ (û · ∇)φ = û · ∇t + û φ = û · û = . ∂u ∂u ∂u (B.92):

∂φ ∂φ ∂φ û · (∇t φ) = û · ∇φ − û = − = 0. ∂u ∂u ∂u 131

(B.93):

∂ ∂φ ∂φ ∂φ ∂φ ∇t · (ûφ) = ∇ − û · (ûφ) = ∇ · (ûφ) − = φ∇ · û + û · ∇φ − = − = 0. ∂u ∂u ∂u ∂u ∂u (B.94):

∂φ ∂ × (ûφ) = ∇ × (ûφ) − (û × û) ∇t × (ûφ) = ∇ − û = φ∇ × û − û × ∇φ ∂u ∂u

∂φ ∂ φ = −û × ∇t φ − û × û = −û × ∇t φ. = −û × ∇t + û ∂u ∂u (B.95):

∂ ∇t × (û × A) = ∇ − û × [û × (At + ûAu )] ∂u

∂ = ∇ − û × (û × At ) ∂u ∂At = ∇ × (û × At ) − û × û × ∂u

∂At ∂At + = ∇ × (û × At ) − û û · ∂u ∂u ∂At = ∇ × (û × At ) + ∂u ∂At = û∇ · At − (û · ∇)At + ∂u

∂At ∂At ∂ · At − + = û ∇t + û ∂u ∂u ∂u = û∇t · At .

(B.96): û × (∇t × A) = û × [∇t × (ûAu + At )] = û × [−û × ∇t Au + (∇t × At )] = −û × (û × ∇t Au ) + û × (∇t × At ) = −û(û · ∇t Au ) + ∇t Au (û · û) = ∇ t Au .

132

(B.97):

∂ × At û × (∇t × At ) = û × ∇ − û ∂u

∂At = û × (∇ × At ) − û × û × ∂u

∂At ∂At + (û · û) = ∇(û · At ) − (û · ∇)At − û û · ∂u ∂u ∂At ∂At =− + ∂t ∂t = 0.

(B.98): û · (û × A) = A · (û × û) = 0.

(B.99): û × (û × A) = û(û · A) − A(û · û) = −[A − û(û · A)] = −At .

5.10. (B.100):

∂φ ∂ φ = ∇t φ + û . ∇φ = ∇t + û ∂u ∂u (B.101):

∂ ∇ · A = ∇t + û · (At + ûAu ) ∂u ∂ ∂Au = ∇t · A t + (û · At ) + ∇t · (ûAu ) + ∂u ∂u ∂Au = ∇t · A t + . ∂u (B.102):

∂ ∇ × A = ∇t + û × (At + ûAu ) ∂u ∂At ∂Au = ∇t × At + û × + ∇t × (ûAu ) + (û × û) ∂u ∂u

∂At − ∇ t Au . = ∇t × At + û × ∂u (B.103):

∂φ ∇ φ = ∇ · (∇φ) = ∇ · ∇t φ + û ∂u 2

∂ = ∇t · (∇t φ) + ∂u

133

∂φ ∂u

= ∇2t φ +

∂2φ . ∂u2

(B.104):

∂At ∇ × (∇ × A) = × ∇t × At + û × − ∇ t Au ∂u

∂At ∂At + ∇t × û × = ∇t × (∇t × At ) + û × ∇t × − ∇ t Au ∂u ∂u

2 ∂Au ∂ At + û × û × − ∇t .

2 ∂u ∂u Use

∂ ∇t + û ∂u

∇t × û × and

∂At − ∇ t Au ∂u

= û ∇t ·

∂At − ∇ t Au ∂u

2

2 ∂Au ∂ At ∂Au ∂ At − ∇ = − − ∇ û × û × t t ∂u2 ∂u ∂u2 ∂u

to get, finally, ∇ × (∇ × A) = ∇t × (∇t × At ) −

∂ 2 At ∂Au ∂ 2 − ∇ − û (∇ · A ) − ∇ A t t t t u . ∂u2 ∂u ∂u

5.11. Take û × ∂/∂u of (5.90): −û ×

∂ Ẽu û × ∇t ∂u

!

Use −û ×

∂ Ẽu û × ∇t ∂u

+ û ×

!

∂ 2 Ẽt û × ∂u2

∂ Ẽu = ∇t ∂u

! = −jω µ̃û ×

and û ×

∂ H̃t ∂ J̃imt − û × . ∂u ∂u

∂ 2 Ẽt û × ∂u2

! =−

∂ 2 Ẽt ∂u2

to get ∂ H̃t ∂ J̃imt ∂ Ẽu ∂ 2 Ẽt − = −jω µ̃û × − û × . ∂u ∂u2 ∂u ∂u Now multiply (5.92) by −jω µ̃: ∇t

û × jω µ̃∇t H̃u − jω µ̃û ×

∂ H̃t = k 2 Et − jω µ̃J̃it . ∂u

Add (*) and (**): ∇t or

∂ Ẽu ∂ 2 Ẽt ∂ J̃imt 2 i − + û × jω µ̃∇ H̃ = k Ẽ − jω µ̃ J̃ − û × t u t t ∂u ∂u2 ∂u

∂2 ∂ Ẽu ∂ J̃imt 2 + jω µ̃û × ∇ H̃ + û × + jω µ̃J̃it . + k Ẽ = ∇ t t t u ∂u2 ∂u ∂u

134

(*)

(**)

5.12. Begin with ∇ × E = −Jm − µ

∂H , ∂t

∇×H=J+

Write ∇ × E = −Jmt − ûJmu − µ

∂E . ∂t

∂Ht ∂Hu − µû . ∂t ∂t

Use (B.102) to get ∇t × Et + û ×

∂Et ∂Ht ∂Hu − û × ∇t Eu = −Jmt − ûJmu − µ − µû . ∂u ∂t ∂t

Equate transverse components: û ×

∂Ht ∂Et − û × ∇t Eu = −Jmt − µ . ∂u ∂t

Equate longitudinal components: ∇t × Et = −ûJmu − µû

∂Hu . ∂t

Do the same with Ampere’s law: ∇t × Ht + û ×

∂Et ∂Eu ∂Ht − û × ∇t Hu = Jt + ûJu + + û , ∂u ∂t ∂u

û ×

So

∂Ht ∂Et − û × ∇t Hu = Jt + , ∂u ∂t ∂Eu . ∇t × Ht = ûJu + û ∂t

û ×

∂Et ∂Ht − û × ∇t Eu = −Jmt − µ ∂u ∂t

(1)

û ×

∂Ht ∂Et − û × ∇t Hu = −Jt + . ∂u ∂t

(2)

∂ 2 Et ∂Eu ∂Jmt ∂ 2 Ht + ∇ = −û × − µû × . t ∂u2 ∂u ∂u ∂t∂u

(3)

∂ 2 Ht ∂Hu ∂Jt ∂ 2 Et − µû × ∇t =µ + µ 2 . ∂t∂u ∂u ∂u ∂t

(4)

and

Apply û × ∂/∂u to (1): − Apply µ ∂/∂t to (2): µû ×

Subtract (4) from (3) to get 2

∂ 1 ∂2 ∂Eu ∂Hu ∂Jmt ∂Jt − 2 2 Et = ∇ t + µû × ∇t + û × +µ .

2 ∂u v ∂t ∂u ∂u ∂u ∂u Next, apply û × ∂/∂u to (2): −

∂ 2 Ht ∂Jt ∂ 2 Et + ∇ H = û × + û × . t u ∂u2 ∂u ∂t∂u 135

(5)

Apply ∂/∂t to (1): û ×

∂ 2 Et ∂Eu ∂Jmt ∂ 2 Ht − û × ∇t = − − µ 2 . ∂t∂u ∂t ∂t ∂t

Add (5) and (6) to get

2 ∂Eu ∂ 1 ∂2 ∂Jmt ∂Jt − 2 2 Ht = ∇t Hu − û × ∇t + − û × .

2 ∂u v ∂t ∂t ∂t ∂u To get the wave equation for Eu , start with (2.259) and use Gauss’s law: ∇2 E − µ

∂2E ∂J 1 = ∇ × Jm + µ + ∇ρ. ∂t2 ∂t

Apply û· to both sides to obtain ∇2 Eu − µ or

∂ 2 Eu ∂Ju 1 ∂ρ = û · (∇t × Jmt ) + µ + , 2 ∂t ∂t ∂u

1 ∂2 ∇ − 2 2 v ∂t 2

Eu =

1 ∂ρ ∂Ju +µ + û · (∇t × Jmt ). ∂u ∂t

The wave equation for Hu is obtained similarly, starting with (2.260).

5.13. Examine the first equation from Problem 5.12 with u = z:

2

2 1 ∂2 ∂ 2 Hz ∂ 2 Hz ∂ Ez ∂ 2 Ez ∂ − (x̂Hx + ŷHy ) = x̂ + ŷ − ẑ × x̂ + ŷ . ∂z 2 v 2 ∂t2 ∂x∂z ∂y∂z ∂x∂t ∂y∂t Equate x-components to obtain: 2

∂ 1 ∂2 ∂ 2 Hz ∂ 2 Ez − H = +

. x ∂z 2 v 2 ∂t2 ∂x∂z ∂y∂t Equate y-components to obtain: 2

∂ 1 ∂2 ∂ 2 Hz ∂ 2 Ez − H = −

. y ∂z 2 v 2 ∂t2 ∂y∂z ∂x∂t Examine the second equation from Problem 5.12 with u = z: 2

2

∂ 1 ∂2 ∂ 2 Ez ∂ 2 Ez ∂ Hz ∂ 2 Hz − (x̂Ex + ŷEy ) = x̂ + ŷ + µẑ × x̂ + ŷ . ∂z 2 v 2 ∂t2 ∂x∂z ∂y∂z ∂x∂t ∂y∂t Equate x-components to obtain: 2

1 ∂2 ∂ 2 Ez ∂ 2 Hz ∂ − E = − µ . x ∂z 2 v 2 ∂t2 ∂x∂z ∂y∂t Equate y-components to obtain: 2

∂ 1 ∂2 ∂ 2 Ez ∂ 2 Hz − E = + µ . y ∂z 2 v 2 ∂t2 ∂y∂z ∂x∂t The third and fourth equations from Problem 5.12 with u = z yield the wave equations.

136

(6)

5.14. Let Πe = ûΠe and Πh = 0. Then

∂2 2 ∇ − µ 2 ûΠe = 0. ∂t Use (B.105): ∇2 (ûΠe ) = û∇2 Πe , so

1 ∂2 2 ∇ − 2 2 Πe = 0. v ∂t

(*)

We also have E = ∇ × (∇ × Πe ) = ∇ × (∇ × ûΠe ).

Use (B.104): E = ∇t

∂Πe − û∇2t Πe . ∂u

By (*) we have ∂ 2 Πe 1 ∂ 2 Πe = , ∂u2 v 2 ∂t2 2

∂Πe ∂ Πe 1 ∂ 2 Πe E = ∇t + û − 2 . ∂u ∂u2 v ∂t2 ∇2 Πe = ∇2t Πe +

so

Finally, H = ∇ ×

∂ ∂ (ûΠe ) = ∇ × (ûΠe ). ∂t ∂t

Use (B.102): H=

∂ ∂Πe (−û × ∇t Πe ) = − û × ∇t . ∂t ∂t

5.15. Let Πh = ûΠh and Πe = 0. Then

∂2 ∇2 − µ 2 ûΠh = 0. ∂t Use (B.105): ∇2 (ûΠh ) = û∇2 Πh , so

1 ∂2 2 ∇ − 2 2 Πh = 0. v ∂t

We also have H = ∇ × (∇ × Πh ) = ∇ × (∇ × ûΠh ).

Use (B.104): E = ∇t

∂Πh − û∇2t Πh . ∂u

By (*) we have ∇2 Πh = ∇2t Πh +

∂ 2 Πh 1 ∂ 2 Πh = , ∂u2 v 2 ∂t2

137

(*)

so ∂Πh H = ∇t + û ∂u

2 ∂ Πh 1 ∂ 2 Πh . − 2 ∂u2 v ∂t2

Finally, from (5.54), ∂ ∂ (ûΠh ) = −µ ∇ × (ûΠh ). ∂t ∂t

E = −µ∇ × Use (B.102): E = −µ

∂ ∂Πh (−û × ∇t Πh ) = µû × ∇t . ∂t ∂t

5.16. From Problem 5.14, E has no u-component when ∂ 2 Πe 1 ∂ 2 Πe = . ∂u2 v 2 ∂t2 So E = ∇t

∂Πe ∂u

(*)

and H = − û × ∇t

∂Πe . ∂t

The wave equation is

1 ∂2 ∇2 − 2 2 Πe = 0. v ∂t Use (*):

∂2 2 ∇ − 2 Πe = 0 =⇒ ∇2t Πe = 0. ∂u 5.17. From Problem 5.15, H has no u-component when 1 ∂ 2 Πh ∂ 2 Πh = . ∂u2 v 2 ∂t2 So H = ∇t

∂Πh ∂u

and E = µû × ∇t

(*) ∂Πh . ∂t

The wave equation is

1 ∂2 2 ∇ − 2 2 Πh = 0. v ∂t Use (*):

∂2 ∇ − 2 ∂u 2

Πh = 0 =⇒ ∇2t Πh = 0.

5.18. Start with Ẽ = x̂Ẽ0 e−jkz ,

H̃ = ŷ

138

Ẽ0 −jkz e . η

(a) TEy , use (5.96): ∂ Ẽy + jω µ̃ŷ × ∇t H̃y ∂y ! ∂ Ẽ0 −jkz = jω µ̃ŷ × ẑ e ∂z η

k 2 Ẽt = ∇t

= jω µ̃x̂(−jk) = k 2 ηx̂

Ẽ0 −jkz e η

Ẽ0 −jkz e . η

So Ẽt = x̂Ẽ0 e−jkz . (b) TMx , use (5.95): ∂ H̃x ∂ Ẽx − jω˜ c x̂ × ẑ ∂x ∂z

c ∂ −jkz = jω˜ ŷ Ẽ0 e ∂z = ω˜ c kŷẼ0 e−jkz

k 2 H̃t = ∇t

=

k2 ŷẼ0 e−jkz . η

So H̃t = ŷ

Ẽ0 −jkz e . η

(c) Use (5.124): Ẽ = jω µ̃ŷ × ∇t Π̃h ∂ = jω µ̃ŷ × ẑ ∂z

Ẽ0 −jkz e ηk 2

!

Ẽ0 −jkz e ηk 2 = x̂Ẽ0 e−jkz .

= ω µ̃kx̂

Use (5.125): ∂ Π̃h + ŷ H̃ = ∇t ∂y

∂2 Ẽ0 2 + k Π̃h = ŷk 2 Π̃h = ŷ e−jkz .

2 ∂y η

∂ Π̃e Ẽ = ∇t + x̂ ∂x

∂2 2 + k Π̃e = x̂k 2 Π̃e = x̂Ẽ0 e−jkz . ∂x2

(d) Use (5.122):

139

Use (5.123): H̃ = −jω˜ c x̂ × ∇t Π̃e ∂ = −jω˜ c x̂ × ẑ ∂z = ω˜ c kŷ = ŷ

Ẽ0 −jkz e k2

!

Ẽ0 −jkz e k2

Ẽ0 −jkz e . η

(e), (f) Similar.

5.19. Using the given identity, we have Z

2

I

2

(ψ̃m ∇ ψ̃n − ψ̃n ∇ ψ̃m ) dS = Γ

S

∂ ψ̃n ∂ ψ̃m ψ̃m + ψ̃n ∂n ∂n

! dl.

2 ψ̃ : Use ∇2 ψ̃n = ∇2t ψ̃n = −kc,n n

Z

2 2 2 2 (−kc,n ψ̃n ψ̃m + kc,m ψ̃m ψ̃n ) dS = (kc,m − kc,n )

S

Z

I ψ̃n ψ̃m dS =

S

Γ

∂ ψ̃m ∂ ψ̃n − ψ̃n ψ̃m ∂n ∂n

! dl.

The boundary conditions are that ψ̃e = 0 on Γ for TM modes and ∂ ψ̃h /∂n = 0 on Γ for TE modes.

Since kc,m 6= kc,n , we must have Z ψ̃n ψ̃m dS = 0.

S

So

Z

2 2 (kc,n ψ̃n )(kc,m ψ̃m ) dS = 0.

S 2 ψ̃ = e For TM modes, kc,n n z,n . In this case

Z ẽz,n ẽz,m dS = 0 S

so that

Z

ẽz,n e±jkz,n z

ẽz,m e±jkz,m z dS = 0

S

or

Z Ẽz,n Ẽz,m dS = 0.

S

TE modes are treated similarly.

5.20. (a) For TM modes we write Ẽ = Ẽt + ẑẼz ,

Ẽm · Ẽn = Ẽtm · Ẽtn + Ẽzm Ẽzn .

140

Let the modal indices for mode m be α, β. Let the modal indices for mode n be γ, δ.

