Electromagnetics (Solutions, Instructor Solution Manual) Third Edition [3 ed.] 1498796567, 9781498796569

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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 −∞

Z

0 ∞ Z 2π

= ρ0

0

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

ˆ Jρ (ρ, t). 1.5. (a) Write J = ρ 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 ˆ a dφ dz = −2πLaJρ (a, t), = − J · dS = − Jρ (a, t)ˆ ρ·ρ 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, ˆ ρ0 J(ρ, t) = ρ

ˆ ρ0 ρ

βρ3 −βt e = ρ0 ρ2 e−βt v, 4

we obtain ˆ v=ρ

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

ˆ ∇ρ = ρ

∂  2 −βt  ρ0 ρ e = 2ˆ ρρ0 ρe−βt , ∂ρ

Dρ βρ β ˆ = −βρ0 ρ2 e−βt + ρ · 2ˆ ρρ0 ρe−βt = − ρ0 ρ2 e−βt Dt 4 2

as desired. 1.6. (a) Z

2π

π

Z

2

Z

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

Q= 0

0

0

(b) Z

2 Z 2π

∞

Z

4 cos2 φ δ(ρ − 2)ρ dρ dφ dz = 50.3 C.

Q= 0

0

0

(c) 2Z ∞Z ∞

Z Q= 0

0

4z 3 δ(x)δ(y) dx dy dz = 16 C.

0

(d) Z Q=

2 Z 2π

Z

(e) Z

Z

∞

0

Z

π

0

Z

∞

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

ρ dV = −∞

1.7.

∞

4ρzδ(ρ − 3)ρ dρ dφ dz = 144π C.

ρ dV = 0

Q=

Z

0

0

∂ρ = −16x2 e−4t , ∂t

ˆ 8xe−4t . ∇ρ = x

∂ρ dρ = + v · ∇ρ dt ∂t ˆ Ax · x ˆ 8xe−4t = −16x2 e−4t + x = 8x2 e−4t (A − 2). We have dρ/dt = 0 when A = 2. 1.8.

∂ρ = −4βr2 e−βt , ∂t

∇ρ = ˆr8re−βt .

dρ ∂ρ = + v · ∇ρ = −4βr2 e−βt + ˆr2r · ˆr8re−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¯ ¯ ¯ · B. P − 0 I · E + L P= c 

Put H =

B µ0

(1)

− M: 

 1¯ ¯ ¯ · [µ0 H + µ0 M], P= P − 0 I · E + L c   1¯ ¯ ¯ · H = P − [µ0 L] ¯ · M. P − 0 I · E + [µ0 L] c

(*)

Next examine (2.100):   1¯ ¯ ¯ M = −M · E − cQ − I · B. µ0 Put H =

B µ0

(2)

− M:   1¯ ¯ ¯ M = −M · E − cQ − I · [µ0 H + µ0 M], µ0 ¯ ¯ ¯ ¯ − ¯I) · M. M · E + (µ0 cQ − I) · H = −M − (µ0 cQ

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 ∇0 · D0 = ρ0 = ργ − γ

v Jx . c2

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 B = µ0 − A = µ0 β 2

C=

1 1 − n2 , c 1 − β 2 n2

1 − n2 . 1 − β 2 n2

Writing ˆ β ˆ · H) = ββ · H Hk = β( β2 we get B = AH + BHk + Cβ × E   B ββ 0 A¯ I+ 0 2 ·H−Ω×E (Ω = Cβ) =µ µ0 µ β   n2 − 1 1 − β2 ¯ n2 − 1 β I − = µ0 ββ · H − ×E 1 − β 2 n2 1 − β 2 n2 1 − n2 β 2 c ¯ ·H−Ω×E = µ0 A where

  1 − β 2 ¯ n2 − 1 ¯ A= I− 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 ¯ · E + Ω × H. D = 0 A 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 B+n

2v

  × (v × B) v × (v × H) v×E 2 0 =µ H+ − (n − 1). 2 2 c c c2

Note that

v × (v × B) v 2 ∼ |B|  |B| c2 c2

and

v × (v × E) v 2 ∼ |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 = But M=

Bk − Hk . µ0 B − H, µ0

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

But

2 v v  |M| ˆ × (ˆ v × 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

so

0 0 v × (v × B) 0 v 2 χ χe = χe v ˆ × (ˆ v × B)  m B , 2 2 0 c c µr χ0 M = m0 B − 0 µ0 µ r



 χ0m 0 + χe v × E. µ0r

2.12. Start with P = Pk + P⊥ ,

Pk = P0k .

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 P=

P0k

  1 1 0 1 0 + P⊥ + β × M − β × cP⊥ . γ c γ ⊥

11

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⊥ ) c0 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

χ0e

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

χ0 + m µ0r

 v × B.

2.13. Start with (2.102): ∇×E=−

∂B . ∂t

Integrate over a surface S: Z

Z

∂B · dS. ∂t

(∇ × E) · dS = − S

S

Use (2.127): Z

∂B d · dS = ∂t dt

S

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 −

S

I

d [E + v × B] · dl = − dt Γ

Therefore

I

Z

d E · dl = − dt Γ ∗

(v × B) · dl, Γ

B · dS

where

Z 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 S

∂E d · dS = ∂t dt

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 (ˆ n × E) dS = −

V

S

∂B dV. ∂t

Use (2.132): Z

d ∂B dV = ∂t dt

V

So

Z

I B dV −

ˆ ) dS. B(v · n

V

S

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

I

or

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

∂E d dV = ∂t dt

V

So

I (ˆ n × B) dS + S

or

1 c2

Z

I E dV −

V

ˆ ) dS. E(v · n S

I

Z ˆ ) dS = µ0 E(v · n

S

Z 1 d E dV, c2 dt V Z 1 d + JP ] dV + 2 E dV. c dt V

[J + JM + JP ] dV + V

 I  Z 1 ˆ )E dS = µ0 [J + JM (ˆ n × B) + 2 (v · n c S V

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 S

d ∂B · dS = ∂t dt

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 (ˆ n × E) dS = −

S

Z Jm dV −

V

V

∂B dV. ∂t

(B)

Use (2.132): Z

d ∂B dV = ∂t dt

V

Z

I B dV −

V

ˆ ) dS. B(v · n S

So (B) becomes I

Z (ˆ n × E) dS = −

S

or

V

Z

d Jm dV − dt

I

ˆ ) dS B(v · n

B dV + V

Z ˆ )B] dS = − [ˆ n × E − (v · n

S

I

V

S

d Jm dV − dt

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 ˆ 1 × H1 dS + ˆ 2 × H2 dS + ˆ 3 × H dS = n n n dV. S1 S2 S3 V ∂t ˆ 1 = −ˆ Choose δ = k∆ so that most of the source lies within V . As ∆ → 0, we have S1 = S2 , n n2 = ˆ 12 , and V → 0. Thus n Z ∂D dV → 0 V ∂t and

Z ˆ 12 × (H1 − H2 ) dS = 0 n S1

or ˆ 12 × (H1 − H2 ) = 0. n Next integrate (2.141) over V : Z Z ˆ 1 × E1 dS + n S1

 Z  ∂B ˆ 2 × E2 dS + ˆ 3 × E dS = n n −Jm − dV. ∂t S2 S3 V Z

15

ˆ 1 = −ˆ Choose δ = k∆ so that most of the source lies within V . As ∆ → 0, we have S1 = S2 , n n2 = ˆ 12 , and V → 0. So n Z Z ˆ 12 × (E1 − E2 ) dS = − Jm dV n V

S1

Z

Z

δ/2

=−

Jm dS dx S1 −δ/2 Z δ/2

=−

Z Jms dS.

f (x, ∆) dx −δ/2

But Z lim

δ/2

∆→0 −δ/2

so

S1

f (x, ∆) dx = 1,

Z

Z ˆ 12 × (E1 − E2 ) dS = − n

Jms dS

S1

S1

or ˆ 12 × (E1 − E2 ) = −Jms . n Next integrate (2.143) over V : Z Z B1 · dS + S1

Z

Z

B2 · dS +

S2

B · dS =

ρm dV.

S3

V

ˆ 1 = −ˆ ˆ 12 . Therefore As ∆ → 0, we have S1 = S2 , n n2 = n Z Z ˆ 12 dS = (B1 − B2 ) · n 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 ˆ 12 = ρms . (B1 − B2 ) · n Integration of (2.144) with ρ = 0 gives ˆ 12 = 0. (D1 − D2 ) · n 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 ˆ )D] dS + ˆ )D] dS = [ˆ n × H − (v · n [ˆ n × H + (v · n J dV + D dV. (A) dt V1 S1 S10 V1 16

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

d J dV + dt V2

S20

S2

Z D dV.

(B)

V2

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

Z Z d ˆ 10 )D] dS D dV − [ˆ n10 × H + (v · n J dV − dt V1 +V2 S10 V1 +V2 Z ˆ 20 )D] dS = 0. − [ˆ n20 × H + (v · n

S1 +S2

S20

ˆ 10 = −ˆ ˆ 12 . Then If δ is very small, then S1 + S2 = S, V1 + V2 = V , and n n20 = n Z Z Z Z Z d ˆ ˆ 12 )(D1 −D2 ) dS. ˆ [ˆ n ×H+(v· n)D] dS − J dV − (v· n D dV = n12 ×(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 ˆ )D] dS − [ˆ n × H + (v · n J dV − D dV = Js dS. (D) dt V S V S10 Consistency between (C) and (D) requires that ˆ 12 × (H1 − H2 ) + (v · n ˆ 12 )(D1 − D2 ) = Js . n 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):   ∂¯ ∂ ¯ ¯ · H − Jm , ∇+ ζ ·E=− µ ∂t ∂t

  ∂¯ ∂ ¯ ∇ − ξ · H = ¯ · E + J. ∂t ∂t

Define a new dyadic ¯−1 by ¯−1 · ¯ = ¯ · ¯−1 = ¯I. Then (2.252) becomes   ∂E ∂¯ −1 ¯ = ¯ · ∇ − ξ · H − ¯−1 · J. ∂t ∂t Next, differentiate (2.252) with respect to time:   ∂¯ ∂E ∂2 ∂Jm ¯ ¯ · H) − ∇+ ζ · = − 2 (µ . ∂t ∂t ∂t ∂t Lastly, substitute (*) to get        ∂¯ ∂¯ ∂2 ∂¯ ∂Jm −1 ¯ ¯ ¯ ¯ · H = ∇ + ξ · ¯−1 · J − ∇ + ζ · ¯ · ∇ − ξ + 2 µ . ∂t ∂t ∂t ∂t ∂t ¯ · H and D = ¯ · E, so ξ¯ = ζ¯ = 0. Then we get For an anisotropic medium we have B = µ   ∂2 −1 ¯ ¯ ¯ · ¯−1 · J − ∂Jm . ¯ ·H=∇ ∇ · ¯ · ∇ + 2 µ ∂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 ˆ ∇ E − µ 2 + ∇ E · z = 0, ∂t  ∂z

ˆ ∇ = x

∂ . ∂z

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

Now use 

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

∇2 H − µ

∂ 2 H 1 ∂ ˆ × (∇ × H) = 0. + z ∂t2  ∂z

ˆ Ey (z, t) and H(r, t) = x ˆ Hx (z, t) then If E(r, t) = y   ∂Hx ∂Hx ˆ × (∇ × H) = z ˆ× y ˆ z = −ˆ x , ∂z ∂z