βπ βπx απy δπ δπx γπy Ẽtm · Ẽtn = −kzm kzn Aαβ Aγδ cos sin · cos sin a a b a a b

απ βπx απy γπ δπx γπy ∓jkzn z ∓jkzm z + e e sin cos · sin cos b a b b a b Z Z b βπ δπ a βπx δπx απy γπy cos Ẽtm · Ẽtn dS = kzm kzn Aαβ Aγδ e e cos dx sin sin dy a a 0 a a b b CS 0

Z Z b απ γπ a βπx απy δπx γπy + sin cos sin dx cos dy b b 0 a a b b 0

Z

∓jkzn z ∓jkzm z

But

so

Z b απy βπx δπx γπy sin cos cos dx = sin dy = 0, a a b b 0 0 Z Ẽtm · Ẽtn dS = 0.

Z a

CS

Next, βπx απy δπx γπy sin sin sin a b a b Z Z a Z b δπx απy γπy βπx 2 2 sin dx sin dy = 0.

Ẽzm Ẽzn dS = kcn kcm Aαβ Aγδ e∓jkzn z e∓kzm z sin sin a a b b CS 0 0 2 2 Ẽzm Ẽzn = kcn kcm Aαβ Aγδ e∓jkzn z e∓kzm z sin

Thus

Z Ẽm · Ẽn dS = 0.

CS

(b) Write H̃m · H̃n = H̃tm · H̃tn

βπx απy γπ δπx γπy απ sin cos · sin cos b a b b a b

βπx απy δπ δπx γπy βπ + cos sin · cos sin , a a b a a b ∓jkzn z ∓jkzm z

= −kzm kzn Yem Yen Aαβ Aγδ e

Z

e

The treatment for TE modes is similar.

141

απ γπ b b

Z a

βπx δπx H̃m · H̃n dS = kzn kzm Yem Yen Aαβ Aγδ e e sin sin dx a a CS 0

Z Z b βπ δπ a βπx δπx απy γπy + cos dx sin sin dy = 0. cos a a 0 a a b b 0 ∓jkzn z ∓jkzm z

Z b cos 0

απy γπy cos dy b b

5.21. ẑ · (ět × ȟ∗t ) = ẑ · [Zh (ẑ × ȟt ) × (−jkz∗ )∇t ψ̌h∗ ] = −kz kz∗ Zh ẑ · [(ẑ × ∇t ψ̌h ) × ∇t ψ̌h∗ ] = |kz |2 Zh ẑ · [∇t ψ̌h∗ × (ẑ × ∇t ψ̌h )] = |kz |2 Zh ẑ · [ẑ(∇t ψ̌h∗ ) · (∇t ψ̌h ) − ∇t ψ̌h (ẑ · ∇t ψ̌h∗ )] = |kz |2 Zh (∇t ψ̌h ) · (∇t ψ̌h∗ ).

1 Re 2

Pav =

Z

Z

1 Re{Zh }|kz |2 2

ẑ · (ět × ȟ∗t ) dS =

CS

(∇t ψ̌h ) · (∇t ψ̌h∗ ) dS.

CS

Use (B.35): Z

(∇t a · ∇t b + a∇2t b) dS =

I a Γ

S

∂b dl. ∂n

Let a = ψ̌h and b = ψ̌h∗ . Then Z

(∇t ψ̌h · ∇t ψ̌h∗ + ψ̌h ∇2t ψ̌h∗ ) dS =

I

CS

ψ̌h Γ

∂ ψ̌h∗ dl. ∂n

Since ∂ ψ̌h /∂n = 0 on Γ and ∇2t ψ̌h = −kc2 ψ̌h , we have Z

(∇t ψ̌h ) · (∇t ψ̌h∗ ) dS = −

Z

CS

ψ̌h ∇2t ψ̌h∗ dS = kc2

CS

Z

ψ̌h ψ̌h∗ dS.

CS

Therefore

Z 1 2 2 Pav = Re{Zh }|kz | kc ψ̌h ψ̌h∗ dS. 2 CS For a lossless guide, kz = β and Zh = ω̌µ/β. In this case, Z Z 1 ω̌µ 2 2 1 ∗ 2 Pav = β kc ψ̌h ψ̌h dS = ω̌µβkc ψ̌h ψ̌h∗ dS. 2 β 2 CS CS

5.22. We have

hwe i = 4 so

Z lZ 0

hwe i

= l 4

Ě · Ě∗ dS dz,

CS

Z

Ě · Ě∗ dS.

CS

Substitute from (5.199) and (5.200): Z hwe i

= |Zh |2 |kz |2 (ẑ × ∇t ψ̌h ) · (ẑ × ∇t ψ̌h∗ ) dS l 4 CS

Z ω̌µ 2 2 β (ẑ × ∇t ψ̌h ) · (ẑ × ∇t ψ̌h∗ ) dS = 4 β CS Z

2 = (ω̌µ) ẑ · [∇t ψ̌h∗ × (ẑ × ∇t ψ̌h )] dS 4 ZCS

2 = (ω̌µ) ∇t ψ̌h∗ · ∇t ψ̌h dS 4 CS Z

2 2 = (ω̌µ) kc ψ̌h ψ̌h∗ dS 4 CS 142

(cf., Problem 5.21).

5.23. For TM modes, 1 Pav = ω̌ βkc2 2 Use ψ̌e (x, y) = Anm sin

Z

ψ̌e ψ̌e∗ dS.

CS

nπx a

sin

mπy b

to get Z a Z b

mπy 1 2 nπx 2 2 sin (Pav )nm = ω̌ βnm kc,nm |Anm | dx sin2 dy 2 a b 0 0 1 a b 2 = ω̌ βnm kc,nm |Anm |2 · · .

2 2 2 5.24. For TE modes, 1 Pav = ω̌µβkc2 2 Use ψ̌h (x, y) = Bnm cos

Z

ψ̌h ψ̌h∗ dS.

CS

nπx a

cos

mπy b

to get Z b Z a nπx mπy 1 2 cos2 dx dy |Bnm |2 cos2 (Pav )nm = ω̌µβnm kc,nm 2 a b 0 0 1 a b 2 = ω̌µβnm kc,nm |Bnm |2 · · .

2 2 2 5.25. Assume µ, constant and that the region is source-free. Write ∇×E=−

∂B ∂t

and equate the θ components:

1 1 ∂Er ∂ ∂Hθ − (rEφ ) = −µ . r sin θ ∂φ ∂r ∂t Write ∇×H=

(1)

∂E ∂t

and equate the φ components:

∂Eφ 1 ∂ ∂Hr (rHθ ) − = . r ∂r ∂θ ∂t Now multiply (2) by r and take ∂/∂r:

∂Eφ ∂ ∂ ∂Hr ∂ (rHθ ) − = r , ∂r ∂r ∂θ ∂r ∂t 143

(2)

or

∂2 ∂ 2 Hr ∂ (rH ) − = θ ∂r2 ∂r∂θ ∂r

∂Eφ r . ∂t

(3)

Multiply (1) by r and take ∂/∂t: ∂ 2 Er ∂ − sin θ ∂φ∂t ∂r

∂Eφ ∂2 r = −µ 2 (rHθ ). ∂t ∂t

Substitute (3) into (4): ∂2 ∂ 2 Hr ∂2 ∂ 2 Er − 2 (rHθ ) + = −µ 2 (rHθ ) sin θ ∂φ∂t ∂r ∂r∂θ ∂t or

∂2 1 ∂2 ∂ 2 Er ∂ 2 Hr − (rH ) = + . θ ∂r2 v 2 ∂t2 sin θ ∂φ∂t ∂r∂θ Thus, Hθ can be found from Er and Hr by solving this differential equation. Similarly,

2 1 ∂2 1 ∂ 2 Hr ∂ 2 Er ∂ − + , (rH ) = − φ ∂r2 v 2 ∂t2 ∂θ∂t sin θ ∂r∂φ 2

∂ 1 ∂2 1 ∂ 2 Er ∂ 2 Hr − (rE ) = + µ , φ ∂r2 v 2 ∂t2 sin θ ∂φ∂r ∂θ∂t and

2 1 ∂2 ∂ 2 Er 1 ∂ 2 Hr ∂ − (rE ) = + µ . θ ∂r2 v 2 ∂t2 ∂θ∂r sin θ ∂φ∂t Now we can perform a TE–TM decomposition.

A. TE case: Er = 0.

∂2 1 ∂2 ∂ 2 Hr − (rH ) = , θ ∂r2 v 2 ∂t2 ∂r∂θ

2 1 ∂2 1 ∂ 2 Hr ∂ − (rH ) = , φ ∂r2 v 2 ∂t2 sin θ ∂r∂φ 2

∂ 1 ∂2 ∂ 2 Hr − (rE ) = µ , φ ∂r2 v 2 ∂t2 ∂θ∂t

and

B. TM case: Hr = 0.

∂2 1 ∂2 − ∂r2 v 2 ∂t2

(rEθ ) = µ

1 ∂ 2 Hr . sin θ ∂φ∂t

∂2 1 ∂2 ∂ 2 Er − (rHθ ) = , 2 2 2 ∂r v ∂t sin θ ∂φ∂t 2

∂ 1 ∂2 ∂ 2 Er − (rH ) = − , φ ∂r2 v 2 ∂t2 ∂θ∂t 2

∂ 1 ∂2 1 ∂ 2 Er − (rE ) = , φ ∂r2 v 2 ∂t2 sin θ ∂φ∂r

and

∂2 1 ∂2 − ∂r2 v 2 ∂t2

(rEθ ) =

144

∂ 2 Er . ∂θ∂r

(4)

5.26. The backscatter direction is θ = π. Since the incident electric field is polarized in the x-direction, only the x-component of Ẽs exists along the direction θ = π. This is given by Ẽxs

= −Ẽθs θ=π

θ=π φ=0

.

In the far zone we have e Ẽθs = Ẽ0

−jkr

kr

cos φ

∞ X

j

n+1

n=1

1 10 1 bn sin θPn (cos θ) − cn P (cos θ) . sin θ n

To evaluate this along θ = π, we use Pn1 (cos θ) (−1)n =− n(n + 1), θ→π sin θ 2 d 1 (−1)n lim Pn (cos θ) = n(n + 1). θ→π dθ 2 lim

(1) (2)

From (2) we also have d 1 Pn (cos θ) = lim (− sin θ)Pn10 (cos θ) θ→π θ→π dθ lim

where

Pn10 (x) =

dPn1 (x) . dx

So

∞ X (−1)n (−1)n e−jkr n+1 cos(0) j − n(n + 1)bn + n(n + 1)cn Ẽx = −Ẽ0 kr 2 2 θ=π n=1

=

∞ e−jkr X

Ẽ0 2 kr

j n+1 n(n + 1)(cn − bn ).

n=1

Use cn = −

Jˆn (ka) (2)

Ĥn (ka)

bn = −

an ,

Jˆn0 (ka) (2)0

Ĥn (ka)

an

to get " cn − bn =

Jˆn0 (ka) (2)0

Ĥn (ka)

−

Jˆn (ka) (2)

Ĥn (ka)

#

" an =

(2) (2)0 Jˆn0 (ka)Ĥn (ka) − Ĥn (ka)Jˆn (ka) (2)0

(2)

Ĥn (ka)Ĥn (ka)

# an .

Use d Jˆn0 (x) = [xjn (x)] = jn (x) + xjn0 (x), dx d (2) (2)0 Ĥn(2)0 (x) = [xh(2) n (x)] = hn (x) + xhn (x), dx and (E.99) to get Jˆ0 (x)Ĥ (2) (x) − Ĥ (2)0 (x)Jˆn (x) n

n

n = [jn (x) + xjn0 (x)][xjn (x) − jxnn (x)] − [jn (x) − jnn (x) + x(jn0 (x) − jn0n (x))][xjn (x)] = x[jn2 (x) − jjn (x)nn (x) − jn2 (x) + jnn (x)jn (x)] + x2 [jn (x)jn0 (x) − jjn0 (x)nn (x) − jn (x)jn0 (x) + jn0n (x)jn (x)] = jx2 [jn (x)n0n (x) − jn0 (x)nn (x)]

= j.

145

So cn − bn = j

an (2)0 (2) Ĥn (ka)Ĥn (ka)

an = j −n

where

2n + 1 . n(n + 1)

Therefore ∞

Ẽ0 e−jkr X n+1 j −n (2n + 1) 1 = j (−1)n n(n + 1)j Ẽx (2)0 2 kr n(n + 1) Ĥn (ka)Ĥn(2) (ka) θ=π =−

n=1 ∞ −jkr X e

Ẽ0 2 kr

(−1)n

n=1

2n + 1 (2)0 (2) Ĥn (ka)Ĥn (ka)

.

Now use (5.167):

s 2 2 |E | σ = lim 4πr , r→∞ |Ẽi |2

σ = lim 4πr

2

Ẽ0 2

2

e−jkr kr

2

∞ X

2n + 1

n

(−1)

(2)0

2

(2)

Ĥn (ka)Ĥn (ka)

n=1

|Ẽ0 |2 |e−jkz |2

r→∞

1 = 4π 4

Ẽi = x̂Ẽ0 e−jkz .

λ 2π

2 X ∞

∞

2

(−1)n (2n + 1)

(2)0 (2) n=1 Ĥn (ka)Ĥn (ka) 2 (−1)n (2n + 1)

λ2 X = (2)0 (2) 4π Ĥn (ka)Ĥn (ka)

.

n=1

5.27. Start with ∞

λ2 X (−1)n (2n + 1) σ= (2)0 (2) 4π Ĥn (ka)Ĥn (ka)

2

.

n=1

Use Ĥn(2) (x) = Jˆn (x) − j N̂n (x) = xjn (x) − jxnn (x).

For x 1, (E.59) and (E.60) give jn (x) ≈

2n n! xn , (2n + 1)!

So Ĥn(2) (x) ≈ −j N̂n (x) = j

nn (x) ≈ −

(2n)! −n x , 2n n!

0

Ĥn(2) (x)Ĥn(2) (x) ≈ n Therefore

(2n)! −(n+1) x .

2n n!

0

Ĥn(2) (x) ≈ −nj

(2n)! 2 −2n−1 x .

2n n!

2

∞ X (−1)n (2n + 1) 2n+1 σ≈ . h i2 (ka) 4π (2n)! n=1 n 2n n!

λ2

146

(2n)! −n−1 x , 2n n!

Since ka 1, the first term in the series dominates: 6 2+1 2 1 1 2 σ∼λ ∼λ ∼ λ−4 . λ λ 2

5.28.

∇×

B µ

= Ji + 0

∂E , ∂t

1 ∂E 1 = Ji + 0 ∇×B−B×∇ , µ µ ∂t

∂ ∂A 1 1 = Ji + 0 − ∇ × ∇ × A − (∇ × A) × ∇ − ∇φ , µ µ ∂t ∂t

∂2A 1 ∂ 1 1 = Ji − 0 2 − 0 ∇φ. ∇(∇ · A) − ∇2 A − (∇ × A) × ∇ µ µ µ ∂t ∂t ∂φ ∇ · A = −µ 0 , ∂t

1 1 ∂2A − ∇2 A − (∇ × A) × ∇ = Ji − 0 2 , µ µ ∂t

2 ∂ A 1 ∇2 A − µ 0 2 + µ∇ × A + ∇ = Ji . ∂t µ Because

∇µ 1 =− 2, ∇ µ µ

we can also write ∇2 A − µ 0

∂2A ∇µ − (∇ × A) × 2 = Ji .

2 ∂t µ

5.29.

Z µ0 e−jkR 0 0 J̃(r , ω) A(r, t) = F dV 4π V R Z i µ0 1 −1 h 0 = F J̃(r , ω)e−jω(R/c) dV 0 4π V R Z µ0 1 = J(r0 , t − R/c) dV 0 .

4π V R −1

This is a “retarded potential” because the value of A at time t is produced by values of J at time t − R/c (i.e., they are retarded in time). Note that R/c is the time it takes a wave to propagate from the source point to the field point.

5.30. Write G(ρ|ρ0 ) =

1 (2π)2

ZZ ∞

G(λ|ρ0 )ejλ·ρ d2 λ,

−∞

where λ = x̂λx + ŷλy ,

|λ|2 = λ2 = λ2x + λ2y , 147

d2 λ = dλx dλy .

Also δ(ρ − ρ0 ) =

1 (2π)2

ZZ ∞

0

ejλ·(ρ−ρ ) d2 λ.

−∞

Then ∇2

ZZ ∞ 1 0 jλ·ρ 2 G(λ|ρ )e d λ (2π)2 −∞

ZZ ∞ ZZ ∞ 1 1 0 0 jλ·ρ 2 2 G(λ|ρ )e d λ =− ejλ·(ρ−ρ ) d2 λ. +k 2 2 (2π) (2π) −∞ −∞

But

ZZ ∞ 1 0 jλ·ρ 2 ∇ G(λ|ρ )e d λ (2π)2 −∞ ZZ ∞ 1 = G(λ|ρ0 )∇2 ejλ·ρ d2 λ (2π)2 −∞ 2

ZZ ∞ ∂ ∂2 ∂2 1 0 G(λ|ρ ) + 2 + 2 ej(λx x+λy y) d2 λ = 2 (2π)2 ∂x ∂y ∂z Z Z−∞ ∞ 1 = (−λ2x − λ2y )G(λ|ρ0 )ejλ·ρ d2 λ (2π)2 −∞ 2

so

1 (2π)2

ZZ ∞ h −∞

i 0 (−λ2x − λ2y + k 2 )G(λ|ρ0 ) + e−jλ·ρ ejλ·ρ d2 λ = 0.

By the Fourier transform theorem this implies 0

(k 2 − λ2x − λ2y )G(λ|ρ0 ) = e−jλ·ρ , hence

0

e−jλ·ρ .

G(λ|ρ0 ) = − 2 k − λ2

Therefore ZZ ∞

0

e−jλ·ρ jλ·ρ 2 e d λ 2 2 −∞ k − λ "Z # Z ∞ ∞ jλy (y−y 0 ) 1 e 0 =− dλx ejλx (x−x ) dλy .

2 2 2 (2π)2 −∞ −∞ k − λx − λy

G(ρ|ρ0 ) = −

1 (2π)2

Write k 2 − λ2x − λ2y =

p p

k 2 − λ2x − λy k 2 − λ2x + λy .

See Figure 37. Because −π/2 ≤ ∠k ≤ 0, k lies in the fourth quadrant of the complex λ-plane.

Hence k 2 liesp in the third or fourth quadrant. So k 2 − λ2x lies in the third or fourth quadrant. If λx is real, then k 2 − λ2x lies in the fourth quadrant. We now carry out the integration over λy .

Case 1: y < y 0 . Close in lower half plane. See Figure 38. No contribution from the integral over C∞ , and the residue evaluation at the pole gives √ 2 2 Z ∞ 0 0 ejλy (y−y ) e−j k −λx |y−y | p dλ = −jπ . y 2 2 2 k 2 − λ2x −∞ k − λx − λy 148

Figure 37

Figure 38 Case 2: y > y 0 . Close in upper half plane. See Figure 39. No contribution from the integral over C∞ , and the residue evaluation at the pole gives the same expression as above.

Therefore √ " # Z ∞ −j k2 −λ2x |y−y 0 | 1 e 0 G(ρ|ρ0 ) = − dλx ejλx (x−x ) −jπ p (2π)2 −∞ k 2 − λ2x √ " Z # 0 0 2 2 j 1 ∞ ejλx (x−x ) e−j k −λx |y−y | p = dλx .

4 π −∞ k 2 − λ2x Finally, use the integral representation (see, for example, Morse and Feshbach) Z ∞

0

√

2

2

0

ejα(x−x ) e−j k −α |y−y | √ dα π −∞ k 2 − α2

1 (2) H0 (k|ρ − ρ0 |) = − to finish the problem.

149

Figure 39 5.31. See Figure 40.

Figure 40

H̃y = −

∂ ψ̃ , ∂x

Ẽx = −

kz ∂ ψ̃ , ω˜ ∂x

Ẽz =

kx2 ψ̃. jω˜

Region 1: ψ̃1 = C1 sin kx1 xe−jkz z .

Region 0: ψ̃0 = (C2 sin kx0 x + C3 cos kx0 x)e−jkz z .

Boundary condition: Ẽz = 0 at x = d2 =⇒ ψ̃ = 0 at x = d2 : C2 sin kx0 d2 + C3 cos kx0 d2 = 0 "

sin kx0 d2

=⇒

C3 = −C2

sin kx0 d2 cos kx0 d2

.

#

cos kx0 x = C̄2 sin kx0 (x − d2 ). cos kx0 d2 ( −kx0 C̄2 cos kx0 (x − d2 ), region 0, ∂ ψ̃ H̃y = − = e−jkz z ∂x −kx1 C1 cos kx1 x, region 1.

ψ̃0 = C2 sin kx0 x − C2

150

BC: H̃y continuous at x = a2 a kx0 C̄2 cos kx0 ( a−d 2 ) = kx1 C1 cos kx1 2 .

BC: Ẽz continuous at x = a2 2 2 kx1 kx0 ) = C1 sin kx1 a2 .

C̄2 sin kx0 ( a−d 2 jω 0 jω˜ 1

Divide the two preceding equations: kx0 kx1 tan kx0 ( a−d tan kx1 a2 . )= 2 0 ˜1

151

Chapter 6 6.1. Consider the geometry shown in Figure 6.1. We will calculate H̃(r) for all r ∈ V by placing a magnetic dipole at location r and using the Lorentz reciprocity theorem Z I (1) (Ẽpm · J̃ − H̃pm · J̃m ) dV 0 . − n̂0 · (Ẽ × H̃pm − Ẽpm × H̃) dS 0 = V0

S

Here n̂ is the normal directed into the region V , and Ẽpm and H̃pm are the fields due to a magnetic dipole. Ẽ and H̃ are the desired unknowns, and we have excluded a small volume element Vδ around the point determined by r.

Let us use

jω ∇ × (p̃m ψ̃), 4π and calculate each term in (1).

Ẽpm = −

H̃pm =

1 ∇ × ∇ × (p̃m ψ̃).

4π µ̃

(a)

1 0 ∇ × ∇ × (p̃m ψ̃) dS 0 n̂ · (Ẽ × H̃pm ) dS = n̂ · Ẽ × 4π µ̃ S S I 1 2 = n̂0 · {Ẽ × [∇0 (∇0 · [p̃m ψ̃]) − ∇0 [p̃m ψ̃]} dS 0 4π µ̃ S I 1 2 n̂0 · {Ẽ × [∇0 (p̃m · ∇0 ψ̃ + ψ̃∇0 · p̃m ) − ∇0 [p̃m ψ̃]} dS 0 . = 4π µ̃ S

I

0

0

I

0

Note that 2

2

2

2

∇0 [p̃m ψ̃] = x̂∇0 [p̃mx ψ̃] + ŷ∇0 [p̃my ψ̃] + ẑ∇0 [p̃mz ψ̃] 2

2

2

= (x̂p̃mx ∇0 ψ̃) + (ŷp̃my ∇0 ψ̃) + (ẑp̃mz ∇0 ψ̃) 2

= (x̂p̃mx + ŷp̃my + ẑp̃mz )∇0 ψ̃ 2

= p̃m ∇0 ψ̃.

So RHS =

1 4π µ̃

I

2

n̂0 · {Ẽ × [∇0 (p̃m · ∇0 ψ̃) − p̃m ∇0 ψ̃]} dS 0 .

S

But ∇0 2 ψ̃ = −k 2 ψ̃ for r 6= r0 , so 1 RHS = 4π µ̃

I

n̂0 · {Ẽ × [∇0 (p̃m · ∇0 ψ̃) + p̃m k 2 ψ̃]} dS 0 .

S

Next use A × ∇φ = φ∇ × A − ∇ × (φA) to write I 1 RHS = n̂0 · {(Ẽ × p̃m )k 2 ψ̃ + (p̃m · ∇0 ψ̃)(∇0 × Ẽ) − ∇0 × [(p̃m · ∇0 ψ̃)Ẽ]} dS 0 4π µ̃ S I 1 = {n̂0 · (Ẽ × p̃m )k 2 ψ̃ + (p̃m · ∇0 ψ̃)n̂0 · (−J̃m − jω µ̃H̃)} dS 0 − third term 4π µ̃ S I 1 = {p̃m · (n̂0 × Ẽ)k 2 ψ̃ − (p̃m · ∇0 ψ̃)n̂0 · (J̃m + jω µ̃H̃)} dS 0 − third term 4π µ̃ S 152

where

X 1 I n̂0 · ∇0 × [(p̃m · ∇0 ψ̃)Ẽ] dS 0 . third term = 4π µ̃ Sn n

Split each surface along lines of discontinuity of Ẽ. By Stokes’s theorem, I I 0 0 0 (dl0 · Ẽ)(p̃m · ∇0 ψ̃). n̂ · ∇ × [(p̃m · ∇ ψ̃)Ẽ] dS = Γna +Γnb

Sn =Sna +Snb

Note that if Ẽ is continuous the two line integrals cancel and the third term is zero.

So I

) (I XI 1 0 0 0 2 0 0 0 n̂ ·(Ẽ×H̃pm ) dS = (dl · Ẽ)∇ ψ̃ . p̃m · [(n̂ × Ẽ)k ψ̃ − n̂ · (J̃m + jω µ̃H̃)∇ ψ̃] dS − 4π µ̃ Γna +Γnb S S n 0

0

(b) I

I jω n̂ · (Ẽpm × H̃) dS = − n̂0 · {[∇0 × (p̃m ψ̃)] × H̃} dS 0 4π S I S jω n̂0 · [H̃ × (∇0 × [p̃m ψ̃])] dS 0 = 4π S I jω (n̂0 × H̃) · (∇0 × [p̃m ψ̃]) dS 0 = 4π S I jω = (n̂0 × H̃) · {∇0 ψ̃ × p̃m + ψ̃∇0 × p̃m } dS 0 4π S I jω p̃m · [(n̂0 × H̃) × ∇0 ψ̃] dS 0 = 4π S I jω = p̃m · (n̂0 × H̃) × ∇0 ψ̃ dS. 4π S 0

Combining terms (a) and (b) gives the LHS of (1) as I LHS = − n̂0 · (Ẽ × H̃pm − Ẽpm × H̃) dS 0 S (I ) XI 1 0 2 0 0 0 0 0 =− p̃m · [(n̂ × Ẽ)k ψ̃ − n̂ · (J̃m + jω µ̃H̃)∇ ψ̃] dS − (dl · Ẽ)∇ ψ̃ 4π µ̃ Γna +Γnb S n I jω + p̃m · (n̂0 × H̃) × ∇0 ψ̃ dS 4π S

I 1 =− p̃m · [−jω µ̃(n̂0 × H̃) × ∇0 ψ̃ + k 2 (n̂0 × Ẽ)ψ̃ − n̂0 · (J̃m + jω µ̃H̃)∇0 ψ̃] dS 0 4π µ̃ S

XI 0 0 − (dl · Ẽ)∇ ψ̃ . n

Γna +Γnb

Now calculate the RHS of (1).

153

(c) Z jω [∇0 × (p̃m ψ̃)] · J̃ dV 0 Ẽpm · J̃ dV = − 4π V 0 V0 Z jω =− J̃ · [∇0 ψ̃ × p̃m + ψ̃∇0 × p̃m ] dV 0 4π V 0 Z jω [p̃m · (J̃ × ∇0 ψ̃] dV 0 =− 4π V 0 Z jω = − p̃m · (J̃ × ∇0 ψ̃) dV 0 .

4π 0 V

Z

0

(d) Z V0

Z 1 {∇0 × [∇0 × (p̃m ψ̃)]} · J̃m dV 0 4π µ̃ V 0 Z 1 2 = [∇0 (∇0 · [p̃m ψ̃]) − ∇0 (p̃m ψ̃)] · J̃m dV 0 4π µ̃ V 0 Z 1 2 J̃m · [∇0 (p̃m · ∇0 ψ̃ + ψ̃∇0 · p̃m ) − p̃m ∇0 ψ̃] dV 0 = 4π µ̃ V 0 Z 1 = J̃m · [∇0 (p̃m · ∇0 ψ̃) + p̃m k 2 ψ̃] dV 0 4π µ̃ V 0 Z 1 = [∇0 · (J̃m [p̃m · ∇0 ψ̃]) − (p̃m · ∇0 ψ̃)∇0 · J̃m + k 2 (p̃m · J̃m )ψ̃] dV 0 4π µ̃ V 0 Z I 1 [k 2 (p̃m · J̃m )ψ̃ + jω ρ̃m (p̃m · ∇0 ψ̃)] dV 0 − J̃m [p̃m · ∇0 ψ̃] · n̂0 dS 0 = 4π µ̃ V 0 S Z

I 1 2 0 0 0 0 0 = p̃m · [k J̃m ψ̃ + jω ρ̃m ∇ ψ̃] dV − (n̂ · J̃m )∇ ψ̃ dS .

4π µ̃ V0 S

H̃pm · J̃m dV 0 =

Combining, we get the RHS of (1): Z

I 1 RHS = p̃m · [−jω µ̃(J̃ × ∇0 ψ̃) − k 2 ψ̃ J̃m − jω ρ̃m ∇0 ψ̃] dV 0 + (n̂0 · J̃m )∇0 ψ̃ dS 0 .

4π µ̃ V0 S Equating both sides of (1) gives

Z I 1 0 0 0 0 2 0 0 − p̃m · [jω µ̃(J̃ × ∇ ψ̃) + k ψ̃ J̃m + jω ρ̃m ∇ ψ̃] dV − (n̂ · J̃m )∇ ψ̃ dS 4π µ̃ V0 S I 1 =− p̃m · [−jω µ̃(n̂0 × H̃) × ∇0 ψ̃ + k 2 (n̂0 × Ẽ)ψ̃ − n̂0 · (J̃m + jω µ̃H̃)∇0 ψ̃] dS 0 4π µ̃ S

XI 0 0 − (dl · Ẽ)∇ ψ̃ , n

hence Z

Γna +Γnb

[jω µ̃J̃ × ∇0 ψ̃ + k 2 ψ̃ J̃m + jω ρ̃m ∇0 ψ̃] dV 0 I I 0 0 2 0 0 0 0 = [−jω µ̃(n̂ × H̃) × ∇ ψ̃ + k (n̂ × Ẽ)ψ̃ − jω µ̃(n̂ · H̃)∇ ψ̃] dS −

V0

S

Γna +Γnb

154

(dl0 · Ẽ)∇0 ψ̃,

Z

ρ̃m 0 [J̃ × ∇0 ψ̃ − jω˜ ψ̃ J̃m + ∇ ψ̃] dV 0 µ̃ 0 V I I = [−(n̂0 × H̃) × ∇0 ψ̃ − jω˜ (n̂0 × Ẽ)ψ̃ − (n̂0 · H̃)∇0 ψ̃] dS 0 − S

(dl0 · Ẽ)∇0 ψ̃

Γna +Γnb

1 . jω µ̃

(2)

Now single out the contributions due to the surface Σ and the volume Vδ . Since Ẽ is continuous on Σ we have I (dl0 · Ẽ)∇0 ψ̃ = 0. ΓΣa +ΓΣb

Also, I lim

[−(n̂0 × H̃) × ∇0 ψ̃ − jω˜ (n̂0 × Ẽ)ψ̃ − (n̂0 · H̃)∇0 ψ̃] dS 0 # I " R̂ R̂ × Ẽ R̂ (R̂ × H̃) × 2 + jω˜ = lim

+ (R̂ · H̃) 2 dS 0 δ→0 Σ δ δ δ " # Z 2π Z π 1 1 R̂ × Ẽ R̂ 2 = lim H̃(R̂ · R̂) 2 − R̂(R̂ · H̃) 2 + jω˜

+ (R̂ · H̃) 2 δ sin θ0 dθ0 dφ0 δ→0 0 δ δ δ δ 0 Z 2π Z π = lim [H̃(r) + jω˜ δ(R̂ × Ẽ)] sin θ0 dθ0 dφ0

δ→0 Σ

δ→0 0

0

= 4π H̃(r), # 1 ρ̃m R̂ 2 R̂ lim J̃m + δ sin θ dδ dθ dφ J̃ × 2 − jω˜ δ→0 Vδ δ δ µ̃ δ 2

Z δ ρ̃m = lim 4π J̃ × R̂ − jω˜ δ J̃m + R̂ dδ δ→0 0 µ̃ = 0.

Z

"

Substitution back into (1) gives the desired result for H̃(r).

6.2. Derive H̃ from Ẽ using Faraday’s law ∇ × Ẽ = −J̃m − jω µ̃H̃.

Calculate ∇ × Ẽ term by term: Z ρ̃ 1 ∇ × Ẽ = ∇ × [−J̃m × ∇0 ψ̃ + ∇0 ψ̃ − jω µ̃J̃ψ̃] dV 0 4π ˜ V Z X 1 + ∇× [(n̂0 × Ẽ) × ∇0 ψ̃ + (n̂0 · Ẽ)∇0 ψ̃ − jω µ̃(n̂0 × H̃)ψ̃] dS 0 4π n Sn I X 1 1 − ∇× (dl0 · H̃)∇0 ψ̃.

4π n jω˜

Γna +Γnb

155

(a) Z

0

Z

0

∇ × (J̃m × ∇ψ̃) dV 0

[−J̃m × ∇ ψ̃] dV =

∇×

ZV

V

=

∇ × [ψ̃∇ × J̃m − ∇ × (ψ̃ J̃m )] dV 0

VZ

=− ZV =− ZV =− ZV =−

∇ × ∇ × (ψ̃ J̃m ) dV 0 [∇(∇ · [ψ̃ J̃m ]) − ∇2 (ψ̃ J̃m )] dV 0 [∇(∇ · [ψ̃ J̃m ]) − J̃m ∇2 ψ̃] dV 0 [∇(J̃m · ∇ψ̃ + ψ̃∇ · J̃m ) − J̃m ∇2 ψ̃] dV 0 .