∇Ez = 0, and

ˆ ∇2 E = y

∂ 2 Ey , ∂z 2

ˆ ∇2 H = x

∂ 2 Hx . ∂z 2

With these the wave equations become ∂ 2 Ey ∂ 2 Ey ˆ − y µ = 0, ∂z 2 ∂t2 ∂ 2 Hx 1 ∂ ∂Hx ∂ 2 Hx ˆ ˆ ˆ x − x µ −x = 0, ∂z 2 ∂t2  ∂z ∂z ˆ y

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 = 0. The equivalent sources are

∇ × (∇ × E) + µ when D = E, B = µH, and Jm 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  = 810 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, ˆ dS = D·n 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) = ˆr

Q U (r − vt). 4π0 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) = − ˆr 0 0 ∂r   1 ∂ Q vt − r = − ˆr U (r − vt) 0 ∂r 4πr vt       Q ∂ vt − r vt − r ˆr U (r − vt) =− + δ(r − vt) 4π0 ∂r vtr vtr    Q 1 −r − vt + r ˆr =− U (r − vt) 4π0 vt r2 Q = ˆr U (r − vt) 4π0 r2

E(r, t) = −

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

Q Q δ(r − vt)[vˆr] = ˆr δ(r − vt). 4π(vt)2 4πvt2

22

Compare this to (2.298): 

 ∂ψ ∂ ∂Er J=∇ = (∇ψ) = −ˆr0 ∂t ∂t ∂t   Q ∂ U (r − vt) = −ˆr0 ∂t 4π0 r2 Q ∂ U (r − vt) = −ˆr 4πr2 ∂t Q = ˆr vδ(r − vt) 4πr2 Q = ˆr vδ(r − vt) 4π(vt)2 Q = ˆr 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 ∂ ∂ ˆ ˆ ∇×E=z − =z [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 ˆE0 cos = ∇ × E = −ˆ xE0 β sin x sin(ωt − βz) + z x cos(ωt − βz), ∂t a a a

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

π  π  E0 E0 π ˆ β sin x cos(ωt − βz) − z 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) ˆ 300 x2 y 2 e−αt U (t). D = E = y

23

(b) Z

t

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

χe (r, t − t0 )E(r, t0 ) dt0

−∞ 2 −αt

ˆ 100 x e =y

Z

t

0

ˆ 300 x2 y 2 U (t − t0 )U (t0 )e−αt dt0 y

U (t) +

ˆ 100 x2 e−αt U (t) + =y

−∞ t

Z

0

ˆ 300 x2 y 2 U (t)e−αt dt0 y 0   e−αt 1 2 −αt 2 2 ˆ 100 x e U (t) + y ˆ 300 x y U (t) =y . − α α

(c) D = ¯ · E ˆx ˆ + 20 x ˆy ˆ + 40 z ˆy ˆ ) · [ˆ = (30 x y10x2 e−αt U (t)] ˆ 200 x2 e−αt U (t) + z ˆ400 x2 e−αt U (t). =x (d) D = (E + β∇ × E) ˆ200xe−αt U (t)] = 70 [ˆ y10x2 e−αt U (t) + z ˆ20]. = 700 xe−αt U (t) [ˆ yx + z 2.27. I

I

∗

Z

J · dS =

2π

Z

π

ˆ dS = (J − ρv) · n

S

S

0

−ρ0 v0 ˆr · ˆrR2 (t) sin θ dθ dφ

0

= −4πρ0 v0 R2 (t) = −4πρ0 v03 t2 . Z

Z ρ dV =

V

∴

2π

Z

π

Z

R(t)

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 = −ˆr · ˆr ρP sb = ˆr · ˆr

31.87 × 10−12 = −7.97 × 10−8 (2 × 10−2 )2 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) ρP = −∇ · P = −

1 ∂ 2 1 ∂ (r Pr ) = − 2 (31.87 × 10−12 ) = 0. r2 ∂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) ˆ ∂ φ ∂B ˆ β E0 sin(ωt − βr), = (rEθ ) = φ ∇×E=− ∂t r ∂r r Z E β E 0 ˆ β 0 sin(ωt − βr) dt = φ ˆ B=φ cos(ωt − βr), r ω r H(r, t) =

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

(b) I ˆ dS [E(r, t) × H(r, t)] · n  Z 2π Z π  ˆ E0 cos(ωt − βr) × φ ˆ β E0 cos(ωt − βr) · ˆr r2 sin θ dθ dφ = θ r ωµ0 r 0 0 β = 4πE02 cos2 (ωt − βr). ωµ0

Psphere (t) =

2.33. (a) ∂E ∂Hz = −ˆ y = −ˆ yβH0 sin(ωt − βx), ∂t ∂x β ˆ E(r, t) = y H0 cos(ωt − βx). ω0

∇ × H = 0

26

(b) ∇ × E = −µ0 so ˆ H=z

∂Ey ∂H β2 ˆ ˆ sin(ωt − βx), =z =z ∂t ∂x ω0

β2 ˆH0 cos(ωt − βx) H0 cos(ωt − βx) = z 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 ˆ dx + ˆ cos ωt · x ˆ dx. ˆ cos(ωt − βb) · x x x x v0 t+a x v0 t Z Z v0 t+a Z b ∂B ˆ µ0 ω sin(ωt − βz) · φ ˆ dz dx. − · dS = − −φ ∂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) ˆ v0 v×B=z I

µ0 cos(ωt − βz). ηρ

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 Z

b



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

Z

v0 t+a Z b

ˆ dz dx ˆ µ0 cos(ωt − βz) · φ φ x=v0 t z=0 ηx   µ0 v0 t + a =− [sin(ωt − βb) − sin ωt]. ln ηβ v0 t

B · dS =

d − dt

Z

  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. ˆ E(r, θ, t) = θ

1 A(t − r/v), r sin θ

ˆ 1 1 A(t − r/v). H(r, θ, t) = φ η r sin θ

(a) Z

π−θ0

V (r, t) = θ0

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

(b)   1 1 1 1 ˆ ˆ ˆ ×H=θ× φ A(t − r/v) = ˆr A(t − r/v), Js = n η r sin θ η r sin θ Z 2π 1 1 2π ˆr I(r, t) = A(t − r/v) · ˆrr 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 ˆx + y ˆy = ρ ˆ ρ and r0 = z ˆz 0 , hence the displacement vector is R = r − r0 = ρ ˆρ − z ˆz 0 We have r = x 2 1/2 2 0 with |R| = (ρ + z ) . 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 ˆ ˆ E(r) = − z ρl (z 0 ) dl = ρ ρ 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) = −ˆ z = −ˆ z ·2 . 2 3/2 2 0 4π 2πρ (ρ + z ) 0 3.2. The ring is shown in Figure 4 along with source and field point position vectors r0 and r. ˆz and r0 = ρ ˆ0a = x ˆ a cos φ0 + y ˆ a sin φ0 . Hence We have r = z ˆz − x ˆ a cos φ0 − y ˆ a sin φ0 R = r − r0 = z 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 ˆ a cos φ0 − y ˆ a sin φ0 ˆz − x 1 z ρl (φ0 ) a dφ0 = 3/2 2 2 4π 0 (z + a )  Z 2π  Z 2π 2 a 0 0 0 0 0 0 ˆ ˆ =− x ρ (φ ) cos φ dφ + y ρ (φ ) sin φ dφ l l 4π(z 2 + a2 )3/2 0 0 Z 2π az ˆ +z ρl (φ0 ) dφ0 . 4π(z 2 + a2 )3/2 0 29

Figure 4 When ρl (φ) = ρ0 sin φ, we obtain E = −ˆ y When ρl (φ) = ρ0 cos2 φ, we get ˆ E=z

a2 ρ0 . 4(z 2 + a2 )3/2 azρ0 . + a2 )3/2

4(z 2

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

Figure 5 ˆz and r0 = ρ ˆ 0 ρ0 = x ˆ ρ0 cos φ0 + y ˆ ρ0 sin φ0 , we obtain Writing r = z ˆz − x ˆ ρ0 cos φ0 − y ˆ ρ0 sin φ0 R = r − r0 = z 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 ˆz − x ˆ ρ0 cos φ0 − y ˆ ρ0 sin φ0 0 0 0 1 1 0 R 0 0 0 z E= dS = ρs (r ) ρ (ρ , φ ) ρ dφ dρ s 4π S |R|3 4π 0 (z 2 + ρ0 2 )3/2 0 30

so that 1 Ex = − 4π

Z 0

∞ Z 2π 0 ∞ Z 2π

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

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 ˆz and Writing r = z ˆ a sin θ0 cos φ0 + y ˆ a sin θ0 sin φ0 + z ˆa cos θ0 , r0 = ˆr0 r0 = x we obtain ˆ(z − a cos θ0 ) − x ˆ a sin θ0 cos φ0 − y ˆ a sin θ0 sin φ0 R = r − r0 = z 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 ˆ(z − a cos θ0 ) − x ˆ a sin θ0 cos φ0 − y ˆ a sin θ0 sin φ0 2 z 1 ρs (θ0 ) a sin θ0 dθ0 dφ0 , = 2 + a2 − 2az cos θ 0 )3/2 4π 0 (z 0 hence Z Ex = Ey = 0

since 0

and

a2 Ez = 2

Z 0

π

2π



 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 = −√ 2z 2 a2 + z 2 + 2az a2 + z 2 − 2az " # a2 ρ0 a+z a−z p = −p 2z 2 (a + z)2 (a − z)2   a2 ρ0 a + z a−z = − 2z 2 |a + z| |a − z| a2 ρ0 = [sgn(a + z) − sgn(a − z)] 2z 2   0, |z| < a,     a2 ρ 0 ˆ 2 , z > a, = z z    a2 ρ0  −ˆ z 2 , z < a. z 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) = 2z 2 a2 + z 2  2   a ρ0 a   √ ˆ , |z| < a, z 2 1 − z a2 + z 2 = 3 a ρ0   −ˆ z √ , |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 = (ˆ n × 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 (ˆ n × E) dS = 0, (ˆ n × E) dS + S10 S1 Z Z (ˆ n × E) dS = 0. (ˆ n × E) dS + S20

S2

Add to obtain Z

Z

Z

(ˆ n20 × E2 ) dS = 0.