V

Continuing, we use the fact that ∇2 ψ̃ + k 2 ψ̃ = −4πδ(r − r0 ): Z Z 0 0 ∇ × [−J̃m × ∇ ψ̃] dV = − [∇(J̃m · ∇ψ̃) − J̃m (−k 2 − 4πδ(r − r0 ))]] dV 0 V V Z = −4π J̃m (r) − [∇(J̃m · ∇ψ̃) + ψ̃k 2 J̃m ] dV 0 ZV = −4π J̃m (r) − [−∇(J̃m · ∇0 ψ̃) + ψ̃k 2 J̃m ] dV 0 ZV = −4π J̃m (r) − [∇(ψ̃∇0 · J̃m − ∇0 · [J̃m ψ̃]) − ψ̃k 2 J̃m ] dV 0 ZV Z 2 0 = −4π J̃m (r) − [−jω∇(ψ̃ ρ̃m ) − ψ̃k J̃m ] dV + ∇ ∇0 · [J̃m ψ̃] dV 0 V V Z I 2 0 = −4π J̃m (r) − [−jω(ρ̃m ∇ψ̃ + ψ̃∇ρ̃m ) − ψ̃k J̃m ] dV − ∇ n̂0 · (J̃m ψ̃) dS 0 S ZV I = −4π J̃m (r) − [jω ρ̃m ∇0 ψ̃ + ψ̃k 2 J̃m ] dV 0 − ∇{n̂0 · (J̃m ψ̃)} dS 0 V S Z I 0 2 0 = −4π J̃m (r) − [jω ρ̃m ∇ ψ̃ + ψ̃k J̃m ] dV − n̂0 · J̃m ∇ψ̃ dS 0 ZV IS 0 2 0 = −4π J̃m (r) − [jω ρ̃m ∇ ψ̃ + ψ̃k J̃m ] dV + (n̂0 · J̃m )∇0 ψ̃ dS 0 .

V

S

(b) Z

ρ̃ 0 ∇ ψ̃ dV 0 = ∇×

˜ V

Z ρ̃ ρ̃ ∇ × − ∇ψ̃ dV 0 = − ∇ × ∇ψ̃ dV 0 = 0.

˜

˜ V V

Z

156

(c) Z

Z

0

∇ × [J̃ψ̃] dV 0

[−jω µ̃J̃ψ̃] dV = −jω µ̃

∇×

ZV

V

= −jω µ̃

[∇ψ̃ × J̃ + ψ̃∇ × J̃] dV 0

ZV = −jω µ̃

J̃ × ∇0 ψ̃ dV 0 .

V

(d) Z ∇×

(n̂0 × Ẽ) × ∇0 ψ̃ dS 0 = −∇ ×

Z

(n̂0 × Ẽ) × ∇ψ̃ dS 0

ZSn

Sn

[ψ̃∇ × (n̂0 × Ẽ) − ∇ × {ψ̃(n̂0 × Ẽ)}] dS 0 Z =∇×∇× ψ̃(n̂0 × Ẽ) dS 0 . = −∇ ×

Sn

Sn

(e) Z ∇×

0

0

Z

0

(n̂0 · Ẽ)(∇ × ∇0 ψ̃) dS 0

(n̂ · Ẽ)∇ ψ̃ dS = Sn

Sn

Z =−

(n̂0 · Ẽ)(∇ × ∇ψ̃) dS 0

Sn

= 0. (f) Z ∇×

[−jω µ̃(n̂0 × H̃)ψ̃] dS 0 = −jω µ̃

Sn

Z ZSn

= −jω µ̃

∇ × [(n̂0 × H̃)ψ̃] dS 0 [ψ̃∇ × (n̂0 × H̃) + ∇ψ̃ × (n̂0 × H̃)] dS 0

Sn

Z = −jω µ̃

(n̂0 × H̃) × ∇0 ψ̃ dS 0 .

Sn

(g) Z 1 1 0 0 ∇× (dl · H̃)∇ ψ̃ = [∇(dl0 · H̃) × ∇0 ψ̃ + (dl0 · H̃)∇ × ∇0 ψ̃]

jω˜ Γna +Γnb Γna +Γnb jω˜ Z 1 =− (dl0 · H̃)∇ × ∇ψ̃ jω˜ Γna +Γnb Z

= 0.

157

Now, let’s calculate H̃. ∇ × Ẽ J̃m − jω µ̃ jω µ̃

Z J̃m ρ̃m 0 J̃m 1 =− + + ∇ ψ̃ − jω˜ ψ̃ J̃ dV 0 jω µ̃ jω µ̃ 4π V µ̃ Z X 1 Z 1 0 0 0 (n̂ · J̃m )∇ ψ̃ dS + − J̃ × ∇0 ψ̃ dV 0 jω µ̃4π 4π Sn V n X 1 Z − ∇ × ∇ × [ψ̃(n̂0 × Ẽ)] dS 0 ω µ̃4π Sn n Z X 1 + (n̂0 × H̃) × ∇0 ψ̃ dS 0 4π n Sn

Z ρ̃m 0 1 0 = ∇ ψ̃ − jω˜ ψ̃ J̃m + J̃ × ∇ ψ̃ dV 0 4π V µ̃ Z 1 X + (n̂0 × H̃) × ∇0 ψ̃ dS 0 4π n Sn X 1 Z − {∇ × ∇ × [ψ̃(n̂0 × Ẽ)] + (n̂0 · J̃m )∇0 ψ̃} dS 0 . jω µ̃4π Sn n

H̃ = −

The volume integral is exactly what we need, but the surface integral still requires manipulation.

Z Z 0 0 0 n̂0 · (−∇0 × Ẽ − jω µ̃H̃)∇0 ψ̃ dS 0 (n̂ · J̃m )∇ ψ̃ dS = Sn Sn Z n̂0 · (∇0 × Ẽ + jω µ̃H̃)ψ̃ dS 0 =∇ ZSn =∇ n̂0 · (∇0 × [Ẽψ̃] − ∇0 ψ̃ × Ẽ + jω µ̃H̃ψ̃) dS 0 .

Sn

Use Stokes’s theorem on the first term: I Z Z 0 0 0 RHS = ∇ (dl · Ẽ)ψ̃ − ∇ ∇ ψ̃ × Ẽ dS + ∇ jω µ̃(n̂0 · H̃)ψ̃ dS 0 Γna +Γnb Sn Sn I Z Z =− (dl0 · Ẽ)∇0 ψ̃ − ∇ (n̂0 × Ẽ) · ∇ψ̃ dS 0 − jω µ̃ (n̂0 · H̃)∇0 ψ̃ dS 0 Sn IΓna +Γnb Z Z Sn =− (dl0 · Ẽ)∇0 ψ̃ − ∇∇ · (n̂0 × Ẽ)ψ̃ dS 0 − jω µ̃ (n̂0 · H̃)∇0 ψ̃ dS 0 . Γna +Γnb

Sn

158

Sn

Thus Z Z 0 0 0 0 (n̂0 × Ẽ)ψ̃ dS 0 {∇ × ∇ × [ψ̃(n̂ × Ẽ)] + (n̂ · J̃m )∇ ψ̃} dS = [∇ × ∇ × −∇∇·] Sn Sn I Z − (dl0 · Ẽ)∇0 ψ̃ − jω µ̃ (n̂0 · H̃)∇0 ψ̃ dS 0 Γna +Γnb Sn Z (n̂0 × Ẽ)ψ̃ dS 0 + remaining terms = −∇2 Sn Z (n̂0 × Ẽ)∇2 ψ̃ dS 0 + remaining terms =− Z Sn = (n̂0 × Ẽ)k 2 ψ̃ dS 0 + remaining terms.

Sn

Back-substitution gives the desired result for H̃.

6.3. Consider two separate cases.

1. Do not exclude Vm , but do exclude all regions Vn with n 6= m. Then the magnetic field external to the excluded region is

Z ρ̃i H̃(r, ω) = J̃i × ∇0 G̃ + m ∇0 G̃ − jω˜ c J̃im G̃ dV 0 µ̃ V +Vm i XZ h (n̂0 × H̃) × ∇0 G̃ + (n̂0 · H̃)∇0 G̃ + jω˜ c (n̂0 × Ẽ)G̃ dS 0 + n6=m Sn

+

X n6=m

1 jω µ̃

I

(dl0 · Ẽ)∇0 G̃,

r ∈ V + Vm .

Γna +Γnb

Note that the valid region for this expression includes the volume Vm .

2. Exclude all of space except for Vm :

Z ρ̃im 0 i 0 c i H̃(r, ω) = J̃ × ∇ G̃ + ∇ G̃ − jω˜ J̃m G̃ dV 0 µ̃ Vm Z h i (n̂0 × H̃) × ∇0 G̃ + (n̂0 · H̃)∇0 G̃ + jω˜ c (n̂0 × Ẽ)G̃ dS 0 − Sm I 1 − r ∈ Vm . (dl0 · Ẽ)∇0 G̃, jω µ̃ Γma +Γmb Note the minus signs introduced by reversal of the normal vector.

Now subtract the two equations for points within Vm :

Z ρ̃im 0 i 0 c i 0= J̃ × ∇ G̃ + ∇ G̃ − jω˜ J̃m G̃ dV 0 µ̃ V N Z h i X + (n̂0 × H̃) × ∇0 G̃ + (n̂0 · H̃)∇0 G̃ + jω˜ c (n̂0 × Ẽ)G̃ dS 0

+

n=1 Sn N X n=1

1 jω µ̃

I

(dl0 · Ẽ)∇0 G̃,

Γna +Γnb

159

r ∈ Vm .

This is (6.8) evaluated within Vm .

6.4. We wish to show

Z ρi ρi 0 i 0 ∇ · E(r) = ∇ · −Jm × ∇ G + ∇ G dV 0 = .

V Moving the divergence through the integral sign we need

∇ · −Jim × ∇0 G =∇ · Jim × ∇G

=∇ · G∇ × Jim − ∇ · ∇ × Jim G

=∇ · ∇ × Jim G =0.

Here we have used ∇ × Jim = 0 since Jim depends on r0 and we are differentiating with respect to r. Thus i

Z ρ 0 ∇ · E(r) = ∇· ∇ G dV 0

V i

Z ρ =− ∇· ∇G dV 0

V i Z i ρ ρ ∇ · ∇G + ∇G · ∇ dV 0 . =−

V The second term is zero since the argument of the derivative depends on r0 . Using ∇ ·∇G = ∇2 G = −δ(r − r0 ) we have Z i ρ ρi ∇ · E(r) = δ(r − r0 )dV 0 = .

V 6.5. By (6.20) we have H̃φ = j

i I˜ h −jkR1 e + e−jkR2 − (2 cos kl)e−jkr .

4πρ

Here R1 = (ρ2 + (z − l)2 )1/2 = (ρ2 + z 2 − 2lz + l2 )1/2 = (r2 − 2lz + l2 )1/2

1/2

2lz l2 lz 1/2 lz z ≈r 1−2 2 ≈r 1− 2 =r− l =r 1− 2 + 2 r r r r r and we can use z = r cos θ to write R1 ≈ r − l cos θ.

Similarly, R2 ≈ r + l cos θ.

160

Using ρ = r sin θ, we obtain i I˜ h −jkr jkl cos θ e e + e−jkr e−jkl cos θ − (2 cos kl)e−jkr 4πρ h i I˜ =j e−jkr ejkl cos θ + e−jkl cos θ − 2 cos kl 4πr sin θ I˜ e−jkr cos(kl cos θ) − cos kl =j 2π r sin θ I˜ e−jkr cos(kl cos θ) − cos kl =j F (θ, kl) where F (θ, kl) = .

2π r sin θ

H̃φ = j

Next, by (6.21),

η I˜ z − l e−jkR1 z + l e−jkR2 z e−jkr Ẽρ = j + − (2 cos kl) .

4π ρ R1 ρ R2 ρ r Here we will use

z−l z ≈ , ρ ρ

z+l z ≈ , ρ ρ

where z = r cos θ and ρ = r sin θ. Also, 1 1 1 = ≈ , R1 r − l cos θ r

1 1 ≈ .

R2 r

So

η I˜ cos θ e−jkr ejkl cos θ cos θ e−jkr e−jkl cos θ cos θ e−jkr + − (2 cos kl) 4π sin θ r sin θ r sin θ r h i η I˜ cos θ e−jkr jkl cos θ =j e + e−jkl cos θ − 2 cos kl 4π sin θ r η I˜ e−jkr =j cos θ F (θ, kl).

2π r

Ẽρ ≈ j

By (6.22),

η I˜ e−jkR1 e−jkR2 e−jkr Ẽz = −j + − (2 cos kl) 4π R1 R2 r

η I˜ e−jkr ejkl cos θ e−jkr e−jkl cos θ e−jkr + − (2 cos kl) ≈ −j 4π r r r η I˜ e−jkr = −j sin θF (θ, kl).

4π r Therefore Ẽ = ρ̂Ẽρ + ẑẼz = j

η I˜ e−jkr F (θ, kl)[ρ̂ cos θ − ẑ sin θ].

2π r

Let V = ρ̂ cos θ − ẑ sin θ = [x̂ cos φ + ŷ sin φ] cos θ − ẑ sin θ.

161

Then Vθ = θ̂ · V = (θ̂ · x̂) cos φ cos θ + (θ̂ · ŷ) sin φ cos θ − (θ̂ · ẑ) sin θ = (cos θ cos φ) cos φ cos θ + (cos θ sin φ) sin φ cos θ + sin2 θ = cos2 θ + sin2 θ = 1, Vr = r̂ · V = (r̂ · x̂) cos φ cos θ + (r̂ · ŷ) sin φ cos θ − (r̂ · ẑ) sin θ = (sin θ cos φ) cos φ cos θ + (sin θ sin φ) sin φ cos θ − cos θ sin θ = sin θ cos θ(cos2 φ + sin2 φ) − cos θ sin θ = 0, Vφ = φ̂ · V = (φ̂ · x̂) cos φ cos θ + (φ̂ · ŷ) sin φ cos θ − (φ̂ · ẑ) sin θ = (− sin φ) cos φ cos θ + (cos φ) sin φ cos θ − (0) sin θ = 0.

Finally then, Ẽ = θ̂j

η I˜ e−jkr F (θ, kl).

2π r

6.6. We have ˜ −jk|z| δ(x)δ(y), J̃i = ẑIe and R = (ρ2 + (z − z 0 )2 )1/2 , Now

  R(z 0 = l) = R1 = (ρ2 + (z − l)2 )1/2 , R(z 0 = −l) = R2 = (ρ2 + (z + l)2 )1/2 ,  R(z 0 = 0) = r = (ρ2 + z 2 )1/2 .

µ̃I˜ Ãe = ẑ 4π

Z l

0

e−jk|z |

−l

e−jkR 0 dz R

and H̃ =

1 1 ∂ Ãez ∇ × Ãe = −φ̂ . µ̃ µ̃ ∂ρ

Therefore Z 0 Z l −jkR −jkR I˜ ∂ I˜ ∂ 0e 0e ejkz dz 0 − e−jkz dz 0 4π ∂ρ −l R 4π ∂ρ 0 R Z Z 0 0 I˜ 0 ∂ e−jk(R−z ) 0 I˜ l ∂ e−jk(R+z ) 0 =− dz − dz .

4π −l ∂ρ R 4π 0 ∂ρ R

H̃φ = −

162

Use

0

0

e−jk(R±z ) ∂ ∂ e−jk(R±z ) = ±ρ 0 ∂ρ R ∂z R[R ∓ (z − z 0 )]

to get " # " # 0 0 0 l e−jk(R−z ) e−jk(R+z ) I˜ I˜ −ρ ρ H̃φ = − − 4π R[R + (z − z 0 )] −l 4π R[R − (z − z 0 )] 0

I˜ e−jkr I˜ e−jkR2 e−jkl e−jkr e−jkR1 e−jkl =− −ρ − ρ −ρ +ρ 4π r[r + z] r[r − z] 4π R2 [R2 + (z + l)] R1 [R1 − (z − l)]

−jkR2

−jkr ˜ ˜ 1 I −jkl e R2 − (z + l) e−jkR1 R1 + (z − l) I e 1 − − · 2 + · 2 =− ρ − ρe 4π r r+z r−z 4π R2 R1 R2 − (z + l)2 R1 − (z − l)2

I˜ r e−jkr z+l z−l I˜ −jkl −jkR2 = e + e−jkR1 − e−jkR2 + e−jkR1 − e 2π ρ r 4πρ R2 R1

−jkr ˜ ˜ I −jkl −jkR2 z+l z−l I re − e e 1− + e−jkR1 1 + . = 2π ρ r 4πρ R2 R1 To find Ẽ, use " # ∂ H̃φ 1 1 1 1 ∂ Ẽ = ∇ × H̃ = ∇ × (φ̂H̃φ ) = −ρ̂ + ẑ (ρH̃φ ) , jω˜

jω˜

jω˜

∂z ρ ∂ρ

and so on.

For kr 1 we make the approximation 2

2

2 1/2

R1 = (ρ + z − 2lz + l )

2

2 1/2

= (r − 2lz + l )

2lz l2 =r 1− 2 + 2 r r

1/2

lz 1/2 lz z ≈r 1−2 2 ≈ r 1 − 2 = r − l = r − l cos θ. r r r and, similarly, R2 ≈ r + l cos θ. Note also that ρ = r sin θ. Hence

I˜ 1 e−jkr I˜ −jkl −jkr −jkl cos θ z+l z−l −jkr jkl cos θ H̃φ ≈ − − e e e 1− +e e 1+ 2π sin θ r 4πρ r + l cos θ r − l cos θ h i −jkr −jkr ˜ ˜ I 1 e I 1 e ≈− − e−jkl cos θ (1 − cos θ) + ejkl cos θ (1 + cos θ) .

2π sin θ r 4π sin θ r To find Ẽ, cross r̂ into both sides of H̃ ≈

r̂ × Ẽ η

to obtain ηr̂ × H̃ = r̂ × (r̂ × Ẽ) = r̂(r̂ · Ẽ) − Ẽ(r̂ · r̂) = −Ẽ or Ẽ = −ηr̂ × H̃.

Since r̂ × φ̂ = −θ̂, we compute i I˜ 1 e−jkr h −jkl cos θ I˜ 1 e−jkr jkl cos θ +η e (1 − cos θ) + e (1 + cos θ) .

Ẽθ ≈ η 2π sin θ r 4π sin θ r 163

Figure 41 6.7. Refer to Figure 41.

By (6.31) we have Z ǎe =

0

J̌i (r0 , ω)ejk0 r̂·r dV 0

ZVπ =

0 ˇ 0 )ejk0 r̂·r0 a dφ0 where I(φ) ˇ φ̂ I(φ = Iˇ0 e−jk0 a|φ| .

−π

But r̂ · r0 = (x̂ sin θ cos φ + ŷ sin θ sin φ + ẑ cos θ) · (x̂ cos φ0 + ŷ sin φ0 )a = a(sin θ cos φ cos φ0 + sin θ sin φ sin φ0 ) = a sin θ cos(φ − φ0 ) and 0

φ̂ = −x̂ sin φ0 + ŷ cos φ0 = −(r̂ sin θ cos φ + θ̂ cos θ cos φ − φ̂ sin φ) sin φ0 + (r̂ sin θ sin φ + θ̂ cos θ sin φ + φ̂ cos φ) cos φ0 = r̂(− sin θ cos φ sin φ0 + sin θ sin φ cos φ0 ) + θ̂(− cos θ cos φ sin φ0 + cos θ sin φ cos φ0 ) + φ̂(sin φ sin φ0 + cos φ cos φ0 ) = r̂ sin θ sin(φ − φ0 ) + θ̂ cos θ sin(φ − φ0 ) + φ̂ cos(φ − φ0 ).

Therefore

Z π ǎθ = a cos θ

ˇ 0 ) sin(φ − φ0 )ejka sin θ cos(φ−φ0 ) dφ0 I(φ

−π

and

Z π ǎφ = a

ˇ 0 ) cos(φ − φ0 )ejk0 a sin θ cos(φ−φ0 ) dφ0 .

I(φ

−π

164

ˇ Substituting for I(φ), we find ǎθ = cos θ ˇ I0 a and

ǎφ = ˇ I0 a

Z π

0

0

sin(φ − φ0 )e−jk0 a[|φ |−sin θ cos(φ−φ )] dφ0

−π

Z π

0

0

cos(φ − φ0 )e−jk0 a[|φ |−sin θ cos(φ−φ )] dφ0 .

−π

We note that these integrals cannot be computed in closed form.

To compute the radiation resistance, we must first find the radiated power. Using (6.34) in (6.33) we have Z 2π Z π 2

k0 η0 ∗ PR = ǎθ ǎθ + ǎφ ǎ∗φ sin θ dθ dφ.

2 0 0 32π Thus, 2PR = η0 Rr = |Iˇ0 |2

k0 a 4π

2 Z 2π Z π " 0

0

# ǎθ 2 ǎθ 2 + sin θ dθ dφ.

Iˇ0 a Iˇ0 a

Figure 42 shows the radiation resistance computed using numerical integration.

600

Rr (Ω)

500 400 300 200 100 0

0

0.5

1

1.5

2

2.5

3

3.5

4

k0a Figure 42 Radiation resistance of a traveling-wave loop antenna.

6.8. From (6.75) we have ǎφ ≈ Iˇ0 a

Z π

φ0 2

4 cos(φ − φ0 )dφ0 = Iˇ0 a cos φ, 3

φ0 2

4 sin(φ − φ0 )dφ0 = Iˇ0 a sin θ sin φ, 3

cos −π

ǎθ ≈ Iˇ0 a sin θ

Z π cos −π

where we have approximated the exponential function as unity. We have from the previous problem the following formula for the radiation resistance: #

Z Z " 2PR k0 a 2 2π π ǎθ 2 ǎθ 2 Rr = = η0 + sin θ dθ dφ.

4π |Iˇ0 |2 Iˇ0 a Iˇ0 a 0 0 165

Thus

Rr = η0

k0 a 4π

2 2 Z 2π Z π

2

4 10π d 2 2 2 cos φ + sin φ sin θ sin θ dθ dφ = η0 , 3 27 λ 0 0

where d = 2a is the loop diameter. The radiation resistance of a small short-circuited antenna is from Example 6.9

π5 d 4 Rr = η0 .

6 λ Thus, the ratio of the radiation resistance of the short-circuited loop to that of the open-circuited loop is

2 π5 d 4 27 4 d 2 d 6 λ = π = 43.8 .

2 10π d 60 λ λ 27

λ

Note that for small d/λ the ratio varies as diameter squared, and thus the radiation resistance of the short-circuited loop becomes small compared to that of the open-circuited loop. The angularlyvarying current in the open-circuited loop produces less cancellation of the far-zone fields than does the constant current in the short-circuited loop.

6.9. Refer to Figure 43.

Figure 43 The field in the aperture is Ẽa = Ẽ0 x̂e−jkz z=0 = Ẽ0 x̂.

We have J̃eq ms = −2n̂ × Ẽa = −2ẑ × (Ẽ0 x̂) = −2ŷẼ0 , Z 0 ãh = J̃ms (r0 )ejkr̂·r dS 0 , 0

0

S 0

r̂ · r = r̂ · (x̂x + ŷy ) = x0 sin θ cos φ + y 0 sin θ sin φ.

166

Therefore Z a/2 Z a/2 ãh = −2ŷẼ0 −a/2

= −2ŷẼ0

0

0

ejkx sin θ cos φ ejky sin θ sin φ dx0 dy 0

−a/2

0 ejkx sin θ cos φ a/2

0

ejky sin θ sin φ a/2 jk sin θ cos φ −a/2 jk sin θ sin φ −a/2

a a 2 2j sin(k 2 sin θ cos φ) 2j sin(k 2 sin θ sin φ) = −2ŷẼ0 a · j k a2 sin θ cos φ j k a2 sin θ sin φ

= −8ŷẼ0 a2 where X=

sin πX sin πY · πX πY

a sin θ cos φ, λ

Y =

a sin θ sin φ. λ

Then, by (6.55), jk r̂ × Ãh 0 −jkr e jk 2 sin πX sin πY · 0 . = − 8(r̂ × ŷ)Ẽ0 a 0 πX πY 4πr

Ẽ =

Here r̂ × ŷ = r̂ × (r̂ sin θ sin φ + θ̂ cos θ sin φ − φ̂ cos φ) = φ̂ cos θ sin φ + θ̂ cos φ, so Ẽ = −

2jk0 e−jkr sin πX sin πY Ẽ0 a2 (φ̂ cos θ sin φ + θ̂ cos φ) · . π r πX πY

Finally, H̃ =

2jk0 e−jkr sin πX sin πY r̂ × Ẽ =− Ẽ0 a2 (φ̂ cos φ − θ̂ cos θ sin φ) · . η ηπ r πX πY

6.10. Refer to Figure 44.

The field in the aperture is Ẽa (r) = ρ̂

Ṽ0 Ṽ0 e−jkz = ρ̂ . ρ ln(b/a) ρ ln(b/a) z=0

We have

Z ãh =

0

0 jkr̂·r J̃eq dS 0 ms (r )e

S

where J̃eq ms = −2n̂ × Ẽa = −2ẑ × ρ̂

Ṽ0 Ṽ0 = −2φ̂ . ρ ln(b/a) ρ ln(b/a)

Here r0 = ρ0 ρ̂0 = ρ0 x̂ cos φ0 + ρ0 ŷ sin φ0

167

Figure 44 so r̂ · r0 = (r̂ · x̂)ρ0 cos φ0 + (r̂ · ŷ)ρ0 sin φ0 = ρ0 cos φ0 sin θ cos φ + ρ0 sin φ0 sin θ sin φ = ρ0 sin θ cos(φ − φ0 ).

Also, 0

φ̂ = −x̂ sin φ0 + ŷ cos φ0 = −(r̂ sin θ cos φ + θ̂ cos θ cos φ − φ̂ sin φ) sin φ0 + (r̂ sin θ sin φ + θ̂ cos θ sin φ + φ̂ cos φ) cos φ0 = r̂(− sin θ cos φ sin φ0 + sin θ sin φ cos φ0 ) + θ̂(− cos θ cos φ sin φ0 + cos θ sin φ cos φ0 ) + φ̂(sin φ sin φ0 + cos φ cos φ0 ) = r̂ sin θ sin(φ − φ0 ) + θ̂ cos θ sin(φ − φ0 ) + φ̂ cos(φ − φ0 ).

Hence

Z 2π Z b ãθ = 0

Ṽ0 0 0 −2 cos θ sin(φ − φ0 ) 0 ejkρ sin θ cos(φ−φ ) ρ0 dρ0 dφ0 = 0 ρ ln(b/a) a

168

(because the integrand is odd in u = φ − φ0 ) and Z 2π Z b

Ṽ0 0 0 −2 cos(φ − φ0 ) 0 ejkρ sin θ cos(φ−φ ) ρ0 dρ0 dφ0 ρ ln(b/a) 0 a

Z b Z 2π 2Ṽ0 jkρ0 sin θ cos u =− cos u e du dρ0 ln(b/a) a 0 Z b

2Ṽ0 =− 2πjJ1 (kρ0 sin θ) dρ0 . . . by (E.83) ln(b/a) a Z b 1 4πj Ṽ0 = J0 (kρ0 sin θ) . . . since J1 (x) dx = −J0 (x) ln(b/a) k sin θ a

ãφ =

=

4πj Ṽ0 J0 (kb sin θ) − J0 (ka sin θ) . ln(b/a) k sin θ

Then jk r̂ × Ãh ˜ −jkr jk e = r̂ × ˜ ah ˜ 4πr

Ẽ =

=

jk e−jkr 4πj Ṽ0 J0 (kb sin θ) − J0 (ka sin θ) (r̂ × φ̂)˜

˜ 4πr ln(b/a) k sin θ

= θ̂

and H̃ =

e−jkr Ṽ0 J0 (kb sin θ) − J0 (ka sin θ) r ln(b/a) sin θ

r̂ × Ẽ e−jkr Ṽ0 J0 (kb sin θ) − J0 (ka sin θ) = φ̂ . η r η ln(b/a) sin θ

6.11.

Ãe (r) = ẑ = ẑ

e−jkr 4πr

Z L/2

e−jkr

"Z

4πr

˜ 0 )ejkz 0 cos θ dz 0 I(z

−L/2 0

I˜0 e

jkz 0

e

jkz 0 cos θ

−L/2

dz 0 +

Z L/2

# −jkz 0

I˜0 e

jkz 0 cos θ

e

0

" # e−jkr ˜ 1 − e−jk(cos θ+1)L/2 1 − ejk(cos θ−1)L/2 I0 − . = ẑ 4πr jk(cos θ + 1) jk(cos θ − 1) 6.12. Refer to Figure 45.

J̃ms = −2ẑ × ẼA = −2ẑ × x̂Ẽ0 = −ŷ2Ẽ0 , Z e−jkr 0 Ãh = J̃ms (r0 )ejkr̂·r dS 0 4πr S 169

dz 0

Figure 45 r̂ · r0 = r̂ · (x̂x0 + ŷy 0 ) = x0 sin θ cos φ + y 0 sin θ sin φ, e−jkr Ãh = −ŷ2Ẽ0 4πr

Z a e

jkx0 sin θ cos φ

0

Z a

dx

0

ejky sin θ sin φ dy 0

−a

−a

e−jkr sin(ka sin θ cos φ) sin(ka sin θ sin φ) = −ŷ2Ẽ0 2 2 4πr k sin θ cos φ k sin θ sin φ Ẽ = jkr̂ × J̃msT

=⇒

Ẽφ = jk J˜msθ

ŷ = r̂ sin θ sin φ + θ̂ cos θ sin φ + φ̂ cos φ, e−jkr sin(ka sin θ cos φ) sin(ka sin θ sin φ) πr k 2 sin2 θ sin φ cos φ cos θ e−jkr = −2j Ẽ0 sin(ka sin θ cos φ) sin(ka sin θ sin φ). πrk sin2 θ cos φ

Ẽφ = jk cos θ sin φ(−2Ẽ0 )

170

Chapter 7 7.1. (a) Begin with (7.14) and (7.17): Z z dṼ (z) 2 Ṽ (t) dt + C1 , =γ dz −l Z z 2 (z − t)Ṽ (t) dt + C1 z + C2 .

Ṽ (z) = γ

(1)

−l

To find C1 and C2 , apply the boundary conditions. At z = −l we have ˜ ˜ = − 1 dṼ (z) . + Ṽ (−l) where I(z) Vg = Zg I(−l) Z dz This gives us Vg = −

Zg 0 Ṽ (−l) + Ṽ (−l) Z

or Vg = −

Zg C1 − C1 l + C2 .

Z

(2)

At z = 0 we have ˜ Ṽ (0) = ZL I(0) which gives us −γ

2

Z 0 tṼ (t) dt + C2 = −γ

2 ZL

Z 0

Z

−l

−l

Ṽ (t) dt − C1

ZL .

Z

(3)

Solving, we obtain C1 =

Z γ2 Zl + ZT

Z 0 tṼ (t) dt − −l

γ 2 ZL Zl + ZT

Z 0 Ṽ (t) dt − −l

Z Vg Zl + ZT

where ZT = Zg + ZL , and C2 =

ZT − Zg Zg + Zl 2 Vg + γ ZT + Zl ZT + Zl

Z 0 tṼ (t) dt − −l

Zg + Zl 2 ZL γ ZT + Zl Z

Z 0 Ṽ (t) dt. −l

Substituting into (1) and simplifying, we get the integral equation for Ṽ (z): Z z Z 0 Z 0 Zg (t − ZL /Z) Zt − ZL Zz − ZL 2 2 2 Ṽ (z) = γ (z−t)Ṽ (t) dt− Vg +γ (l+z) Ṽ (t) dt+γ Ṽ (t) dt, Zl + ZT Zl + ZT −l −l Zl + ZT −l holding for −l ≤ z ≤ 0. We see that if Zg = 0, then ZT = ZL , the last term vanishes, and the equation reverts to (7.21) from the text. (b) Now we solve the integral equation using the MoM. Expand Ṽ (z) =

N X n=1

171

an Pn (z)

and point match at zm : N X

an Pn (z) =

n=1

N X

Z zm Zzm − ZL (zm − t)Pn (t) dt − Vg an γ 2 Zl + ZT −l n=1

Z 0 N X Zt − ZL 2 + an γ (l + zm ) Pn (t) dt −l Zl + ZT n=1 Z 0

N X Zg (t − ZL /Z) Pn (t) dt . + an γ 2 Zl + ZT −l n=1

Write

  m < n, 0, 2 1 (t − zm )Pn (t) dt = Umn = − 8 γ 2 ∆2 , γ m = n,  −l  2 γ ∆(zn − zm ), m > n, Z 0 Zt − ZL Zzn − ZL −γ 2 (l + zm ) Pn (t) dt = Wmn = −γ 2 (l + zm )∆ , Zl + ZT −l Zl + ZT Z 0 Zg (t − ZL /Z) Zg (zn − ZL /Z) 2 −γ Pn (t) dt = Zmn = −γ 2 ∆ , Zl + ZT Zl + ZT −l Z zm

and

Zzm − ZL Vg = bm .

Zl + ZT Then we have the set of linear equations −

N X

an [δmn + Umn + Wmn + Zmn ] = bm .

n=1

Solution is indicated in Figure 46. The theoretical voltage may be found in many elementary textbooks: −γz

Z0 V g e + Γeγz Ṽ (z) = Z0 + Zg eγl − ΓΓg e−γl where Γ=

ZL − Z0 , ZL + Z0

Γg =

Zg − Z0 .

Zg + Z0

For the special case of Zg = Z0 , this reduces to Ṽ (z) =

Vg −γl −γz e e + Γeγz .

2

Note that the MoM solution matches well with the theoretical voltage.

7.2.

Z zZ u

Z u z Z z Z u d F (t) dt du = u F (t)dt − u F (t) dt du du a a a a a u=a Z z Z z Z z Z z Z z =z F (t) dt − uF (u) du = z F (t) dt − tF (t) dt = (z − t)F (t) dt.

Z z

F (t) dt du = a

a

a

du du

Z u

a

a

172

a

a

1.5

magnitude 1.0

V(z)/V g

0.5 0.0 -0.5

real imaginary

-1.0 -1.5

Analytic Integral equation

-2.0 -2.0

-1.5

-1.0

-0.5

0.0

z/λ

Figure 46 7.3. Use Leibniz’s rule: Z b(z) Z b(z) db(z) da(z) ∂f (t, z) d f (t, z) dt = f (b(z), z) − f (a(z), z) + dt dz a(z) dz dz ∂z a(z) to differentiate

Z z g(z) = −

(z − t)kz2 (t)g(t) dt + C1 z + C2

0

twice with respect to z. The results are 0

Z z

g (z) = −

kz2 (t)g(t) dt + C1

0

and g 00 (z) = −kz2 (z)g(z).

173

7.4. Let D = sin kz d − akz cos kz d. The first term on the RHS of (7.41) is Z z (z − t)g(t) dt T1 = −kz2 0

2Ẽ0 = −kz2 D

Z z

2Ẽ0 = −kz2

D

(z − t) sin kz (d − t) dt 0

z−t sin kz (d − t) cos kz (d − t) − kz kz2

z 0

2Ẽ0 = [sin kz (d − z) + zkz cos kz d − sin kz d].