(ˆ n10 × E1 ) dS −

(ˆ n × E) dS −

S20

S10

S1 +S2

ˆ 10 = −ˆ ˆ 12 , S1 + S2 = S, S10 = S20 . So Now let δ → 0 so that n n20 = n Z I [ˆ n12 × (E1 − E2 )] dS. (ˆ n × E) dS =

(*)

S10

S

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

(**)

S10

ˆ 12 × (E1 − E2 ) = 0. To have (**) give the same result as (*), we must have n (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 ˆ dS = D·n

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 ˆ dS + ˆ dS = 0, D·n D·n ZS1 ZS10 ˆ dS + ˆ dS = 0. D·n D·n S2

Add to obtain

S20

Z

Z ˆ dS − D·n

S1 +S2

Z ˆ 10 dS − D1 · n

S10

ˆ 20 dS = 0. D2 · n S20

ˆ 10 = −ˆ ˆ 12 , S1 + S2 = S, S10 = S20 . So Now let δ → 0 so that n n20 = n I Z ˆ dS = D·n [ˆ n12 × (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 ˆ dS = D·n ρs dS. (**) S

S10

ˆ 12 · (D1 − D2 ) = ρs . To have (**) give the same result as (*), we must have n 33

Figure 7 3.6. The geometry appears in Figure 7. We have ρsp

ˆ · P =n

z=0

Z Φ1 (r) = V

∂Φ ˆ · (D − 0 E) z=0 = ( − 0 )ˆ n · E z=0 = −( − 0 ) =n . ∂n z=0

ρ(r0 ) G1 (r|r ) dV 0 = 1 0

Z

G1 (r|r0 )

V

Qδ(z 0 − h)δ(x0 )δ(y 0 ) 0 0 0 dx dy dz 1

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 −∞ ˆ ky ) · (ˆ ˆy + z ˆz) = kx x + ky y kρ · r = (ˆ xkx + y xx + 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

∞

e−kρ (h−z) +

1 −2 −kρ (h+z) 1 +2 e

2kρ

0

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

2π−φ

e

jkρ ρ cos x

Z

So Q 1 Φ1 = 1 2π

Z

∞

π

dx =

−φ

0

ejkρ ρ cos(ξ−φ) kρ dkρ dξ

ejkρ ρ cos x dx = 2πJ0 (kρ ρ)

−π

e−kρ (h−z) +

1 −2 −kρ (h+z) 1 +2 e

2

0

J0 (kρ ρ) dkρ

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

Z

∞

kρ e−kρ (h−z) +

1 −2 −kρ (h+z) 1 +2 kρ e

2

0

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 22 = 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, ∞

22 e−kρ (h−z) j(kx x+ky y) e dkx dky 2kρ −∞ 1 + 2 Z 2π Z ∞ −kρ (h−z) Q 1 22 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 ˆ=z ˆ =⇒ n = = 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

ZZ

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 0

p

ZZ

ρ(r0 ) dV 0  0

∞ −∞

e−kρ |z−z | jkρ ·(r−r0 ) 2 e d kρ . 2kρ

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

∞

ZZ

i h 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

Φ

z=d

1 = 0 =⇒ (2π)2 kρ d

=⇒ Ae

ZZ

∞

( Aekρ d + Be−kρ d +

−∞

+ Be

−kρ d

Z V

Z + V

0

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

(*)

) ejkρ ·r d2 kρ = 0

0

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

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 ¯ ρ )ekρ z + B(k ¯ ρ )e−kρ z ejkρ ·r d2 kρ , Φs2 = A(k (2π)2 −∞ ZZ ∞ h i s ∂Φ2 1 ¯ ρ )ekρ z − B(k ¯ ρ )e−kρ z ejkρ ·r kρ d2 kρ . = A(k ∂z (2π)2 −∞ Φs1

Boundary condition (c) gives ¯ ρ ) + B(k ¯ ρ ) = 0, A(k

¯ ρ ) = −A(k ¯ ρ ). so B(k 39

Therefore Φs2 = ∂Φs2 ∂z

=

1 (2π)2

ZZ

∞

1 (2π)2

ZZ

−∞ ∞

Φp1

¯ ρ )2 sinh kρ z ejkρ ·r d2 kρ , A(k ¯ ρ )2 cosh kρ z ejkρ ·r kρ d2 kρ . A(k

−∞

Boundary condition (a) reads + = Φs2 at z = d. # Z Z ∞ −kρ (z 0 −d) Z " ρ(r0 ) e 1 0 jkρ ·(r−r ) 2 e d k dV 0 ρ 2 2kρ 1 −∞ V1 (2π) ZZ ∞ ZZ ∞ 1 1 −kρ d jkρ ·r 2 ¯ ρ )2 sinh kρ d ejkρ ·r d2 kρ , + B(kρ )e e d kρ = A(k 2 (2π)2 (2π) −∞ −∞ hence

Φs1

0

Z V1

e−kρ z ekρ d −jkρ ·r0 ρ(r0 ) ¯ ρ ) sinh kρ d. e dV 0 + B(kρ )e−kρ d = 2A(k 2kρ 1

Boundary condition (b):

∂Φp1 ∂Φs1 2 ∂Φs2 + = , ∂z ∂z 1 ∂z 0

Z V1

e−kρ z ekρ d −jkρ ·r0 ρ(r0 ) 2 ¯ e dV 0 − B(kρ )e−kρ d = 2A(k ρ ) cosh kρ d. 2kρ 1 1

Calling

0

Z F = V1

e−kρ z ekρ d −jkρ ·r0 ρ(r0 ) e dV 0 2kρ 1

we have so far ¯ ρ ) sinh kρ d, F + B(kρ )e−kρ d = 2A(k 2 ¯ F − B(kρ )e−kρ d = 2A(k ρ ) cosh kρ d. 1 Adding (*) and (**), we obtain ¯ ρ ) sinh kρ d + 2 2A(k ¯ ρ ) cosh kρ d 2F = 2A(k 1 or

¯ ρ ) = 1 F A(k S Subtracting instead, we obtain −kρ d

2B(kρ )e

where

S = 1 sinh kρ d + 2 cosh kρ d.

  2 ¯ ρ ) sinh kρ d − cosh kρ d , = 2A(k 1

¯ ρ )T 1 B(kρ )e−kρ d = A(k



where

T = 1 sinh kρ d − 2 cosh kρ d,

¯ ρ ) = A(k ¯ ρ )ekρ d T = 1 F ekρ d T = F ekρ d T . B(k 1 S 1 S 40

(*) (**)

So Z Φ1 = V1

ρ(r0 ) 1 1 (2π)2

ZZ

∞

(

−∞

0

0

e−kρ |z−z | e−kρ z ekρ d kρ d −kρ z T + e e 2kρ 2kρ S

) 0

ejkρ ·(r−r ) d2 kρ dV 0 .

This implies 1 G1 (r|r ) = (2π)2 0

Also Z Φ2 = V1

ZZ

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

ZZ

1 (2π)2

ZZ

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

−∞

so that G2 (r|r0 ) =

0

∞

ρ(r0 ) 1 1 (2π)2

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

Φp2

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 −∞ 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

ZZ

∞

  ¯ ρ ) + B(k ¯ ρ ) ejkρ ·r d2 kρ + A(k

−∞

Z

"

V2

Let Z FA = V2

1 (2π)2

0

ZZ

∞ −∞

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

e−kρ z −jkρ ·r0 ρ(r0 ) e dV 0 . 2kρ 2 41

Then

1 (2π)2

ZZ

∞

  ¯ ρ ) + B(k ¯ ρ ) + FA ejkρ ·r d2 kρ = 0 A(k

−∞

so ¯ ρ ) + B(k ¯ ρ ) = −FA . A(k 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 = A(k )e + B(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) −∞ ¯ ρ )e−kρ d + e−kρ d FB ¯ ρ )ekρ d + B(k B(kρ )e−kρ d = A(k where Z FB = V2

0

ekρ z −jkρ ·r0 ρ(r0 ) e dV 0 2kρ 2

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

i ∞ h ¯ ρ )ekρ d − B(k ¯ ρ )e−kρ d ejkρ ·r kρ d2 kρ B(kρ )e e A(k −∞ −∞ # 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

ZZ

i h ¯ ρ )ekρ d − B(k ¯ ρ )e−kρ d kρ − 2 e−kρ d FB kρ −1 B(kρ )e−kρ d kρ = 2 A(k We have so far: ¯ = −FA A¯ + B ¯ kρ d + Be ¯ −kρ d + e−kρ d FB Be−kρ d = Ae 1 −kρ d ¯ kρ d + Be ¯ −kρ d + e−kρ d FB Be = −Ae 2 Subtracting (***) from (**), we obtain Be

−kρ d

  1 ¯ kρ d 1− = 2Ae 2

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

−kρ d

  1 ¯ −kρ d + 2FB e−kρ d 1+ = 2Be 2 42

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

Divide to get 1+ 1−

1 2 1 2

¯ −kρ d + 2FB e−kρ d 2Be ¯ kρ d 2Ae

=

or ¯ kρ d Q = Be ¯ −kρ d + FB e−kρ d Ae

where

Q=

2 + 1 . 2 − 1

(**’)

So far then, ¯ = −FA A¯ + B ¯ 2kρ d − B ¯ = FB AQe

(*) (**’)

Adding (*) and (**’), we get ¯ + Qe2Kρ d ) = −FA + BB A(1   2 + 1 kρ d −kρ d ¯ = [−FA + FB ]e−kρ d A 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 ¯ A[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 A¯ = e dV 0 S 2k  ρ 2 V2 ¯ = −FA , A¯ + B ¯ −A¯ + B

1 1 −2kρ d e = − FB e−2kρ d . Q Q

Adding (*) and (**”) we get     ¯ 1 + 1 e−2kρ d = − FA + 1 FB e−2kρ d , B Q Q and substitution leads to ¯=− 1 B 2kρ

Z V2

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

Now use Be

  1 ¯ kρ d 1− = 2Ae 2

−kρ d

to get B= or B= So Z Φ1 = V2

22 ekρ d S

ρ(r0 ) 1 2 (2π)2

ZZ

2 e2kρ d 2A¯ 2 − 1

Z V2 ∞

−∞



sinh kρ z 0 jkρ ·r0 ρ(r0 ) e dV 0 . 2kρ 2  22 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

∞

−∞

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

 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 ) =

∞

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

ZZ



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

  q ˆr · (r − r0 )  a  ˆr · (r − r0i ) ˆ · D = ˆr · E r=a = − . ρs = n 4π |r − r0 |3 r0 |r − r0i |3

Here r0 = ˆr0 r0 , so that

r = ˆra,

r0i = ˆr0 ri0 ,

  r0 ˆr · (r − r0 ) = ˆr · ˆra − a ˆr0 = a(1 − k 0 ˆr · ˆr0 ), a

k0 =

r0 a

and |r − r0 |3 = a3 |ˆr − k 0 ˆr0 |3 . Let γ be the angle between ˆr and ˆr0 . Then 2

|ˆr − k 0 ˆr0 |3 = (1 + k 0 − 2k 0 cos γ)3/2 , Hence q ρs (r) = ρs (γ) = 4πa2

"

ˆr · (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 0 0 1 − k 0 cos γ k 0 − cos γ = pk + 1 − pk − 1 = 1 − 1 = 0 p = (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  dV 0 . Φ(r) = − 2 0  4π |r − r | a r − r0 V r2

To check this, evaluate the potential at r = a:   Z 0 a/a  ρ(r ) 1  1 dV 0 = 0. − 2 Φ(r = aˆr) = 0  4π |r − r | a r − r0 V a2 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 E0 a

π

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 ˆ 1 ∂Φs Es (r) = −∇Φs (r) = −ˆr −θ ∂r r ∂θ   ∂ −2 ˆ 1 ∂ 3 cos θ = −E0 a ˆr r + θ 3 ∂r r ∂θ  a 3 ˆ sin θ]. = E0 [2ˆr cos θ + 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 = ˆr · D r=a  a 3 ˆ sin θ)] (2ˆr cos θ + θ = ˆr · [ˆ zE0 + E0 r=a r ˆ + 2 cos θ] = E0 [ˆr · z = 3E0 cos θ. Total surface charge: Z

2π

Z

π

3E0 cos θa2 sin θ dθ dφ 0 0 Z π 2 = 6πE0 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 − 2r 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 = −2r B1 a−3 − r E0 .

(6)

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

1 − r . 1 + 2r

1 − r A1 = E0 − E0 = −E0 1 + 2r

So Φs (r) =

E0 a3 1 − r cos θ, r2 1 + 2r



3r 1 + 2r

Φ1 (r) = −E0 r

 .