D The second term on the RHS of (7.41) is Z a − z 2Ẽ0 2 d k (d − t) sin kz (d − z) dt T2 = a−d D z 0

a − z 2Ẽ0 2 d − t sin kz (d − t) d = k cos kz (d − t) − a − d D z kz kz2 0 =

a − z 2Ẽ0 2 k [−dkz cos kz d + sin kz d] . a−d D z

Substituting these into (7.43), we get 2Ẽ0 z−d sin kz (d − z) = T1 + T2 + 2Ẽ0 D a−d 2Ẽ0 = [sin kz (d − z) + zkz cos kz d − sin kz d] D 2Ẽ0 z − d 2Ẽ0 a − z [−dkz cos kz d + sin kz d] + D . + D a−d D z+d Canceling the term on the LHS, we obtain

a−z z−d 0 = zkz cos kz d − sin kz d + [−dkz cos kz d + sin kz d] + [sin kz d − akz cos kz d]. a−d a−d Now multiply through by a − d and rearrange to see that (7.43) is satisfied.

7.5. Begin with (7.29): " # ∂ Ẽy ∂ Ẽy j H̃ = −x̂ + ẑ . ω µ̃(z) ∂z ∂x Substitute into Ampere’s law: "

∂ H̃x ∂ H̃z ∇ × H̃ = ŷ − ∂z ∂x We obtain

# = jω˜ (z)Ẽy (x, z).

" # " # ∂ j ∂ Ẽy ∂ j ∂ Ẽy − − = jω˜ (z)Ẽy , ∂z ω µ̃(z) ∂z ∂x ω µ̃(z) ∂x 174

" # " # ∂ 1 ∂ Ẽy ∂ 1 ∂ Ẽy + = −ω 2 ˜(z)Ẽy , ∂x µ̃(z) ∂x ∂z µ̃(z) ∂z 1 ∂ 2 Ẽy 1 ∂ 2 Ẽy µ̃0 (z) ∂ Ẽz + − = −ω 2 ˜(z)Ẽy , µ̃(z) ∂x2 µ̃(z) ∂z 2 µ̃(z) ∂y so

∂ 2 Ẽy ∂ 2 Ẽy µ̃0 (z) ∂ Ẽy + − + k 2 Ẽy = 0, ∂x2 ∂z 2 µ̃(z) ∂z

k 2 (z) = ω 2 µ̃(z)˜ (z).

Let Ẽy (x, z) = f (x)g(z), substitute to get g(z)

d2 f (x) d2 g(z) µ̃0 (z) dg(z) + f (x) − f (x) + k 2 f (x)g(z) = 0, dx2 dz 2 µ̃(z) dz

and divide by f g: 1 d2 f 1 d2 g 1 µ̃0 (z) dg + − = −k 2 . f dx2 g dz 2 g µ̃(z) dz Now separate variables to obtain 1 d2 g 1 µ̃0 (z) dg − = −kz2 g dz 2 g µ̃(z) dz

1 d2 f = −kx2 , f dx2 where

kx2 + kz2 = k 2 .

So

1 dµ̃(z) dg d2 g − + kz2 g = 0. dz 2 µ̃(z) dz dz

Integrate to obtain an integral equation: Z z Z z 0 0 g (z) − h(t)g (t) dt + f (t)g(t) dt = C̄1 .

0

0

If µ̃(z) is twice differentiable, we can integrate the second term by parts: Z z Z z Z z z 0 0 h(t)g (t) dt = h(t)g(t) 0 − h (t)g(t) dt = h(z)g(z) − h(0)g(0) − h0 (t)g(t) dt.

0

0

Then

Z z

0

g (z) + h(z)g(z) −

0

Z z

0

h (t)g(t) dt + 0

f (t)g(t) dt = C̄1 + h(0)g(0) = C1 0

or g 0 (t) + h(z)g(z) −

Z z

[f (t) − h0 (t)]g(t) dt = C1 .

0

Integrate again: Z z g(z) +

Z z h(t)g(t) dt +

0

(z − t)[f (t) − h0 (t)]g(t) dt = C1 z + C2 .

0

175

Therefore

Z z

Z z h(t)g(t) dt −

g(z) = −

(z − t)[f (t) − h0 (t)]g(t) dt = C1 z + C2 .

0

0

The constants C1 and C2 are determined by applying the boundary conditions g(d) = 0 and g(0) + ag 0 (0) = 2Ẽ0 ,

a=

jη0 . ω µ̃(0) cos φ0

We find C2 =

2Ẽ0 a − C1 h0 h0

where

h0 = 1 − ah(0)

and

Z d 2Ẽ0 h0 C1 = − {h(t) + (d − t)[f (t) − h0 (t)]}g(t) dt. a − dh0 a − dh0 0 Substitution into (*) gives the desired result.

7.6. (a) Start with g(z) = A1 g1 (z) + A2 g2 (z).

Use (7.35) to get A1 g1 (d) + A2 g2 (d) = 0 or A2 = −A1

Then

g(z) = A1

g1 (d) . g2 (d)

g1 (d) g1 (z) − g2 (z) . g2 (d)

Now apply (7.38):

2Ẽ0 = A1

g1 (d) g1 (d) 0 0 g2 (0) + aA1 g1 (0) − g (0) . g1 (0) − g2 (d) g2 (d) 2

We obtain A1 =

2Ẽ0 g2 (d) F + aG

where G = g10 (0)g2 (d) − g1 (d)g20 (0).

F = g1 (0)g2 (d) − g1 (d)g2 (0), Substitution yields g(z) =

2Ẽ0 [g1 (z)g2 (d) − g1 (d)g2 (z)].

F + aG

(b) g(0) Ẽ0 2 = −1 + [g1 (0)g2 (d) − g1 (d)g2 (0)] F + aG 2F = −1 + F + aG F − aG = .

F + aG

Γ = −1 +

176

(*)

7.7. The differential equation is d2 g(z) + kz2 (z)g(z) = 0 dz 2

(1)

where kz2 (z) = k 2 (z) − kx2 . We wish to verify that when ˜(z) = 0 ˜r0 eκz , a solution is g(z) = Bν (λeκz/2 )

(2)

√ where λ = 2k0 µ̃r ˜r0 /κ and ν = 2kx /κ. Use k 2 (z) = ω 2 µ̃˜ (z) = ω 2 µ̃r µ0 0 ˜r0 eκz = k02 µ̃r ˜r0 eκz .

Use

d κ (λeκz/2 ) = λ eκz/2 Bν0 (λeκz/2 ), dz 2 κ 2 κ 2 g 00 (z) = λ eκz/2 Bν0 (λeκz/2 ) + λ eκz Bν00 (λeκz/2 ).

2 2 g 0 (z) = Bν0 (λeκz/2 )

Substitute these: λ

κ2 κ2 κz/2 0 e Bν (λeκz/2 ) + λ2 eκz Bν00 (λeκz/2 ) + [k02 µ̃r ˜r0 eκz − kx2 ]Bν (λeκz/2 ) = 0 4 4

or Bν00 (u) +

1 −κz/2 0 4 e Bν (u) + [k02 µ̃r ˜r0 − kx2 e−κz ] 2 2 Bν (u) = 0 λ λ κ

where u = λeκz/2 . This can be rewritten as 1 Bν00 (u) + Bν0 (u) + u

ν2 1 − 2 Bν (u) = 0. u

From (E.1) we see that this equation is satisfied when Bν (u) is Jν (u) or Nν (u).

7.8. The integral equation is, from (7.41), Z z g(z) = −

(z − t)kz2 (t)g(t) dt +

0

a−z a−d

Z d

(d − t)kz2 (t)g(t) dt + 2Ẽ0

0

z−d a−d

where kz2 (z) = ω 2 µ̃˜ (z) = ω 2 µ0 0 ˜r0 eκz = k02 ˜r0 eκz . (a) Let g(z) =

N X

an Pn (z)

n=1

where

( 1, zn − ∆/2 ≤ z ≤ zn + ∆/2, Pn (z) = 0, elsewhere,

177

1 zn = n − ∆.

2

Substitute the expansion and point match at z = zm : N X

an Pn (zm ) = −

n=1

N X

Z zm an

n=1

+

N X

an

n=1

a − zm a−d

(zm − t)kz2 (t)Pn (t) dt

0

Z d

(d − t)kz2 (t)Pn (t) dt + 2Ẽ0

0

zm − d , a−d

m = 1, 2, . . . , N.

We can write this as a matrix equation N X

an Amn = bm

n=1

where

zm − d , a−d Am = δmn + Umn + Vmn , Z zm Umn = (zm − t)kz2 (t)Pn (t) dt, bm = 2Ẽ0

0

and a − zm Vmn = − a−d

Z d

(d − t)kz2 (t)Pn (t) dt.

0

To compute Umn we consider three cases.

(1) m > n Z zn +∆/2

(zm − t)kz2 (t) dt zn −∆/2 Z zn +∆/2 2 = k0 ˜r0 (zm − t)eκt dt zn −∆/2 zn +∆/2 κt e = k02 ˜r0 2 [(zm − t)κ + 1] κ zn −∆/2

Umn =

(

eκ(zn +∆/2) eκ(zn −∆/2) = k02 ˜r0 [(z − z − ∆/2)κ + 1] − [(zm − zn + ∆/2)κ + 1] m n κ2 κ2 o eκzn n = k02 ˜r0 2 eκ∆/2 [(m − n − 1/2)κ∆ + 1] − e−κ1∆/2 [(m − n + 1/2)κ∆ + 1] . κ

178

)

(2) m = n Z zm

(zm − t)kz2 (t) dt zm −∆/2 zm eκt = k02 ˜r0 2 [(zm − t)κ + 1] K zm −∆/2

Umn =

(

eκzm eκ(zm −∆/2) = k02 ˜r0 − [κ∆/2 + 1] κ2 κ2 o eκzm n = k02 ˜r0 2 1 − e−κ∆/2 [κ∆/2 + 1] . κ

)

(3) m < n Umn = 0.

Next compute a − zm Vmn = − a−d

Z d

(d − t)kz2 (t)Pn (t) dt

0

Z zn +∆/2 a − zm 2 =− k ˜r0 (d − t)eκt dt a−d 0 zn −∆/2 o a − zm 2 eκzn n κ∆/2 =− k0 ˜r0 2 e [(d − zn − ∆/2)κ + 1] − e−κ∆/2 [(d − zn + ∆/2)κ + 1] . a−d κ (b) Let the numerical values be as given in the problem statement. The solution for g(z) yields a numerical value for the reflection coefficient of Γ = 0.7368 − j0.6761, which matches the analytic result to four significant digits.

A plot of |g(z)| is shown in Figure 47 and compared to the analytic result from Problem 7.7.

Excellent agreement is seen.

7.9. We have sin kz d + akz cos kz d sin kz d − akz cos kz d ejkz d − e−jkz d + jakz (ejkz d + e−jkz d ) = jkz d e − e−jkz d − jakz (ejkz d + e−jkz d ) (1 + jakz )ejkz d − (1 − jakz )e−jkz d = (1 − jakz )ejkz d − (1 + jakz )e−jkz d

Γ=

1+jakz −2jkz d 1−jakz − e z −2jkz d 1 − 1+jak 1−jakz e Γ0 − P̃ 2 = where 1 − Γ0 P̃ 2

=

Γ0 =

179

1 + jakz , 1 − jakz

P̃ = e−jkz d .

2.5 MoM N=2000 Analytic

|g(z)|/E 0

2.0

1.5

1.0

0.5

0.0 0.00

0.01

0.02

z-position (m)

Figure 47 Now Γ0 =

1 + jakz 1 − jakz

q k 2 − k02 sin2 φ0 q = 0 1 − j ωµ̃jη k 2 − k02 sin2 φ0 cos φ0 p ω µ̃r µ0 cos φ0 − η0 k02 µ̃r ˜cr − sin2 φ0 p = ω µ̃r µ0 cos φ0 + η0 k02 µ̃r ˜cr − sin2 φ0 q p √ ω µ̃r µ0 cos φ0 − µ 00 ω µ0 0 µ̃r ˜cr − sin2 φ0 q = p √ ω µ̃r µ0 cos φ0 + µ 00 ω µ0 0 µ̃r ˜cr − sin2 φ0 p µ̃r cos φ0 − µ̃r ˜cr − sin2 φ0 p . = µ̃r cos φ0 + µ̃r ˜cr − sin2 φ0 0 1 + j ωµ̃jη cos φ0

180

0.03

7.10. In region 0, Ẽ0 −jk0 z e , η0

Ẽi = ŷẼ0 e−jk0 z ,

H̃i = −x̂

Ẽr = ŷẼr ejk0 z ,

H̃r = x̂

Ẽt = ŷẼt e−jk0 z ,

H̃t = −x̂

Ẽr jk0 z e . η0

In region 2, Ẽt jk0 z e , η0

We can develop an integral equation for the electric field in region 1 by starting with Maxwell’s equations. Faraday’s law: −j ∂ Ẽy . ω µ̃ ∂z

∇ × Ẽ = jω µ̃H̃

=⇒

H̃x =

∇ × H̃ = jω˜ c Ẽ

=⇒

jω˜ c Ẽy =

Ampere’s law

Combine to obtain jω˜ c Ẽy = − or

∂ H̃x . ∂z

j ∂ 2 Ẽy ω µ̃ ∂z 2

∂ 2 Ẽy + k 2 Ẽy = 0, ∂z 2

k 2 = ω 2 µ̃˜ c .

Let g(z) = Ẽy (z). Then we have the differential equation g 00 + k 2 g = 0.

Integrate to obtain 0

g (z) + k

2

Z z g(t) dt = C1 0

and again to obtain g(z) = −k 2

Z z (z − t)g(t) dt + C1 z + C2 .

0

To determine C1 and C2 we apply the boundary conditions. At z = 0, continuity of tangential E gives Ẽ0 + Ẽr = g(0).

(1) Continuity of tangential H gives η0 0 g (0). ω µ̃

(2)

η0 0 g (0). ω µ̃

(3)

−Ẽ0 + Ẽr = −j Subtracting (2) from (1), we obtain 2Ẽ0 = g(0) + j 181

At z = d, continuity of tangential Ẽ gives

Continuity of tangential H gives −j

g(d) + Ẽt e−jk0 d .

(4)

η0 0 g (d) = Ẽt e−jk0 d . ω µ̃

(5)

η0 0 g (d) = 0. ω µ̃

(6)

Subtracting (5) from (4), we obtain g(d) + j

Using (3) and the facts that g(0) = C2 and g 0 (0) = C1 , we get C2 = 2Ẽ0 − j

η0 C1 . ω µ̃

Using (6) and the relationships Z d (d − t)g(t) dt + C1 d + C2 , g(d) = −k 2

g 0 (d) = −k 2

Z d g(t) dt + C1 , 0

0

we get k2 C1 = d

Z d

η0 (d − t)g(t) dt − j ω µ̃ 0

Z d

g(t) dt .

0

Now substitute C1 and C2 back to get the desired integral equation.

7.11. (a) To get a differential equation for Hy (z), use Maxwell’s equations. First, " # ∂ H̃ ∂ H̃ 1 y y ∇ × H̃ = jω˜ c (z)Ẽ =⇒ Ẽ = −x̂ + ẑ , jω˜ c (z) ∂z ∂x so Ẽx = − Then

∂ H̃y 1 , c jω˜ (z) ∂z

Ẽz =

"

#

∂ Ẽx ∂ Ẽz ∇ × Ẽ = ŷ − ∂z ∂x

∂ H̃y 1 . c jω˜ (z) ∂x

= −jω µ̃(z)H̃,

so " # 1 ∂ Ẽx ∂ Ẽz − H̃y = − jω µ̃(z) ∂z ∂x ! !#

" 1 ∂ 1 ∂ H̃y ∂ 1 ∂ H̃y 1 = − − − jω µ̃(z) jω ∂z ˜c (z) ∂z ∂x ˜c (z) ∂x " # 1 ∂˜ c /∂z ∂ H̃y 1 ∂ 2 H̃y 1 ∂ 2 H̃y =− 2 − c 2 + c + c ω µ̃(z) (˜ ) ∂z ˜ ∂z 2 ˜ ∂x2 " # ∂ 2 H̃y ∂ 2 H̃y 1 ˜c0 (z) ∂ H̃y =− 2 + − . k (z) ∂x2 ∂z 2 ˜c (z) ∂z 182

Thus

∂ 2 H̃y ∂ 2 H̃y ˜c0 (z) ∂ H̃y + − + k 2 (z)H̃y = 0. ∂x2 ∂z 2 ˜c (z) ∂z

Seek a product solution. Let H̃y (x, z) = f (x)g(z): g

d2 f d2 g ˜c0 dg + f − cf + k 2 f g = 0. dx2 dz 2 ˜ dz

Divide by f g: 1 d2 f 1 d2 g ˜c0 1 dg + − c = −k 2 . f dx2 g dz 2 ˜ g dz Separate variables: 1 d2 f = −kx2 , f dx2

1 d2 g ˜c0 1 dg − c = −kz2 , g dz 2 ˜ g dz

where kz2 (z) + kx2 = k 2 (z).

Hence the differential equation for g(z) is g 00 (z) −

˜c0 (z) 0 g (z) + kz2 g(z) = 0. ˜c (z)

(b) The incident and reflected fields in region 0 are H̃yi = H̃0 e−jkx0 x e−jkz0 z ,

H̃yr = H̃r e−jkx0 x ejkz0 z ,

Ẽxi = η0 cos φ0 H̃0 e−jkx0 x e−jkz0 z ,

Ẽxr = −η0 cos φ0 H̃0 e−jkx0 x ejkz0 z ,

where kx0 = k0 sin φ0 ,

kz0 = k0 cos φ0 ,

so that 2 2 kx0 + kz0 = k02 = ω 2 µ̃0 0 .

To satisfy the boundary conditions we must have kx = kx0 . Thus, f (x) = e−jkx0 x .

Boundary condition 1, tangential Ẽ continuous at z = 0: η0 H̃0 cos φ0 − η0 H̃r cos φ0 = Boundary condition 2, tangential H̃ continuous at z = 0: H̃0 + H̃r = g(0).

Boundary condition 3, tangential Ẽ is zero at z = d: g 0 (d) = 0.

183

j ω˜ c (0)

g 0 (0).

Thus we have the three equations H̃0 − H̃r =

j g 0 (0), ω˜ c (0)η0 cos φ0

(1)

H̃0 + H̃r = g(0),

(2)

g 0 (d) = 0.

(3)

and Adding (1) and (2), we get an alternative form of the boundary condition 2H̃0 = g(0) + ag 0 (0),

a=

j ω˜ c (0)η0 cos φ0

.

Now return to the differential equation for g(z): ˜c0 (z) . h(z) = − c ˜ (z)

g 00 (z) + h(z)g 0 (z) + kz2 (z)g(z) = 0, Integrate: g 0 (z) +

Z z

Z z

h(t)g 0 (t) dt +

0

kz2 (t)g(t) dt = C1 .

0

Integrate again: Z z

Z z

0

(z − t)h(t)g (t) dt =

g(z) + 0

(z − t)kz2 (t)g(t) dt = C1 z + C2 .

0

To find C1 and C2 we apply (1’) and (3). We obtain Z d C1 =

[h(t)g 0 (t) + kz2 (t)g(t)] dt,

C2 = 2H0 − aC1 .

0

Substitution yields Z z

Z z

0

(z − t)h(t)g (t) dt +

g(z) + 0

(z − t)kz2 (t)g(t) dt

0

= C1 z + 2H̃0 − aC1 = 2H̃0 + C1 (z − a) Z d = 2H̃0 + (z − a) [h(t)g 0 (t) + kz2 (t)g(t)] dt.

0

So the integro-differential equation for g(z) is Z d g(z) =

[(z − a) − (z − t)U (z − t)][kz2 (t)g(t) + h(t)g 0 (t)] dt + 2H̃0 ,

0

holding for 0 ≤ z ≤ d, where U (x) is the unit step function.

184

(1’)

7.12.

Z (m−n+ 1 )∆ 2

Amn = where

Z 2π G(u) = 0

G(u) du

(1)

(m−n− 12 )∆

e−jk0 R(ξ,u) du, R(ξ, u)

q R(ξ, u) = 4a2 sin2 (ξ/2) + u2 .

In (1) let v = −u and use G(−u) = G(u): Z −(m−n+ 1 )∆ Amn =

2

Z −(m−n− 1 )∆ 2

G(v)(−dv) =

−(m−n− 21 )∆

Z (m−n+ 1 )∆ 2

G(v) dv =

−(m−n+ 12 )∆

(m−n− 21 )∆

7.13. Refer to Figure 48.

Figure 48 Use the Schelkunoff equivalence principle of § 6.3.4.

J̃sm (r) = −2n̂ × Ẽ = −2ẑ × ρ̂ Z Ãh (r) =

Ṽ0 /2 Ṽ0 = −φ̂ , ρ ln(b/a) ρ ln(b/a)

0 J̃sm (r0 )

S

e−jkR dS 0 , 4πR

1 ∇ × Ãh 0 Z e−jkR Ṽ0 0 =∇× φ̂ 0 dS 0 ρ ln(b/a) 4πR S

Z −jkR 1 Ṽ0 dS 0 0e = ∇ × φ̂ .

4π ln(b/a) S R ρ0

Ẽ = −

Use (B.49) to write

−jkR 0e

∇ × φ̂

R

−jkR e = −φ̂ × ∇ R 0

185

G(u) du = Anm .

where, by (B.74), ∇

−jkR

−jkR e 1 e = −R + jk .

R R R2

Then Ṽ0 1 Ẽ = 4π ln(b/a) 0

Z

0

φ̂ × R S

1 + jk R

e−jkR dS 0 .

R 2 ρ0

0

But φ̂ × R = φ̂ × (ẑz − ρ̂0 ρ0 ) = ρ̂0 z + ρ0 ẑ for points on the z-axis, so −jkR Z b Z 2π 1 1 Ṽ0 e 0 0 Ẽz = + jk dρ0 ρ dφ 2 4π ln(b/a) 0 R R a −jkR Z b 0 Ṽ0 ρ 1 e = + jk dρ0 .

2 ln(b/a) a R R R Now ∂ ∂ρ0 where

−jkR

−jkR e 1 e ∂R =− + jk R R R ∂ρ0

∂R ∂ = 0 0 ∂ρ ∂ρ

q ρ0 ρ0 ρ0 2 + z 2 = p = .

R ρ0 2 + z 2

Therefore −jkR Z b Ṽ0 ∂ e Ẽz = − dρ0 0 2 ln(b/a) a ∂ρ R −jkR1

Ṽ0 e e−jkR2 = − 2 ln(b/a) R1 R2 √ √ where R1 = a2 + z 2 and R2 = b2 + z 2 .

7.14. For a straight wire aligned along z, use ∂ ∂ = , ∂u ∂z

0

û · û0 = 1,

G(u, u0 ) = 2π

e−jk0 R̄(z−z ) , R̄(z − z 0 )

R̄ =

Thus, from (7.80), the Pocklington integral equation is 2

Z L ∂ 8π 2 k0 i 0 2 0 0 ˜ ) I(z − k G(z − z ) dz = j Ẽ (z), 0 ∂z∂z 0 η0 −L Use

where

to write

p (z − z 0 )2 + a2 .

−L ≤ z ≤ L.

" # ∂G(z − z 0 ) ∂ e−jk0 R̄ R̄(−jk0 ) − 1 −jk0 R̄ ∂ R̄ = 2π = 2π e , ∂z ∂z ∂z R̄ R̄2 ∂ R̄ ∂ p z − z0 = (z − z 0 )2 + a2 = ∂z ∂z R̄ ∂G(z − z 0 ) z − z0 = G1 (z − z 0 ) = −2π (1 + jk0 R̄)e−jk0 R̄ . ∂z R̄3 186

The integral equation becomes Z L Z L 0 8π 2 k0 i 0 2 0 0 0 0 ∂G1 (z − z ) ˜ ˜ dz − k I(z )G(z − z ) dz = j Ẽ (z).

I(z ) 0 ∂z 0 η0 −L −L Use integration by parts in the first integral: L

0

Z L

˜ 0 ) ∂G1 (z − z ) dz 0 = I(z ˜ 0 )G1 (z − z 0 ) I(z − 0 ∂z −L −L

so

˜ 0) ∂ I(z G1 (z − z 0 ) dz 0 , 0 ∂z −L

Z L

# Z L" ˜ 0 ∂ I(z ) 8π 2 k0 i 0 2˜ 0 0 0 Ẽ (z), G (z − z ) + k I(z )G(z − z ) dz = −j 1 0 ∂z 0 η0 −L

˜ ˜ since I(−L) = I(L) = 0. We can solve this integral equation using pulse function expansion and point matching. Let N X ˜ I(z) = an Pn (z) n=1

where ( 1, zn − ∆/2 ≤ z ≤ zn + ∆/2, Pn (z) = 0, elsewhere, We have

∆=

2L , N

zn = −L + (n − 1/2)∆.

N

X ˜ ∂ I(z) = an [δ(z − [zn − ∆/2]) − δ(z − [zn + ∆/2])] ∂z n=1

so that " # Z zn +∆/2 N X 8π 2 k0 i 2 0 0 Ẽ (zm ) an G1 (zm − [zn − ∆/2]) − G1 (zm − [zn + ∆/2]) + k0 G(zm − z ) dz = −j η0 zn −∆/2 n=1

for m = 1, 2, . . . , N . This is a matrix equation of the form N X

an Amn = bm .

n=1

Letting u = zm − z 0 , we can write k02

Z zn +∆/2

0

G(zm − z ) dz zn −∆/2

where G(u) = 2π

0

= k02

ejk0 R̄(u) , R̄(u)

Z zm −zn +∆/2 G(u) du zm −zn −∆/2

R̄(u) =

p u2 + a2 ,

and zm − zn = −L + (m − 1/2)∆ − [−L + (n − 1/2)∆] = (m − n)∆.

187

Thus Amn = G1 ([m − n + 1/2]∆) − G1 ([m − n − 1/2]∆) + k02

Z [m−n+1/2]∆ G(u) du [m−n−1/2]∆

where G1 (u) = −2π

u (1 + jk0 R)e−jk0 R̄ , R̄3

and bm = −j where

8π 2 k0 i Ẽ (zm ) η0

" # e−jk0 R1 (z) e−jk0 R2 (z) V0 , Ẽ (z) = − 2 ln(b/a) R1 (z) R2 (z) p p R1 (z) = z 2 + a2 , R2 (z) = z 2 + b2 . i

The input impedance found by solving the matrix equation is shown in Figure 49 for the case of a = 0.01 m. Comparing this to Figure 7.15, we see that (1) the thin wire approximation breaks down sooner with Hallen’s integral equation, and (2) Pocklington’s equation predicts a somewhat higher input impedance than Hallen’s equation.

200 180

impedance ( Ω)

160 140

resistance

120 100 80 60 40

reactance

20 0 0

50

100

150

200

250

number of pulse functions

Figure 49

188

300

350

400

The input impedance found by solving the matrix equation is shown in Figure 50 for the case of a = 0.0001 m. Comparing this to Figure 7.14, we see that even for N = 2000 pulses the Pocklington solution has not converged to a stable value.

400 350

impedance ( Ω)

300 250

resistance

200 150 100 50

reactance

0 -50 -100 0

200

400

600

800 1000 1200 1400 1600 1800 2000

number of pulse functions

Figure 50 The input impedance found by solving the matrix equation is shown in Figure 51 for the case of a = 0.001 m. This result is somewhat in between the previous two cases. The result takes a large number of pulses to converge, but reaches a stable value by about N = 500 pulses.

7.15. Begin with the Pocklington integral equation Z 2 ˜ 0 )K(u, u0 ) du0 = j 8π k0 Ẽ i (u), I(u η0 Γ where

∂2 2 0 − k0 (û · û ) G(u, u0 ), K(u, u ) = ∂u∂u0 0

u ∈ Γ, 0

e−jk0 R(u,u ) G(u, u ) = .

R(u, u0 ) 0

For the circular loop we have u = bφ,

û = φ̂,

∂ 1 ∂ = , ∂u b ∂φ

û · û0 = cos(φ − φ0 ),

189

G(u, u0 ) = G(φ − φ0 ).

400 350

impedance ( Ω)

300 250 200

resistance

150 100 50

reactance

0 0

50

100 150 200 250 300 350 400 450 500 550 600

number of pulse functions

Figure 51 The Hallen form of the integral equation is, from (7.83), Z 2π 0

2 Z φ

˜ 0 )Π(φ, φ0 ) dφ0 = A sin k0 bφ + B cos k0 bφ − j 8π I(φ η0

Ẽ i (ζ) sin k0 b(φ − ζ) dζ.

0

The kernel is, from (7.84), Π(φ, φ0 ) = cos(φ − φ0 )G(φ − φ0 )

Z φ 1 ∂G(χ − φ0 ) 1 ∂ 0 0 − cos k0 b(φ − χ)b dχ. cos(χ − φ )G(χ − φ ) + b ∂χ b ∂φ0 0 Note that we have chosen the constant lower limit to be zero for convenience in the next step. We

190

can remove the derivatives using integration by parts. Let Z φ

∂ cos(χ − φ0 )G(χ − φ0 ) cos k0 b(χ − φ) dχ 0 ∂χ

φ = cos(χ − φ0 )G(χ − φ0 ) cos k0 b(χ − φ) 0 Z φ

k0 b cos(χ − φ0 )G(χ − φ0 ) sin k0 b(χ − φ) dχ +

I1 =

0

= cos(φ − φ0 )G(φ − φ0 ) − cos φ0 G(φ0 ) cos k0 bφ Z φ

k0 b cos(χ − φ0 )G(χ − φ0 ) sin k0 b(χ − φ) dχ. + 0

Let

Z φ I2 = 0

and use

∂G(χ − φ0 ) cos k0 b(χ − φ) dχ ∂φ0

∂G(χ − φ0 ) ∂G(χ − φ0 ) = − ∂φ0 ∂χ

to write φ I2 = −G(χ − φ ) cos k0 b(χ − φ) 0 − 0

= −G(φ − φ0 ) + G(φ0 ) cos k0 bφ −

Z φ

k0 bG(χ − φ0 ) sin k0 b(χ − φ) dχ

0

Z φ

k0 bG(χ − φ0 ) sin k0 b(χ − φ) dχ.

0

Thus Π(φ, φ0 ) = cos(φ − φ0 )G(φ − φ0 ) − cos(φ − φ0 )G(φ − φ0 ) + cos φ0 G(φ0 ) cos k0 bφ Z φ

− k0 b cos(χ − φ0 )G(χ − φ0 ) sin k0 b(χ − φ) dχ 0 Z φ 0 0 k0 bG(χ − φ0 ) sin k0 b(χ − φ) dχ + G(φ − φ ) − G(φ ) cos k0 bφ + 0

or Π(φ, φ0 ) = G(φ − φ0 ) + G(φ0 ) cos k0 bφ[cos φ0 − 1] Z φ + k0 b G(χ − φ0 ) sin k0 b(χ − φ)[cos(χ − φ0 ) − 1] dχ.

0

7.16. Using r̂ · r0 = r̂ · (ẑz 0 ) = z 0 cos θ and k0 = ω/c, we get Z L µ0 e−jk0 r ˜ 0 , ω)ejk0 r̂·r0 dz 0 Ẽθ (r, θ, φ) = jω sin θ I(z 4π r −L i h Z L 0 ω µ0 −jω rc − z cos 0 c ˜ = sin θ [jω I(z , ω)]e dz 0 .

4πr −L

191

˜ 0 , ω) ↔ I(z 0 , t). Then, by the differentiation and time-shifting theorems, Let I(z

h i 0 θ r z 0 cos θ −jω rc − z cos 0 0 0 c ˜ [jω I(z , ω)]e − ↔ I z ,t − c c where I 0 (t) = dI(t)/dt. So µ0 Eθ (r, θ, t) = sin θ 4πr

Z L

r z cos θ 0 I z ,t − + dz 0 . c c −L

7.17. Assume In (z) = sin

0

nπ 2L

[z − L] .

We have am = Rm /Cm .

Rm = −8π 2 sm 0 sin θi E0 (sm )J0 (−jsm a sin θi /c)

Z L

Im (z)e−γm z cos θi dz.

−L

Let Z L I=

Im (z)e−γm z cos θi dz

−L Z L

=

sin −L

mπ 2L

[z − L] e−γm z cos θi dz

mπ mπ i L [z − L] − cos [z − L] 2L 2L 2L + (γm cos θi )2 −L 2L i h h i γ L cos θ −γ L cos θ m m i i e mπ mπ e − − − cos mπ . =

2 mπ 2 mπ 2L 2L + (γm cos θi )2 + (γm cos θi )2 2L L =

e−γm z cos θi

mπ 2

h

−γm cos θi sin

mπ

So mπ 2L

2

Rm = −8π sm 0 sin θi E0 (sm )J0 (−jsm a sin θi /c)

mπ 2 2L

+ (γm cos θi )2

h i (−1)m eγm L cos θi − e−γm L cos θi .

Then Cm = Cm1 + Cm2 where ∂Im (z) ∂Im (z 0 ) F (z − z 0 , sm ) dz 0 dz 0 ∂z ∂z −L −L Z LZ L mπ

mπ

mπ 2 = cos [z − L] cos [z 0 − L] F (z − z 0 , sm ) dz 0 dz 2L 2L −L −L 2L Z LZ L

Cm1 =

and Z LZ L

2 γm Im (z)Im (z 0 )g(z − z 0 , sm ) dz 0 dz c −L −L Z LZ L 2 mπ

mπ

γm = sin [z − L] sin [z 0 − L] g(z − z 0 , sm ) dz 0 dz.

2L 2L −L −L c

Cm2 =

192

Use the identities 1 cos x cos y ≡ [cos(x − y) + cos(x + y)], 2

1 sin x sin y ≡ [cos(x − y) − cos(x + y)] 2

Then Cm = Cm1 + Cm2

Z Z mπ

2 γm 1 L L mπ 2 0 0 F (z − z , sm ) + = g(z − z , sm ) cos [z − z 0 ] dz 0 dz 2 −L −L 2L c 2L

Z L Z L

mπ

2 1 mπ 2 γ + F (z − z 0 , sm ) − m g(z − z 0 , sm ) cos [z + z 0 − 2L] dz 0 dz.