3r cos θ. 1 + 2r

Inside the cavity, E1 = −∇Φ1 = E0

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

ˆ. Is the field inside the cavity greater than the impressed field? The since ∇(r cos θ) = ∇z = z inequality 3r E0 > E0 1 + 2r 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 4π0 |r − rQ | 4π0 |r − rQi |3

Here p

r − rQ = xˆ x + yˆ y + (z − d)ˆ z = ρˆ ρ + (z − d)ˆ z,

|r − rQ | =

r − rQi = xˆ x + yˆ y + (z + d)ˆ z = ρˆ ρ + (z + d)ˆ z,

|r − rQi | =

ρ2 + (z − d)2 ,

and so that

p ρ2 + (z + d)2 ,

  Q ρˆ ρ + (z − d)ˆ z ρˆ ρ + (z + d)ˆ z E(r) = − . 4π0 [ρ2 + (z − d)2 ]3/2 [ρ2 + (z + d)2 ]3/2

The induced surface charge is given by 2dQ 1 ˆ · E z=0 = − . ρs = 0 z 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=

Z

2π

Z

S

∞

−

ρs dS = 0

0

2dQ 1 4π (ρ2 + d2 )3/2

 ρ dρ dφ = −

50

2dQ (2π) 4π

Z 0

∞

ρ dρ = −Q. (ρ2 + d2 )3/2

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

2d Q . 2 4π0 (ρ + d2 )3/2

For the force computation, refer to Figure 16. The stress tensor is given by ¯ e = 1 (D · E)¯I − DE. T 2 We have     1 1 1 2 1 Q 2 4d2 ¯ ¯ ˆ · Te = (D · E)ˆ ˆ · DE = 0 z ˆ E · E − Ez Ez = −0 z ˆ Ez = −0 z ˆ n z·I−z . 2 2 2 2 4π0 (ρ2 + d2 )3 The force can be calculated using Z

1 1 x dx =− 2 3 2 +a ) 4 (x + a2 )2

(x2 to get Q2 d2 ¯ e dS = z ˆ 2 ˆ ·T F=− n 8π 0 S Z

Z 0

2π

Z 0

∞

ρ dρ Q2 d2 ˆ dφ = z (ρ2 + d2 )3 4π0

Z 0

∞

ρ dρ Q2 ˆ = z . (ρ2 + d2 )3 4π0 (2d)2

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

Q2 4π0 (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) ¯ = Q

Z

2 (3r0 r0 − r0 ¯I)ρ(r0 ) dV 0

V

= 6qr0 r0 − 2qr02¯I − 3q(r0 + d)(r0 + d) + q(r0 + d) · (r0 + d)¯I − 3q(r0 − d)(r0 − d) + q(r0 − d) · (r0 − d)¯I = q[6r0 r0 − 3r0 r0 − 3r0 d − 3dr0 − 3dd − 3r0 r0 + 3r0 d + 3dr0 − 3dd] + q[−2r2 + r2 + 2d · r0 + d2 + r2 − 2d · r0 + d2 ]¯I 0

0 2¯

0

= −6qdd + qd I

Note that the first nonzero moment is independent of r0 as expected. To three terms, the potential

52

expansion is ¯ · ˆr ˆr · p Q 1 ˆr · Q + + 4π0 r 4π0 r2 2 4π0 r3 ¯ · ˆr 1 ˆr · Q =0+0+ 2 4π0 r3   1 ˆr · −6qdd + qd2¯I · ˆr = 3 8π0 r   q = −6(ˆr · d)(d · ˆr) + d2 ˆr · ¯I · r 3 8π0 r   q 2 2 −6(ˆ r · d) + d = 8π0 r3 i qd2 h ˆ 2 . = 1 − 6(ˆ r · d) 8π0 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

Q n cos 2θ0 r0 πa3

But 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 In =

π

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 +

  1 4

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

=⇒



so that 4 In = 3

Z 0

π

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

0

0

0

53

Z 0

π

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 q00 =

and

r q20 =

8 , 15

1 2Q 4π 3

5 2Q 2 a 4π 5

and In = 0 for other n. 



2 − 3

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 2n+1

n X

# qnm Ynm (θ, φ)

m=−n

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 + 4π0 9 r 4π0 8 · 75 r3 4 4      1 2Q 1 2 3 1 2 = − + Qa cos 2θ + 4π0 r 9 4π0 r3 15 4 4 2 2 Q 1 Qa =− + (3 cos 2θ + 1). 9 4π0 r 30 4π0 r3 =

3.17. The ring of current is shown in Figure 18. We have µ0 I B(r) = 4π Beginning with r = xˆ x + yˆ y + zˆ z and

r0

=

ˆ x0 x

+

I

dl0 ×

Γ ˆ, y0y

ˆ R . R2

we write

R = (x − x0 )ˆ x + (y − y 0 )ˆ y + zˆ z. 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 ˆ0 × R dl0 × R = a dφ0 φ ˆ cos φ0 ) × [ˆ ˆ (y − y 0 )z + z ˆz 0 ] = a dφ0 (−ˆ x sin φ0 + y x(x − x0 ) + y ˆ z sin φ0 − z ˆ(x − x0 ) cos φ0 + x ˆ z cos φ0 ] = a dφ0 [−ˆ z(y − y 0 ) sin φ0 + y 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 0

2π

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].

Bz = = = Write

µ0 I a 4π

Z

µ0 I a π

Z

π/2

−π/2 π/2

0

µ0 I 1 a 3/2 π F

a + ρ[1 − 2 sin2 x] 2 dx [z 2 + ρ2 + a2 + 2aρ(1 − 2 sin2 x)]3/2

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

(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. 2ρ k2

B=

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, ˆ · a dφ0 [−ˆ ˆ z sin φ0 − z ˆ(x − x0 ) cos φ0 + x ˆ z cos φ0 ] [dl0 × R]φ = φ z(y − y 0 ) sin φ0 + y = 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 0

2π

[z 2

+

ρ2

sin u du = 0. + a2 − 2aρ cos u]3/2

56

ˆ · a dφ0 [−ˆ ˆ z sin φ0 − z ˆ(x − x0 ) cos φ0 + x ˆ z cos φ0 ] [dl0 × R]ρ = ρ z(y − y 0 ) sin φ0 + y = a dφ0 [z cos φ0 cos φ + z sin φ0 sin φ] = a dφ0 z cos(φ − φ0 ). Therefore µ0 I az Bρ = 4π

2π

Z 0

[z 2

+

ρ2

cos u du . + a2 − 2aρ cos u]3/2

Change of variables: u = π − 2x =⇒ cos u = −[1 − 2 sin2 x], µ0 I Bρ = − az 4π

Z

−π/2

µ0 I az =− π F 3/2 Write

π/2

Z

1 − 2 sin2 x (2 dx) [z 2 + ρ2 + a2 + 2aρ − 4aρ sin2 x]3/2

π/2

0

1 − 2 sin2 x dx. [1 − k 2 sin2 x]3/2

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 k2 − 2 = k2 − 1

4aρ F 4aρ F

−2 −1

=

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

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 ˆ 1 − k K(k ) − E(k ) . =φ 2π ρ 2

ˆ µ0 I A(r) = φ πk

k=

2(aρ)1/2 F 1/2

∂Aφ 1 ∂ ˆ φ (ρ, z)] = −ˆ ˆ B = ∇ × A = ∇ × [φA ρ +z (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

k2 =

and

     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 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 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)ˆ z + Bρ (ρ, z)ˆ ρ.

60

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

F2 = Γ2

where ˆ × [Bz (a2 , d)ˆ dl2 × B = a2 dφφ z + Bρ (a2 , d)ˆ ρ] = a2 dφˆ ρBz (a2 , d) − a2 dφˆ zBρ (a2 , d). So Z

2π

Z ˆ sin φ] dφ − Bρ (a2 , d)a2 [ˆ x cos φ + y

F2 = Bz (a2 , d)a2

= −ˆ zBρ (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 ˆ 2 =⇒ B = J=z πa Maxwell’s stress tensor:

ˆ dφ z 0

0

( ˆ µ0 Iρ2 , ρ ≤ a, φ 2πa ˆ µ0 I , ρ ≥ a. φ 2πρ

¯ m = 1 (B · H)¯I − BH. T 2

Magnetic force: Z Fm = −

¯ m · dS. T

S

Thus, the force on the half of the wire occupying y > 0 is Z Z ¯m · y ˆ dx dz. Fm = T x

z

61

2π

We have

   2 2 ˆ 0 Iρ 2 cos φ, ρ ≤ a,  µ0 Iρ 2 y ˆ − φµ 2 2πa 2πa ¯m · y  2   ˆ= T  µ0 I 2 y ˆ 0 I ˆ − φµ cos φ ρ ≥ a. 2 2πρ 2πρ

ˆ =y ˆ cos φ holds. So Note: in the plane y = 0 the relation φ (
Ix 2 ˆ , x ≤ a, − µ20 2πa y 2 ¯ ˆ= Tm · y
µ0 I 2 ˆ , x ≥ a. − 2 2πx y Therefore Z Fm = −ˆ y

"Z

l

a

dy 0

−a

µ0 2



I 2πa2

2

Z

2

−a

x dx + −∞

µ0 2



I 2π

2

dx + x2

Z a

∞

µ0 2



I 2π

2

# x3 a 1 −a 1 ∞ − − 3a4 −a x −∞ x a    µ0 I 2 2 1 1 = −ˆ y + + 8π 2 3a a a    µ0 I 2 8 µ0 I 2 = −ˆ y = −ˆ y 2 . 2 8π 3a 3π a

Fm = −ˆ y l



µ0 I 2 8π 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 ˆ · D = ρs . Using n ˆ = −ˆ Boundary condition at z = d is n z and ρs = Q/A, we obtain E = −∇Φ = −ˆ z

ˆ] = −ˆ z · [−C1 z or C1 = Therefore E=−

Q ˆ, z 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 . 2A

(b) 1 W = 2

Z V

1 D · E dV = 2

Z 0

dZ

Z 

y

x

   Q Q Q2 d ˆ · − z ˆ dx dy dz = − z . A A 2A

(c) dW d F = −ˆ z = −ˆ z dz dz



Q2 z 2A

 = −ˆ z

Q2 . 2A

(d) See Figure 20.

Figure 20

  1 ¯e · n ˆ= T (D · E)¯I − DE · (−ˆ z) 2 1 = − D · Eˆ z + DEz 2  2  2  Q Q ˆ+z ˆ =− z 2 A A 2 Q ˆ, = z 2A2 Z Z I Q2 Q2 ¯ ˆ ˆ. Fe = − Te · dS = − z dx dy = − z 2 2A S x y 2A 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 = −ˆ z = −ˆ z . 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 ˆ = −ˆ Note that on the top plate we have n z so that   V0 V0 ˆ · D = −ˆ ρs = n z · −ˆ z = , 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 ˆ · − z ˆ dx dy dz = − z 2 0 y x d d =

AV02 . 2d

(c) dW d F = −ˆ z = −ˆ z dz dz



AV02 2z

64



z=d

= −ˆ z

AV02 . 2d2

(d) Refer to Figure 20 again.   1 ¯ ¯ ˆ= Te · n (D · E)I − DE · (−ˆ z) 2 1 z + DEz = − D · Eˆ 2    2  V0 2 V0 ˆ+z ˆ =− z 2 d d 2 V ˆ, = 02 z 2d I Z Z AV02 V02 ¯ ˆ ˆ. z dx dy = − z Fe = − Te · 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 =

Z V

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

N N 1 XX

2

i=1

! Vi

n=1

N N 1 XX cmn Vn Vm . cin Vn Vi = 2 m=1 n=1

i=1 n=1

3.25. The flux ψm through loop m is Z

Z ˆ dS = B(r) · n

ψm =

N X

ˆ dS, Bn (r) · n

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

ˆ dS [∇ × An (r)] · n

Sm n=1 N XZ

ˆ dS [∇ × An (r)] · n

n=1 Sm N I X

An (r) · dl

n=1 Γm

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

µ0 4π

I

66

Γn

In dl0 |r − r0 |

to get ψm

N I X

 I In dl0 µ0 · dl = 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 1X Im W = 2 m=1

N X

! Lmn In

n=1

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 V

1 |δJ|2 dV ≥ 0, σ

as required. 67

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

∞ X 1 n x n

n=1

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). 4πc z l=0

Hence

∞

Φ(r, θ) =

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

r > c.

l=0

For r < c,

∞

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

r > c.

l=0

3.30. (a) ˆ D = ¯ · E = θ

80 20 + ˆr 2 C/m2 . 2 r r

(b) ˆ 80 C/m2 . P = D − 0 E = θ r2 (c) ρP = −∇ · P = 3.31.

1 E(r) = 4π0

Z V

80 cos θ C/m3 . r3 sin θ

  1 ρ(r ) −∇ dV 0 R 0



    Z Z 1 1 1 1 0 0 0 dV = − ρ(r )∇ · ∇ dV 0 ∇ · E(r) = − ∇· ρ(r )∇ 4π0 R 4π R 0   Z V Z V 1 1 1 0 2 0 =− ρ(r )∇ dV = − ρ(r0 )[−4πδ(r − r0 )] dV 0 4π0 V R 4π0 V 1 = ρ(r). 0 3.32. See Figure 21.

Figure 21

¯ = 1 (D · E)¯I − DE = T 2 69



V02 0 d2



 1¯ ˆz ˆ , I−z 2

  V02 0 1¯ V02 0 ˆ ˆ ˆ I − z z · z = −ˆ z , d2 2 2d2 Z V 2 0 ¯ ·n ˆ dS = −ˆ −F = T z 0 2 A, 2d S

¯ ·n ˆ= T



ˆ F=z

V02 0 A . 2d2

3.33. See Figure 22.

Figure 22

¯ e = 1 (D · E) ¯I − DE, T 2 1 ¯ e ˆE0 · 0 z ˆE0 ¯I − 0 E0 z ˆE0 z ˆ, T = z z=0 2   1 1 ¯ ¯ ˆ= ˆE0 · 0 z ˆE0 I − 0 E0 z ˆE0 z ˆ ·z ˆ = − 0 E02 z ˆ, Te z=0 · n z 2 2 I ¯ e ˆ dS = 50 E02 z ˆ N. Fe = − T ·n z=0 S

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

1 2π

Z

∞

−∞



Write the solutions as



∂2 ∂2 + ∂x2 ∂y 2



˜ y)ejkx dk = 0 Φ(k,

 ∂2 2 ˜ − k Φ(k, y) = 0. ∂y 2

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

Boundary condition on continuity of potential: ˜ 0+ ) = Φ(k, ˜ 0− ) =⇒ A(k) = B(k). Φ(x, 0+ ) = Φ(x, 0− ) =⇒ Φ(k, 70

Other boundary condition: ˆ · [1 E(x, 0+ ) − 2 E(x, 0− )] = ρ0 δ(x) y ∂Φ(x, y) ∂Φ(x, y) = 1 + ρ0 δ(x) =⇒ 2 ∂y y=0− ∂y y=0+ Z ∞ ˜ ˜ ∂Φ ∂Φ =⇒ 2 = 1 + ρδ(x)e−jkx dx. ∂y y=0− ∂y y=0+ −∞ So 2 |k|A(k) = 1 (−|k|)A(k) + ρ0 which gives A(k) = Therefore Φ(x, y) =

1 2π

Z

∞

−∞

ρ0 . |k|(1 + 2 )

ρ0 e∓|k|y ejkx dk |k|(1 + 2 )

for y ≷ 0.

3.35. ρ/ = ∇ · E # " # " e−r/a e−r/a ˆr + ∇ · ˆr =∇· 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 !   ˆr e−r/a e−r/a e−r/a = − 2 · ˆr + e−r/a 4πδ(r) + 2 − r a r a ra2 = 4πδ(r) −

e−r/a . ra2

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



e−ar r

71

e−r/a . ra2



+ ∇2

a 2

 e−ar .

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

  ˆr 1 ∇ = − 2, r r

and

∇(e

−ar

) = −ˆrae

−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

∇2 (e−ar ) = to find that

∇2 Φ =

a3 −ar e − 4πδ(r). 2

So ρ = 4πδ(r) − 

a3 −ar e . 2

3.37. Z 1 r − r0 E= ρ(r0 ) dV 0 4π0 V |r − r0 |3 Z ∞Z ∞Z a 1 (x − x0 )ˆ x + (y − y 0 )ˆ y + (z − z 0 )ˆ z = dx0 dy 0 dz 0 . ρ(z 0 ) 4π0 −∞ −∞ −a [(x − x0 )2 + (y − y 0 )2 + (z − z 0 )2 ]3/2 We have 1 Ex = 4π0

Z

∞

Z

a

0

Z

∞

ρ(z ) −∞

−a

−∞

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



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 4π0 −a [u −∞ Z a Z−∞ ∞ 2 1 = ρ(z 0 )(z − z 0 ) du dz 0 2 0 2 4π0 −a −∞ u + (z − z ) Z a 1 z − z0 = π ρ(z 0 ) dz 0 . 2π0 −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.  20 −a 20 z

3.38. Write

1 Φ(r) = 4π0

Z

η(r0 )

S

1 dS 0 |r − r0 |

where r0 = aˆr0 = aˆ x sin θ0 cos φ0 + aˆ y sin θ0 sin φ0 + aˆ z cos θ0 ,

r = zˆ z,

dS 0 = a2 sin θ0 dθ0 dφ0 .

So r − r0 = −aˆ x sin θ0 cos φ0 − aˆ y sin θ0 sin φ0 + (z − a cos θ0 )ˆ z, |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 4π0 0 0 [a + z − 2az cos θ ] Z π a2 η(θ0 ) sin θ0 dθ0 √ = . 20 0 a2 + z 2 − 2az cos θ0 3.39. J=∇×H     a −ρ/a ˆ = ∇ × H0 e φ ρ = ∇ × (ΦA)

where

Φ = H0 ae−ρ/a and A =

1ˆ φ. ρ

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

 ∇×

1ˆ φ ρ

 =

δ(ρ) ˆ. z ρ

(The latter identity can be verified by writing   Z 2π Z b  Z 2π 1ˆ 1ˆ ˆ ˆρ dρ dφ = ∇× φ ·z φ · φρ dφ = 2π ρ ρ 0 0 0 73

and

2π

Z 0

b

Z 0

δ(ρ) ˆ·z ˆρ dρ dφ = 2π, z ρ

where we have used Stokes’ theorem to obtain the first equality.) We have     1 −ρ/a δ(ρ) −ρ/a ˆ ˆ z J = ∇ H0 ae × φ + H0 ae ρ ρ 1 ˆ + H0 ae−ρ/a δ(ρ) z ˆ = −H0 e−ρ/a (ˆ ρ × φ) ρ ρ     a −ρ/a 1 ˆ δ(ρ) − = H0 e z . ρ a The presence of δ(ρ) indicates a line current along the z-axis. 3.40. Write ˆ = −∇Φpm , Hp = H0 z

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, ˆr · (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 0 cos θ − H0 r cos θ. r µ + 2µ0

So

3µ0 ˆ H0 z µ + 2µ0

H(r ≤ a) = and

 a 3 µ − µ  a 3 µ − µ 0 0 ˆ − H0 z ˆ H(r ≥ a) = −2H0 cos θˆr + H0 sin θθ r µ + 2µ0 r µ + 2µ0  a 3 µ − µ 0 ˆ − H0 z ˆ. (2 cos θˆr − sin θθ) = −H0 r µ + 2µ0 3.41. We have

Z

µ0 A= 4π

S

Js (r0 ) dS 0 |r − r0 |

where ˆ0, r0 = ρ0 ρ

r = zˆ z,

|r − r0 | =

q z 2 + ρ0 2 ,

and ˆ Js = η0 v = η0 ωρφ. So A(0, 0, z) = =

µ0 4π

Z

µ0 4π

Z

0

0

ˆ0 η ωρ0 φ p0 ρ0 dρ0 dφ0 2 2 0 0 z +ρ Z 2π η0 ωρ0 0 0 ˆ 0 dφ0 p ρ dρ φ 2 2 0 0 z +ρ

2π

a

Z

a

= 0.

75

Next, Z

µ0 B= 4π Here

S

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

ˆ 0 × (zˆ ˆ) ˆ 0 ) = η0 ωρ0 (zˆ ρ0 + ρ0 z Js (r0 ) × (r − r0 ) = η0 ωρ0 φ z − ρ0 ρ

so we have µ0 B(0, 0, z) = 4π

Z

2π

0

Z

a

η0 ωρ0

2

0

ˆ zˆ ρ0 + ρ0 z 0 0 2 3/2 dρ dφ 2 0 (z + ρ )

which simplifies to ˆ B(0, 0, z) = z

  µ0 η0 ω 2z 2 + a2 √ − 2|z| . 2 z 2 + a2

3.42. We have 

   Q0 ˆr Q0 ˆr − ∇ · 4π0 r2 4π0 ra     ˆr ˆr Q0 Q0 = ∇· − ∇· 2 4π0 r 4π0 a r   Q0 Q0 1 ∂ = 4πδ(r) − (r) 4π0 4π0 a r2 ∂r ρ = , 0

∇·E=∇·

so

 1 . 4πar2

 ρ = Q0 δ(r) −

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

0

Z 0

π

Z 0

a

1 2 r sin θ dr dθ dφ r2

= 0. 3.43. (a) ˆ · P n

   z 2 ˆ·z ˆ P0 − 1 = 0, =z 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 = 20 −a 20 z Z z Z a 2z 0 2z 0 1 1 P0 2 dz 0 + P0 2 dz 0 =− 20 −a a 20 z a P0 P0 =− (z 2 − a2 ) + (a2 − z 2 ) 2 20 a 20 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

Z

Pm (cos θ) sin θ dθ 0

0

n=0

π/2

Pn (cos θ)Pm (cos θ) sin θ dθ = V0

which gives G0 =

V0 , 2

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)

π

Z

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 4n + 3 Φ(r, θ) = + P2n (0)P2n+1 (cos θ) for r > a. 2 r r 4n + 4 n=0

3.46. The primary potential of the dipole is Φp =

p0 cos θ. 4π0 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) 2π0 a2 n=0 ∞ X

! p0 = − − cos θ + [Bn nan−1 − (n + 1)Cn a−n−2 ]Pn (cos θ) 2π0 a2 n=0 p0 p0 − A1 0 =  − B1 + 2C1 a−3 =⇒ 0 2π0 a3 2π0 a3 p0 =⇒ ( − 0 ) = B1 a3 − 2C1  − A1 0 a3 . 2π0 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) 2π0 b3 n=0 ! ∞ X p0 = −0 − cos θ − Dn (n + 1)b−n−2 Pn (cos θ) 2π0 b3 n=0 p0 3 = B1 b − 2C1 + 20 D1 . =⇒ ( − 0 ) 2π0 Hence we have four equations in four unknowns: A1 a3 = B1 a3 + C1 , F = B1 a3 − 2C1  − A1 0 a3 ,