2 −L −L 2L c 2L Use cos

mπ

mπ

mπ

[z + z 0 − 2L] = cos [z + z 0 ] cos(mπ) + sin [z + z 0 ] sin(mπ) 2L 2L 2L

mπ 0 m [z + z ] = (−1) cos 2L

to get − + Cm = Cm + (−1)m Cm

where 1 ± Cm = 2

Z L Z L −L

−L

mπ

2 mπ 2 γm 0 0 g(z − z , sm ) cos [z ± z 0 ] dz 0 dz.

F (z − z , sm ) ∓ 2L c 2L

7.18. The integral equation for I(φ, s) in the Laplace domain is, by (7.86), Z π 8π 2 ω/c i I(φ0 , s)K(φ − φ0 , s)b dφ0 = j E (φ, s) η0 −π 8π 2 γ i = E (φ, s), −π ≤ φ ≤ π, γ = s/c. η0 Expand K(φ − φ0 , s) =

∞ X

0

αn (s)ejn(φ−φ )

n=−∞

where

γ2 n2 K (s) + [Kn−1 (s) + Kn+1 (s)], n b2 2 Z π 1 Kn (s) = G(ζ, s)e−jnζ dζ, 2π −π Z 2π −γ R̄(ξ,ζ) e G(ζ, s) = dξ.

R̄(ξ, ζ) 0

αn (s) =

Substitute to get ∞ X n=−∞

Z π αn (s)

0

I(φ0 , s)ejn(φ−φ ) dφ0 =

−π

193

8π 2 γ i E (φ, s), bη0

−π ≤ φ ≤ π,

or

∞ X

jnφ

Z π

0

I(φ0 , s)e−jnφ dφ0 =

αn (s)e

−π

n=−∞

Let In (s) =

1 2π

Z π

8π 2 γ i E (φ, s), bη0

−π ≤ φ ≤ π,

0

I(φ0 , s)e−jnφ dφ0

−π

be the Fourier series coefficients of I(φ0 , s) so that I(φ, s) =

∞ X

In (s)ejnφ .

n=−∞

Then

∞ X

αn (s)In (s)ejnφ =

n=−∞

4πγ i E (φ, s), bη0

−π ≤ φ ≤ π.

Now expand In (s) in a pole series: In (s) =

2N X In,m (s) m=1

So

2N X ∞ X

αn (s)

m=1 n=−∞

s − sn,m

.

In,m (s) jnφ 4πγ i e = E (φ, s), s − sn,m bη0

−π ≤ φ ≤ π.

Multiply through by (s − sα,β ) and let s → sα,β . Only one term in the series is nonzero so that αn (sn,m )In,m (sn,m ) = 0 or αn (sn,m ) = 0.

Thus

n2 γ2 K (s) + [Kn−1 (s) + Kn+1 (s)] = 0 n b2 2 defines the natural frequencies sn,m . We see that for each modal index n there is an infinite number of natural frequencies sn,m and modal amplitudes In,m (sn,m ).

7.19. Let Z zm I=

ejk0 u cos θi sin k0 (zm − u) du

−L

Z zm =−

eu(jk0 cos θi ) sin(k0 u − k0 zm ) du.

−L

Use

Z

eax sin(bx + c) dx =

eax [a sin(bx + c) − b cos(bx + c)] a2 + b2

194

to get zm ejk0 u cos θi [jk cos θ sin k (u − z ) − k cos k (u − z )] I=− 2 0 i 0 m 0 0 m k0 − k02 cos2 θi −L

=−

ejk0 zm cos θi e−jk0 L cos θi [−k ] + [−jk0 cos θi sin k0 (L + zm ) − k0 cos k0 (L + zm )].

0 k02 sin2 θi k02 sin2 θi

Thus

jk0 zm cos θi 8π 2 j cos θi sin k0 (L + zm ) cos k0 (L + zm ) e bm = −j − − Ẽ0 sin θi J0 (k0 a sin θi ) η0 k0 sin θi k0 sin2 θi k0 sin2 θi i 8π 2 J0 (k0 a sin θi ) h jk0 zm cos θi = −j Ẽ0 e − j cos θi sin k0 (zm + L) − cos k0 (zm + L) . η0 k0 sin θi 7.20. Note that we can move ρ0 to the origin without loss of generality. In that case, the line source lies along the z-axis. Setting ρ0 = 0 gives G2D (ρ|0) =

1 (2) H (k0 ρ).

4j 0

We have

1 ∂ ∂G2D ∇ G2D = ρ ρ ∂ρ ∂ρ

1 ∂ k0 (2)0 = ρ H0 (k0 ρ) , ρ 6= 0 ρ ∂ρ 4j 2

1 k0 (2)00 k0 (2)0 = ρ H (k0 ρ) + H0 (k0 ρ) ρ 4j 0 4j

2 k0 1 (2)00 (2)0 = H0 (k0 ρ) + H (k0 ρ) .

4j k0 ρ 0 2

So

k2 ∇ G2D + k02 G2D = 0

1 (2)00 (2)0 (2) H0 (k0 ρ) + H (k0 ρ) + H0 (k0 ρ) .

4j k0 ρ 0

2

(2)

But (E.1) shows that H0 (z) satisfies d2 (2) 1 d (2) (2) H0 (z) + H (z) + H0 (z) = 0.

2 dz z dz 0 Thus ∇2 G2D + k02 G2D = 0,

ρ 6= 0.

To examine the behavior of G2D near the origin, integrate the equation ∇2 G2D + k02 G2D = −δ(ρ − ρ0 ) over the disk 0 ≤ ρ ≤ a. Let Z

[∇ · (∇G2D ) + k02 G2D ] dS ZS Z 2 = [ρ̂ · ∇G2D ] dS + k0 G2D dS

I=

Γ

S

195

by (B.26), where Γ is the contour of the disk. But ∇G2D = ρ̂

∂ 1 (2) k0 (2)0 k0 (2) H0 (k0 ρ) = ρ̂ H0 (k0 ρ) = −ρ̂ H1 (k0 ρ) ∂ρ 4j 4j 4j

so Z 2π

Z 2π Z a 2 k0 (2) k0 (2) I=− H1 (ka)a dφ + H0 (k0 ρ)ρ dρ dφ 4j 0 0 δ 4j Z a k0 k02 (2) (2) = −2π aH1 (ka) + 2π H0 (k0 ρ)ρ dρ 4j 4j δ where δ → 0. Use (E.106) to write

δ (2) k0 k02 a (2) (2) H (ka) − H1 (kδ) I = −2π aH1 (ka) + 2π 4j 4j k0 1 k0 k0 (2) = −2π δH1 (kδ).

4j Thus k0 (2) lim δH1 (kδ) 4j δ→0

k0 δ 1 2 k0 −j − = −2π lim δ 4j δ→0 2 π k0 δ = −1

lim I = −2π

δ→0

and

Z

[∇2 G2D + k02 G2D ] dS = −1.

S

Now since

Z −

ZZ δ(ρ) dS = −

δ(x)δ(y) dx dy = −1,

S

we have established that ∇2 G2D + k02 G2D = −δ(ρ − ρ0 ).

7.21. The electric line source produces fields (excitation and scattered) that are TM to z. Thus, the solution to this problem is given in Section 7.4.2.1. Only the excitation field changes, so the matrix entries Amn remain the same. However, the bm change.

The field of the line source is, from (4.343), Ẽzi = −

ω µ̃ ˜ (2) IH0 (k0 |ρ − ρ0 |).

4

For ρ on the strip, we use ρ0 = dŷ and ρ = xx̂ so that p |ρ − ρ0 | = d2 + x2 .

Thus bm =

p 4 ˜ (2) (k0 x2m + d2 ).

Ẽzi (xm ) = −IH 0 ωµ0 196

Consider the numerical solution for the induced current on the strip using the following parameters: f = 300 MHz, w = 1 m, N = 201, I˜ = 1 A. We get the following results. (a) d = w/4. Comparing Figures 52 and 53 to text Figure 7.35, we see that the current magnitude is very strong immediately beneath the line source, when the line source is close to the strip. Also, the phase at the center of the strip is affected significantly.

surface current magnitude (A/m)

3

2

1

0 -1.0

-0.5

0.0

0.5

1.0

x-position (m)

Figure 52 (b) d = w. Comparing Figures 54 and 55 to Figure 7.35, we see that the current magnitude is still strong immediately beneath the line source, but not as strong as in the case of d = w/4. (c) d = 4w. Comparing Figures 56 and 57 to text Figure 7.35, we see that the current magnitude is very similar to that for plane-wave excitation. This is because when the line source is far from the strip, the excitation field is fairly uniform across the strip.

7.22.

1 Amn = k0 1 An+1,n = k0

Z 3 k0 ∆ 2

1 k ∆ 2 0

Z k0 ∆(m−n+ 1 ) 2

k0 ∆(m−n− 12 ) (2) H0 (|u|) du =

197

(2)

H0 (|u|) du.

1 k0

Z 3 k0 ∆ 2

1 k ∆ 2 0

(2)

H0 (u) du.

180

surface current phase (deg)

120

60

0

-60

-120

-180 -1.0

-0.5

0.0

0.5

1.0

x-position (m)

Figure 53 1 An−1,n = k0

Z − 1 k0 ∆ 2

− 32 k0 ∆

Let 1 An+1,n = k0

(2) H0 (|u|) du =

Z 3 k0 ∆ h 2

1 k ∆ 2 0

1 k0

Z 3 k0 ∆

(2) H0 (u) − f0 (u)

198

2

1 k ∆ 2 0

i

(2)

H0 (|x|) dx = An+1,n .

1 du + k0

Z 3 k0 ∆ 2

1 k ∆ 2 0

f0 (u) du.

surface current magnitude (A/m)

3

2

1

0 -1.0

-0.5

0.0

0.5

x-position (m)

Figure 54 Examine I=

1 k0

Z 3 k0 ∆ 2

1 k ∆ 2 0 3 k ∆ 2 0

f0 (u) du

2 1 − j (ln u + γ − ln 2) du 1 π k ∆ 0 2

Z 3 k0 ∆ 2 1 2 1 2 = (k0 ∆) 1 − j (γ − ln 2) + −j ln u du k0 π k0 1 k0 ∆ π 2 3

k ∆ 2 0 1 2 1 2 = (k0 ∆) 1 − j (γ − ln 2) + −j [u ln u − u] k0 π k0 π 1 k ∆ 2 0

2∆ 3 ∆ =∆−j γ − ln 2 − 1 + ln 3 + ln k0 . π 2 2 1 = k0

Z

199

1.0

180

surface current phase (deg)

120

60

0

-60

-120

-180 -1.0

-0.5

0.0

0.5

1.0

x-position (m)

Figure 55 Thus An+1,n = An−1,n =

1 k0

Z 3 k0 ∆ 2

1 k ∆ 2 0

(2)

[H0 (u) − f0 (u)] du

2∆ 3 ∆ +∆−j γ − ln 2 + ln 3 + ln k0 − 1 . π 2 2 7.23. (a) For TE polarization, consider Figure 7.38. The current induced within the conducting strip is x-directed, so the relationship between the incident and scattered fields within the strip is given according to (7.23) and (7.27) as Ẽxs − J˜x (x)Z i (x) = −Ẽxi (x) where Z i (x) =

1 . [σ̃(x) + jω(˜ (x) − 0 )]t

Substitution from (7.156) and (7.155) gives the equation −

jω ∂ 2 Ãsex − jω Ãsex − J˜x (x)Z i (x) = −Ẽ0 sin φ0 ejk0 x cos φ0 2 2 ∂x k0 200

surface current magnitude (A/m)

3

2

1

0 -1.0

-0.5

0.0

0.5

1.0

x-position (m)

Figure 56 or

∂2 k02 ˜ k02 2 s i + k à = − J (x)Z (x) + Ẽ0 sin φ0 ejk0 x cos φ0 . x 0 ex ∂x2 jω jω

The solution to this differential equation is Ãsex (x) = fp (x) + fc (x) where fc (x) = C1 sin k0 x + C2 cos k0 x and 1 fp (x) = k0

Z x x0

Z x 2 1 k02 k jk0 u cos φ0 −j sin φ0 e sin k0 (x − u) du + j 0 J˜x (u)Z i (u) sin k0 (x − u) du. ω k 0 x0 ω

Choose x0 = 0. The first integral is given in the text. With that we find Ẽ0 1 k0 ejk0 x cos φ0 + j fp (x) = −j ω sin φ0 ω

201

Z x 0

Z i (u)J˜x (u) sin k0 (x − u) du.

180

surface current phase (deg)

120

60

0

-60

-120

-180 -1.0

-0.5

0.0

0.5

1.0

x-position (m)

Figure 57 Substituting for Ãsex from (7.145), we write the integral equation as Z µ0 w ˜ 0 (2) Jx (x )H0 (k0 |x − x0 |) dx0 + C1 sin k0 x + C2 cos k0 x 4j −w Z Ẽ0 1 k0 x i jk0 x cos φ0 Z (u)J˜x (u) sin k0 (x − u) du. = −j e +j ω sin φ0 ω 0 Since ωµ0 = k0 η0 , we have Z w (2) J˜x (x0 )H0 (k0 |x − x0 |) dx0 + C1 sin k0 x + C2 cos k0 x −w

4 − η0

Z x 0

4Ẽ0 1 Z i (u)J˜x (u) sin k0 (x − u) du = ejk0 x cos φ0 , η0 k0 sin φ0

holding for −w ≤ x ≤ w. This is Hallen’s integral equation for the lossy strip. (b) To solve the integral equation, expand J˜x (x) in a rectangular pulse series: ( N X 1, xn − ∆/2 ≤ x ≤ xn + ∆/2, J˜x (x) = an Pn (x), Pn (x) = 0, elsewhere, n=1 202

where

1 ∆, xn = −w + n − 2

∆ = 2w/N.

Substitute this and point match at x = xm . At this point let’s consider only the case Z i (u) = Z i = constant.

N X n=1

Z xn +∆/2 an xn −∆/2

(2)

H0 (k0 |xm − x0 |) dx0 + C1 sin k0 xm + C2 cos k0 xm

N

Z xm

n=1

0

4Z i X − an η0

Pn (u) sin k0 (xm − u) du = bm

where bm = Examine

4Ẽ0 1 ejk0 xm cos φ0 . η0 k0 sin φ0

Z xm Pn (u) sin k0 (xm − u) du.

I= 0

(a) n < m

1 1 1 I=− ∆ − cos k0 m − n − ∆ . cos k0 m − n + k0 2 2 (b) n = m I=−

1 ∆ cos k0 − 1 . k0 2

(c) n > m I = 0.

Next, use the substitution u = xm −x0 in the remaining integral. This gives the system of equations N X n=1

or

Z (m−n+ 1 )∆ an

2

(m−n−1/2)∆

(2)

H0 (k0 |u|) du + C1 sin k0 xm + C2 cos k0 xm +

N X

an Fmn = bm

n=1 N X

an [Amn + Fmn ] = bm

n=1

where

 4Z i

1 1   k0 η0 cos k0 m − n + 2 ∆ − cos k0 m − n − 2 ∆ , n < m, i Fmn = k4Z cos k0 ∆ n = m, η 2 −1 , 0   0 0, n > m.

Two additional equation are generated by the conditions J˜x (−w) = 0 and J˜x (w) = 0. This gives an (N + 2) × (N + 2) matrix equation. The matrix entries change from Amn in the PEC case to Amn + Fmn in the lossy case. Everything else remains the same. Note that when Z i = 0, the results for the PEC case are recovered.

203

(c) Let the numerical values be as given in the problem statement. The magnitude of the current density is shown in Figure 58, and is compared to the current for a PEC strip. It is seen that as the strip becomes lossy, the induced current is reduced in strength and the density becomes more uniform. A similar effect was seen in the case of TM polarization.

0.010

surface current magnitude (A/m)

0.009 0.008

PEC

0.007 0.006 0.005 0.004 0.003

Lossy

0.002 0.001 0.000 -1.0

-0.5

0.0

0.5

1.0

x-position (m)

Figure 58 The phase of the current density is shown in Figure 59 and is compared to the current for a PEC strip. It is seen that as the strip becomes lossy, the phase changes to near 180◦ and is very uniform across the strip.

7.24. Let fa (z) = Az 2 + Bz + C. Then fa (0)=C, so we need only to find C. Evaluate:

2

∆ ∆ ∆ fa +B = a1 = A +C 2 2 2

3∆ 3∆ 2 3∆ fa = a2 = A +B +C 2 2 2

5∆ 5∆ 2 5∆ fa = a3 = A +B +C 2 2 2 204

180

surface current phase (deg)

120

Lossy 60

0

PEC -60

-120

-180 -1.0

-0.5

0.0

0.5

x-position (m)

Figure 59 Let ā1 = 4a1 , ā2 = 4a2 , ā3 = 4a3 , Ā = A∆2 , B̄ = B∆, and C̄ = 4C. Then ā1 = Ā + 2B̄ + C̄ ā2 = 9Ā + 6B̄ + C̄ ā3 = 25Ā + 10B̄ + C̄.

Eliminating Ā gives the two equations ā2 − 9ā1 = −12B̄ − 8C̄ ā3 − 25ā1 = −40B̄ − 24C̄.

Eliminating B̄ then gives 60ā1 − 40ā2 + 12ā3 = 32C̄, and so Fa (0) = C =

1 C̄ = [15a1 − 10a2 + 3a3 ] .

4 8

205

1.0