F = ( − 0 )

p0 , 2π0

B1 b3 = D1 − C1 , F = B1 b3 − 2C1 + 20 D1 . Solve these for A1 , B1 , C1 , D1 and substitute into the equations   p0 + A1 r cos θ, Φ1 = 4π0 r2   p0 −2 Φ2 = + B 1 r + C1 r cos θ, 4π0 r2 and

 Φ3 =

 p0 −2 + D1 r cos θ. 4π0 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 ∂Φ ˆ 1 ∂Φ E = −∇Φ = −ˆr −θ    ∂r  r ∂θ    a 3 a 3 ˆ = ˆrE0 2 + 1 cos θ + θE0 − 1 sin θ. r r Now Z

2π

π/2

Z

ˆ·F=− Fz = z 0

0

80

¯ ·n ˆ·T ˆ a2 sin θ dθ dφ z

where   |E|2 ¯ ˆ) − ˆ·T·n ˆ = 0 (ˆ ˆ) z · E)(E · n z (ˆ z·n 2   |E|2 z · E)(E · ˆr) − (ˆ z · ˆr) . = 0 (ˆ 2 So 2π

π/2 

 2 |E| ˆ − Er (Er (ˆ z · ˆr) + Eθ (ˆ z · θ)) Fz = 0 a (ˆ z · ˆr) sin θ dθ dφ 2 0 0 r=a    Z 2π Z π/2  Er2 + Eθ2 2 2 ˆ r Eθ + (ˆ z · θ)E (ˆ z · ˆr) Er − sin θ dθ dφ = 0 a 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

Z

Z

But Eθ (a, θ) = 0 by the boundary condition on tangential electric field, and Er (a, θ) = 3E0 cos θ, so Z Z 90 a2 E02 2π π/2 cos3 θ sin θ dθ dφ Fz = 2 0 0   π/2 90 a2 E02 1 4 = 2π − cos θ 2 4 0 =

9π0 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

z→0− |z| a



1 +

a |z|

=

" lim |z| lim

z→0−

2 z→0− |z| a

  1 = [0] a =0 to give lim

z→0−

Ezs (0, z) 1 =− . E0 2 81

1 +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 ≈− − − E0 π |z| 3 |z|3

 1 |z| a

+

a |z|

 = −1 π

" #! 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

  z 3 + 20 log10 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 ˆPz (ρ, φ). P(r) = z 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 ˆ · P(a, φ) = −ˆ ˆPz (a, φ) = 0 ρP s |ρ=a = n ρ·z and ˆ · P(b, φ) = ρ ˆ·z ˆPz (b, φ) = 0. ρP s |ρ=b = n 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 ˜ 0 b = BI1 (γb) + CK1 (γb) + µ0 H b  γa  0 Da = BI1 (γa) + CK10 (γa) µr  0  A γb ˜ 0b = − + µ0 H BI1 (γb) + CK10 (γb) . b µr

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

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

I1 (γa) − N = M K1 (γa) −

γa 0 µr I1 (γa) . γa 0 µr K1 (γa)

Substituting C = −f B into (**) and (****) and subtracting gives   γb γb 0 0 ˜ 2µ0 H0 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) = ˜ 0b 2µ0 H 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 H0 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), 1 Re ˜ (r, ω) − 0 = − P.V. π

Z

1 = − P.V. π

Z

c

∞

−∞ 0 −∞

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

∞

Z 0

Im ˜c (r, Ω) dΩ. Ω−ω

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 1 Re ˜ (r, ω) − 0 = − π 1 =− π 1 =− π 2 =− π c

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

∞

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 = −j0 ω[(ω 2 − ν 2 − ωc2 ) − 2jνω] ωp2 ωc [(ω 2 − ν 2 − ωc2 ) + 2jνω] = −j0 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 = −j0

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

˜¯ ] [σ jω

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

! ,

σ ˜xx = σ ˜yy = 0

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

˜xy = −˜ yx = 0 σ ˜xy

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

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

ωp2 ν , [ω(ω 2 + ν 2 )]

and ˜zx = ˜xz = ˜zy = ˜yz = σ ˜zx = σ ˜xz = σ ˜zy = σ ˜yz = 0. ˜ ˜¯ ] = −[σ ˜¯ ]T , we see that [˜¯] and [σ ˜ Since [˜¯] and [σ ¯ ] are real matrices, and since [˜¯] = −[˜¯]T and [σ ¯] 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 Im ˜(ω) =

1 P.V. π

Z

∞

Re ˜(ω) − ∞ dΩ Ω−ω

−∞ ∞ s −∞ 1+Ω2 τ 2

Z

1 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

=

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 +

C1

Z

∞

,

= 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, c

Re ˜ − ∞

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

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 Re ˜c − ∞

" # 2j τ1 1 τ (s − ∞ ) ω = · (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 ˜ 1 = ˜¯1 · E ˜ 1 + ˜¯ξ 1 · H ˜ 1, D ¯˜ · E ˜ =ζ ˜ +µ ˜ . ˜¯ · H B 1

1

1

(1) (2)

1

1

For the dual problem we have ˜ 2 = ˜¯2 · E ˜ 2 + ˜¯ξ 2 · H ˜ 2, D ˜¯ · E ˜ =ζ ˜ +µ ˜ . ˜¯ · H B 2

2

2

(3) (4)

2

2

Now substitute the solution to the dual problem (4.181)–(4.184) into (1): ! ˜2 ˜2 B E ˜ 2 ) + ˜¯ξ 1 · = ˜¯ · (−η0 H −η0 η0 or

˜ 2 = (η 2˜¯1 ) · H ˜ 2 + (−˜¯ξ 1 ) · E ˜ 2. B 0

Compare this to (4). We find that ˜ ¯ = −˜¯ξ , ζ 2 1

˜¯ 2 = η02˜¯1 . µ

Next substitute (4.181)–(4.184) into (2): ˜2 E η0

˜¯ · (−η H ˜2 = ζ ˜ ˜¯ 1 · η0 D 0 2) + µ 1 or ˜¯ ) · H ˜ 2 = (−ζ ˜2 + D 1



µ ¯˜ 1 η02



˜ 2. ·E

Comparison to (3) shows that ˜ ¯2 =

˜¯ 1 µ , η02

˜¯ξ = −ζ ˜¯ . 2 1

88

!

4.5. ∇(e

−jk·r

 ∂ ∂ ∂ ˆ ˆ ˆ e−jkx x e−jky y e−jkz z )= x +y +z ∂x ∂y ∂z ∂ ∂ ∂ ˆ e−jky y e−jkz z e−jkx x + y ˆ e−jkx x e−jkz z e−jky y + z ˆe−jkx x e−jky y e−jkz z =x ∂x ∂y ∂z ˆ − jky y ˆ − jkz z ˆ)e−jkx x e−jky y e−jkz z = (−jkx x 

ˆ ky + z ˆkz )e−j(kx x+ky y+kz z) = −j(ˆ xkx + y = −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 = − . ω0 ω r p σ √ c k = ω µ˜  = ω µ0 0 r − j = β − jα. ω0 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 ω0 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 ω0 c 2 ω0 c 2 ω0 c 20 r ω √ 20 . 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 20 r √ 20 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 ˆ cos θi )k1 , ki = (ˆ x sin θi + z

ˆ cos θi )k1 , kr = (ˆ x sin θi − z

ˆ cos θt )k2 . kt = (ˆ x sin θt + z

Use (4.261): ˜i = y ˜0 e−jki ·r = y ˜0 e−jk1 (x sin θi +z cos θi ) , ˆE ˆE E ⊥ ˜r = y ˜0 Γ ˜ ⊥ e−jkr ·r = y ˜0 Γ ˜ ⊥ e−jk1 (x sin θi −z cos θi ) , ˆE ˆE E ⊥ ˜t = y ˜0 T⊥ e−jkt ·r = y ˜0 T˜⊥ e−jk2 (x sin θt +z cos θt ) . ˆE ˆE E ⊥

Use the equations following (4.261): ˜ ˆi × E ˜ i = E0 (ˆ ˜ i = k1 k ˆ cos θi )e−jk1 (x sin θi +z cos θi ) , z sin θi − x H ⊥ ⊥ k1 η1 η1 ˜ ˜ ˆr × E ˜ r = k1 k ˜ r = E0 Γ⊥ (ˆ ˆ cos θi )e−jk1 (x sin θi −z cos θi ) , H z sin θi + x ⊥ ⊥ k1 η1 η1 ˜ ˜ ˆt × E ˜ t = k2 k ˜ t = E0 T⊥ (ˆ ˆ cos θt )e−jk2 (x sin θt +z cos θt ) . H z sin θt − x ⊥ ⊥ k2 η2 η2 91

˜ ⊥ and T˜⊥ are real numbers. Since regions 1 and 2 are lossless, and θi < θc , we know that Γ In region 1 we have ˜ tot = E ˜i + E ˜r E ⊥ ⊥ h i ˜0 e−jk1 x sin θi e−jk1 z cos θi + Γ ˜ ⊥ ejk1 z cos θi ˆE =y h i ˜0 e−jk1 x sin θi (1 + Γ⊥ )e−jk1 z cos θi + 2jΓ⊥ sin(k1 z cos θi ) ˆE =y and, similarly, h i ˜0 E ˜ tot = z ˜ ⊥ )e−jk1 z cos θi + 2j Γ ˜ ⊥ sin(k1 z cos θi ) ˆ e−jk1 x sin θi (1 + Γ H η1 h i ˜0 E ˜ ⊥ )e−jk1 z cos θi + 2j Γ ˜ ⊥ sin(k1 z cos θi ) . ˆ e−jk1 x sin θi (−1 + Γ +x η1 Note the traveling and standing wave terms in each final expression. In region 2 we have ˜ tot = E ˜t = y ˜0 T˜⊥ e−jk2 x sin θt e−jk2 z cos θt ˆE E ⊥ and

˜ ˜ ˜ tot = H ˜ t = E0 T⊥ (ˆ ˆ cos θt )e−jk2 x sin θt e−jk2 z cos θt . H z sin θt − x ⊥ 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 ˇ tot × H ˇ∗ ] Re[E tot 2 2 ˇ0 | 1 |E ˆ = x 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 ˇ0 |2 1 |E ˆ − z 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 ˆ · Sav,1 = − z

ˇ0 |2 1 |E Re[−1 + Γ2⊥ − Γ⊥ e2jk1 z cos θi + Γ⊥ e−2jk1 z cos θi ] cos θi . 2 η1

At z = 0, ˆ · Sav,1 = z

ˇ 0 |2 1 |E Re(1 − Γ2⊥ ) cos θi . 2 η1

In region 2, ˆ · Sav,2 = z

ˇ 0 |2 T 2 1 |E ⊥ cos θt . 2 η2

ˆ · Sav at z = 0 would require that Continuity of z ˇ 0 |2 T 2 ˇ0 |2 1 |E 1 |E ⊥ 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⊥

ˆ · Sav is continuous at z = 0. The case of parallel polarization is similar. We see that z 4.9. For perpendicular polarization we have ˆ cos θi )k1 , ki = (ˆ x sin θi + z

ˆ cos θi )k1 , kr = (ˆ x sin θi − z

where αc =

ˆ sin θi − jˆ kt = x zαc ,

q k12 sin2 θi − k22 .

Use (4.261): ˜i = y ˜0 e−jki ·r = y ˜0 e−jk1 (x sin θi +z cos θi ) , ˆE ˆE E ⊥ ˜r = y ˜0 Γ ˜ ⊥ e−jkr ·r = y ˜0 Γ ˜ ⊥ e−jk1 (x sin θi −z cos θi ) , ˆE ˆE E ⊥ ˜t = y ˜0 T˜⊥ e−jkt ·r = y ˜0 T˜⊥ e−jk2 x sin θt e−αc z . ˆE ˆE E ⊥

Use the equations following (4.261): ˜ ˆi × E ˜ i = k1 k ˜ i = E0 (ˆ ˆ cos θi )e−jk1 (x sin θi +z cos θi ) , H z sin θi − x ⊥ ⊥ k1 η1 η1 ˜ ˜ ˆr × E ˜ r = E0 Γ⊥ (ˆ ˜ r = k1 k ˆ cos θi )e−jk1 (x sin θi −z cos θi ) , z sin θi + x H ⊥ ⊥ k1 η1 η1 ˜ ˜ ˜ t = 1 kt × E ˜ t = E0 T⊥ (ˆ ˆ αc )e−jk2 x sin θt e−αc z . H zk1 sin θi + j x ⊥ ⊥ k2 η2 k2 η2 ˜ ⊥ and T˜⊥ are real numbers. Since regions 1 and 2 are lossless, and θi < θc , we know that Γ In region 1 we have ˜ tot = E ˜i + E ˜r E ⊥ ⊥ h i ˜0 e−jk1 x sin θi e−jk1 z cos θi + Γ ˜ ⊥ ejk1 z cos θi ˆE =y h i ˜0 e−jk1 x sin θi ejφ⊥ /2 e−j(k1 z cos θi +φ⊥ /2) + ej(k1 z cos θi +φ⊥ /2) ˆE =y ˜0 e−jk1 x sin θi ejφ⊥ /2 2 cos(k1 z cos θi + φ⊥ /2). ˆE =y 93

Similarly, ˜0 E ˜ tot = z ˆ e−jk1 x sin θi ejφ⊥ /2 2 cos(k1 z cos θi + φ⊥ /2) sin θi H η1 ˜0 E ˆ e−jk1 x sin θi ejφ⊥ /2 2j sin(k1 z cos θi + φ⊥ /2) cos θi . +x η1 These are pure standing waves. In region 2 we have the evanescent wave expressions ˜ tot = E ˜t = y ˜0 T˜⊥ e−jk2 x sin θi e−αc z ˆE E ⊥ and

˜ ˜ ˜ tot = H ˜ t = E0 T⊥ (ˆ ˆ αc )e−jk2 x sin θt e−αc z . zk1 sin θi + j x H ⊥ k2 η2

The case of parallel polarization proceeds similarly. 4.10. We have 1 ˇ tot × H ˇ∗ ] Re[E tot 2 1 ˇtot,y H ˇ∗ ] = Re[E tot,x 2 1 ˇ0 |2 2 cos(k1 z cos θi + φ⊥ /2)[−j2 sin(k1 z cos θi + φ⊥ /2)] cos θi } = Re{|E 2 = 0.

ˆ · Sav,1 = z ˆ· z

The rest of the problem is similar. 4.11. In region 1 we have ˜1 = 0 1r , In region 2,

(a)

µ ˜1 = µ0 µ1r .

σ ˜c2 = 2r 0 − j , ω

µ ˜2 = µ0 µ2r .

p √ √ µ ˜1 ˜1 = ω µ0 0 µ1r 1r , r p σ √ √ ˜2 ˜c2 = ω µ0 0 µr2 r2 1 − j . k2 = ω µ ω0 2r k1 = β1 = ω

The definition of a good conductor is through the inequality σ  1. ω0 2r So

r

σ 1−j ≈ ω0 2r

r −j

σ 1−j = √ ω0 2r 2

94

r

σ . ω0 2r

This implies r

σ √ 1−j µ2r 2r √ ω0 2r 2 r r p ωσ ωµ ˜2 σ = µ ˜2 (1 − j) = (1 − j) 2 2 √

k2 = β2 − jα2 ≈ ω µ0 0

or

r β 2 = α2 =

ωµ ˜2 σ . 2

Examining the ratio q

ωµ ˜2 σ

β2 2 ≈ √ = β1 ω µ ˜1 ˜1

s

µ ˜2 σ = 2ω µ ˜1 ˜1

r

σ ω0 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, ˜2 = y ˜ i e−jkt ·r , ˆ T˜⊥ E E ⊥

ˆβ2 (1 − j). kt = k02 + jk002 ≈ z 95

So ˜2 ≈ y ˜ i e−jβ2 z e−β2 z , ˆ T˜⊥ E E ⊥

˜ 2 = σE ˜2 = y ˜ i e−jβ2 z e−β2 z . ˆ σ T˜⊥ E J ⊥

The current per unit width is Z

∞

˜2 · y ˆ dz J Z ∞ ˜i e−(1+j)β2 z dz = σ T˜⊥ E ⊥

J˜s =

0

0

σ ˜ i [e−(1+j)β2 z ] ∞ = T˜⊥ E ⊥ 0 −(1 + j)β2 1 σ ˜ ˜i T⊥ E⊥ . = 1 + j β2 But

1 1−j = , 1+j 2

so

˜ =y ˜i 1 − j . ˆ σ T˜⊥ E K ⊥ 2β2

Now ˜i k t T˜⊥ E ⊥ −jβ2 z −β2 z ˜ 2 ≈ −ˆ e e H x z k2 η2 ˜i T˜⊥ E ⊥ −jβ2 z −β2 z ≈ −ˆ x e e . η2 Here s η2 =

µ ˜2 = ˜c2

s

µ ˜2 = ˜2r 0 − j ωσ

r

1 µ ˜2 q 2r 0 1 − j

r σ ω0 2r

≈

µ2r µ0 1 + j √ 2r 0 2

r

ω0 2r = (1 + j) σ

So ˜i T˜⊥ E ⊥ ˜ 2 ≈ −ˆ H x 1+j

r

2σ −jβ2 z −β2 z e e ωµ ˜2 r 2σ ˜i 1 − j = −ˆ xT˜⊥ E ⊥ 2 ωµ ˜2 ˜i 1 − j σ = −ˆ xT˜⊥ E ⊥ 2 β2

since

r β2 =

ωµ ˜2 σ . 2

Thus, at z = 0, ˜ t ≈ −ˆ ˜ =y ˜i 1 − j σ ≈ K ˜i 1 − j . ˆ σ T˜⊥ E −ˆ z×H z × (−ˆ x)T˜⊥ E ⊥ ⊥ 2 β2 2β2 96

r

ωµ ˜2 . 2σ

(c) E ˜y Zs = ˜x H

= z=0

˜i T˜⊥ E 2β2 ⊥ = (1 + j)β2 . = 1−j i ˜ (1 − j)σ T˜⊥ E ⊥ 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 ˇ ×H ˇ ∗ ]|z=0 Re[E 2 

ˇ i × −ˆ ˇi 1 − j σ ˆ T˜⊥ E =y xT˜⊥ E ⊥ ⊥ 2 β2 σ ˇ i |2 ˆ|T˜⊥ E . =z ⊥ 2β2 But

∗

ˇ =y ˇi 1 − j . ˆ σ T˜⊥ E K ⊥ 2β2

So

ˇ ·K ˇ ∗ = σ 2 |T˜⊥ E ˇ i |2 (1 − j)(1 + j) = σ 2 |T˜⊥ E ˇ i |2 1 . K ⊥ ⊥ 2 2β2 β22

Hence

1 ˇ ˇ∗ 1 ˇ ˇ∗ ˆ (K ˆ (K z · K )Rs = z ·K ) 2 2

4.12. From the text we have

With Γ2 = −Γ1 ,



β2 σ



1 ˇ i |2 σ = Sav . ˆ|T˜⊥ E =z ⊥ 2 2β2 β2

˜2 ˜ 1 = b1 = Γ1 + Γ2 P1 . R a1 1 + Γ1 Γ2 P˜12 ˜2 ˜ 1 = Γ1 1 − P1 . R 1 − Γ21 P˜12

Use of the expansion (1 − x)−1 = 1 + x + x2 + x3 + · · · gives ˜ 1 (ω) = Γ1 (1 − P˜12 )(1 + Γ21 P˜12 + Γ41 P˜14 + Γ61 P˜16 + · · · ) R = Γ1 [1 + Γ21 P˜12 − P˜12 − Γ21 P˜14 + Γ41 P˜14 − Γ41 P˜16 + Γ61 P˜16 + · · · ] = Γ1 [1 + (Γ21 − 1)P˜12 + Γ21 (Γ21 − 1)P˜14 + Γ41 (Γ21 − 1)P˜16 + · · · ]. Now use P˜12n (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 ˆ + kzi z ˆ, ki = kxi x

kxi = k sin θi ,

kzi = k cos θi ,

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

Region 0: ˜i = y ˜ i e−j(kxi x+kzi z) , ˆE E ⊥ ⊥ ˜r = y ˜ r e−j(kxi x−kzi z) , ˆE E ⊥ ⊥

˜i ˆkxi E xkzi + z ⊥ −j(kxi x+kzi z) ˜ i = −ˆ e H , ⊥ k η ˜r ˆ kzi + z ˆkxi E ⊥ −j(kxi x−kzi z) ˜r = x H e . ⊥ k η 98

Region 1: kx2 + kz2 = k02 , ˜+ ˆkx E xkz + z ⊥ −j(kx x+kz z) ˜ + = −ˆ e , H ⊥ k0 η0 ˜− ˆ kz + z ˆkx E ⊥ −j(kx x−kz z) ˜− = x H e . ⊥ k0 η0

˜+ = y ˜ + e−j(kx x+kz z) , ˆE E ⊥ ⊥ ˜− = y ˜ − e−j(kx x−kz z) , ˆE E ⊥ ⊥ Region 2: (kxt )2 + (kzt )2 = k 2 ,

˜t ˆkxt E xkzt + z ⊥ −j(kxt x+kzt z) ˜ t = −ˆ e . H ⊥ k η

˜t = y ˜ t e−j(kxt x+kzt z) , ˆE E ⊥ ⊥

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 k¯z where k¯z = k 2 sin2 θi > k02 , or sin2 θi >

k02 1 = k2 r

so that

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

1 θi > sin−1 √ = θc . r

˜r , E ˜t , E ˜ i , and E ˜ + , we apply the boundary conditions. To find E ⊥ ⊥ ⊥ ⊥ (1) Tangential E continuous at z = z1 : ˜ − ejkz z1 , ˜ + e−jkz z1 + E ˜ r ejkzi z1 = E ˜ i e−jkzi z1 + E E ⊥ ⊥ ⊥ ⊥ or ˜i + P1 E ⊥

1 ˜r ˜+ + 1 E ˜− E = Q1 E ⊥ P1 ⊥ Q1 ⊥

where

i

P1 = e−jkz z1 ,

Q1 = e−jkz z1 .

(2) Tangential H continuous at z = z1 : − or

˜+ ˜− 1 ˜i ˜r 1 kzi E ki E kz E kz E ⊥ ⊥ ⊥ P1 + z ⊥ =− Q1 + k η k η P1 k0 η0 k0 η0 Q1

1 Z1⊥

    1 − 1 + r 1 i ˜ ˜ ˜ ˜ E⊥ − P1 E⊥ = E⊥ − Q1 E⊥ P1 Z2⊥ Q1

where Z1⊥ =

kη , kzi

Z2⊥ =

99

k0 η0 . kz

(3) Tangential E continuous at z = z2 : ˜ + e−jkz z2 + E ˜ − ejkz z2 = E ˜ t e−jkzi z2 , E ⊥ ⊥ ⊥ or

˜+ + 1 E ˜t ˜ − = P2 E Q2 E ⊥ ⊥ Q2 ⊥

where

i

P2 = e−jkz z2 ,

Q2 = e−jkz z2 .

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

˜+ ˜− 1 ˜t kz E kz E ki E ⊥ ⊥ Q2 + = − z ⊥ P2 k0 η0 k0 η0 Q2 k η

1 ˜− 1 1 ˜t 1 ˜+ E⊥ Q2 + E⊥ =− E P2 . Z2⊥ Z2⊥ Q2 Z1⊥ ⊥

So we have   1 ˜r − 1 + i ˜ ˜ ˜ P1 E⊥ + E = E⊥ + E⊥ Q1 , P1 ⊥ Q1   Z1⊥ ˜ − 1 1 ˜r + i ˜ ˜ E = E⊥ − E⊥ Q1 , − P1 E⊥ + P1 ⊥ Z2⊥ Q1   ˜− 1 + E ˜ t P2 = E ˜ + Q2 , E ⊥ ⊥Q ⊥ 2   Z1⊥ ˜ − 1 + t ˜ ˜ − E⊥ P2 = E⊥ − E⊥ Q2 . Z2⊥ Q2

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

Subtracting (2) from (1) we get ˜i = E ˜− 1 2P1 E ⊥ ⊥Q 1



Z1⊥ 1− Z2⊥



  Z1⊥ + ˜ + E⊥ Q1 1 + . Z2⊥

Adding (4) to (3) we get Z1⊥  ˜ − ˜ +  ˜ − 1 + E + Q2 , − E⊥ − E⊥ Q2 = E ⊥Q ⊥ Z2⊥ 2 hence

˜ − = −E ˜ + Q22 Z2⊥ − Z1⊥ = −E ˜ + Q22 Γ. E ⊥ ⊥ ⊥ Z2⊥ + Z1⊥

Substitution into (1) yields ˜+ = E ⊥

2Z2⊥ P1 Q1 Z1⊥ + Z2⊥ Q22

1 Q21 Q22

−

Γ2

˜i = τ E ˜i . E ⊥ ⊥

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

 Γ 1− Γ2

−

100

Q21 Q22 Q21 Q22

 ˜i . E ⊥

(*)

From (3) we obtain 2 ˜ t = P1 Q1 12− Γ E ˜i . E ⊥ P2 Q2 Q1 − Γ2 ⊥ Q2 2

Let

  Q21 r Γ 1 − ˜ 2 E Q2 , = P12 R= ⊥ 2 i Q ˜ E Γ2 − 1 ⊥

T =

Q22

˜t E P1 Q1 1 − Γ2 ⊥ = . ˜i P2 Q2 Q21 − Γ2 E ⊥ 2 Q 2

Now use

1 ˇ ×H ˇ ∗ }. Re{E 2

Sav = (a) In region 0 at z = z1 we have

    R i i r i −jkzi z1 jkzi z1 −jkxi x i ˇ ˇ ˇ ˇ ˇ ˆ E⊥ e ˆ E⊥ P1 + e−jkx x , E = E⊥ + E ⊥ = y + Re e =y P1 h
−jki z1
jki i −jki x i i i i ˇr = 1 E ˇ =H ˇi +H ˇ i −ˆ x . z z e ˆ ˆ ˆ H x k + z k e + x k + z k z x z x e ⊥ ⊥ kη ⊥ So       ∗ i 1 ˇ i P1 + R e−jkxi x × kz x ˇ i∗ −P1∗ + R ejkxi x · z ˆE ˆ ˆ Re y E ⊥ 2 P1 kη ⊥ P1∗    ∗  R 1 1 ˇi 2 R 2 = Re − |E | −1 + 2 − + |R| 2 Z1⊥ ⊥ P1 P12 1 1 ˇi 2 |E | (1 − |R|2 ). = 2 Z1⊥ ⊥

ˆ · Sav = z

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

k¯z ˇ i |2 Im{Γ}. |τ |2 |E ⊥ η0 k0

1 1 ˇi 2 ˆ · Sav = |T |2 z |E | . 2 Z1⊥ ⊥

Power is conserved if 1 − |R|2 = |T |2 . 4.14. Begin with the wave equation (4.319): 1 ˜ − ω2µ ˜ = 0. ˜¯ · H ∇ × (∇ × H)  ˜ ˜ 0 e−jkz z where kz = β − jα, we have Assuming H(r) =H   ˜ =∇× H ˜ 0 e−jkz z = −jkz z ˜ 0 e−jkz z ˆ×H ∇×H

101

and   ˜ = −jkz ∇ × z ˜ 0 e−jkz z ˆ×H ∇ × (∇ × H) h i ˜ 0 ) − (ˆ ˜ 0 )e−jkz z = −jkz e−jkz z ∇ × (ˆ z×H z×H ˜ 0 )e−jkz z ˆ × (ˆ = (−jkz )2 z z×H ˜ 0. = k 2 e−jkz z H z

Substitute into (4.319) to get kz2 ˜ −jkz z ˜ 0 e−jkz z ˜¯ · H H0 e = ω2µ ˜ or ˜ 0 = ω 2 ˜µ ˜ 0. ˜¯ · H kz2 H Now let

(1)



 µ ˜xx µ ˜xy 0 ˜ ˜yx µ ˜yy 0  . µ ¯ = µ 0 0 µ ˜0

Write (1) in component form: ˜ 0x = ω 2 ˜µ ˜ 0x + ω 2 ˜µ ˜ 0y , kz2 H ˜xx H ˜xy H ˜ 0y = ω 2 ˜µ ˜ 0x + ω 2 ˜µ ˜ 0y , kz2 H ˜yx H ˜yy H or



˜xx −ω 2 ˜µ ˜xy kz2 − ω 2 ˜µ 2 2 ˜yy −ω ˜µ ˜yx kz − ω 2 ˜µ

    ˜ 0x H 0 ˜ 0y = 0 . H

A nontrivial solution requires the determinant of the coefficient matrix to vanish: (kz2 − ω 2 ˜µ ˜xx )(kz2 − ω 2 ˜µ ˜yy ) − ω 4 ˜2 µ ˜yx µ ˜xy = 0, kz4 + [−ω 2 ˜(˜ µxx + µ ˜yy )]kz2 + ω 4 ˜2 (˜ µxx µ ˜yy − µ ˜xy µ ˜yx ) = 0. But µ ˜xx = µ ˜yy and µ ˜yx = −˜ µxy , so kz4 + [−2ω 2 ˜µ ˜xx ]kz2 + ω 4 ˜2 (˜ µ2xx + µ ˜2xy ) = 0. We get kz2 =

2ω 2 ˜µ ˜xx ±

q 4ω 4 ˜2 µ ˜2xx − 4ω 4 ˜2 (˜ µ2xx + µ ˜2xy ) 2

q = ω 2 ˜µ ˜xx ± ω 2 ˜ −˜ µ2xy = ω 2 ˜(˜ µxx ∓ j µ ˜xy ).

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 µ ˜xx

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

µ ˜xy ≈

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

Thus 

µ ˜xx + j µ ˜xy

 ωM ωM ωM ωM α ≈ 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 µ ˜xx − j µ ˜xy ≈ 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, ˜ ˜z = − j H ˜ z0 H (2) (kρ), ˜φ = −ZT E Hz0 H (2) (kρ), H E 0 1 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 ˆE ˇφ × z ˇ z∗ ˆH Re φ 2 r r     ˇ z0 1 H 2 −j(βρ−3π/4) j ˇ ∗ 2 j(βρ−π/4) −αρ ˆ Re −ZT E =ρ e Hz0 e e 2 4 πkρ 4 πk ∗ ρ   ˇ z0 |2 2 |H 1 −2αρ jπ/2 ˆ Re −ZT E =ρ e je 2 16|k| πρ e−2αρ ˇ 2 ˆ Re{ZT E } =ρ |Hz0 | . 16π|k|ρ

Sav =

103

The power passing through a cylinder of radius ρ and length l is Z ˆ dS Sav · ρ Pav = S Z l Z 2π e−2αρ ˇ 2 Re{ZT E } = |Hz0 | ρ dφ dz 16π|k|ρ 0 0 e−2αρ ˇ 2 = 2πl Re{ZT E } |Hz0 | . 16π|k| Hence

e−2αρ ˇ 2 Pav = Re{ZT E } |Hz0 | . 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

ˇ z0 |2 η|H Pav = . l 8k

4.16. We have ˜zs = E

∞ X

Dn Hn(2) (k0 ρ) cos nφ

n=0

where

˜0 n j −n Jn (k0 a) . Dn = E (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

r 2 −jk0 ρ jπ/4 jnπ/2 −n Jn (k0 a) ˜ E0 n j e e e cos nφ (2) Hn (k0 a) πk0 ρ n=0 r ∞ 2j e−jk0 ρ ˜ X Jn (k0 a) = n (2) cos nφ. √ E0 πk0 ρ Hn (k0 a)

˜s ≈ E z

n=0

Then 

2 πk0

σ2D = 2πρ



P 2 ˜ E0 ∞ n=0 n

2 cos nφ (2) Hn (k0 a) Jn (k0 a)

˜2 ρE 0 ∞ 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 2 4 1 1 π2a σ2D ≈ . 2 = 2 k0 j π ln(0.891k0 a) k0 a ln (0.891k0 a) 4.17. Start with ˆ cos φ)E ˜i = y ˜0 e−jk0 x = (ˆ ˜0 e−jk0 ρ cos φ ˆE E ρ sin φ + φ and

˜0 ˜0 E E ˜i = z ˆ e−jk0 x = z ˆ e−jk0 ρ cos φ . H η0 η0 Internal to the cylinder (ρ < a) we have ˜z = H

∞ X

Bn Jn (kρ) cos nφ,

n=0 ∞

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

∞

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

External to the cylinder (ρ > a) we have ˜s = H z ˜s = E φ ˜ρs = E

∞ 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

˜ z is continuous at ρ = a. Boundary condition 1: H ∞ ∞ X X ˜ (2) ˜i + H ˜s = H ˜ z =⇒ E0 e−jk0 a cos φ + D H (k a) cos nφ = Bn Jn (ka) cos nφ. H n n 0 z z η0 n=0

n=0

Apply orthogonality: Z π ˜ Z π ∞ X E0 −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 ˜0 E (2) 2πm j −m Jm (k0 a) + 2πDm Hm (k0 a) = 2πBm Jm (ka). η0

(1)

˜φ is continuous at ρ = a. Boundary condition 2: E ˜0 e−jk0 a cos φ + cos φE

∞ 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 Z π ∞ X −jk0 a cos φ (2)0 ˜ cos φ cos mφe dφ + Dn jη0 Hn (k0 a) E0 −π

=

Bn jZT E Jn0 (ka)

n=0

Z

cos mφ cos nφ dφ

−π

n=0 ∞ X

π

π

cos mφ cos nφ dφ, −π

which gives 0 (2)0 0 ˜0 j2πj −m Jm m E (k0 a) + Dm jη0 2πHm (k0 a) = Bm jZT E 2πJm (ka).

Solving (1) and (2) simultaneously, we get Dm = −

0 (ka) − ˜0 Jm (k0 a)Jm E m j −m (2) η0 Hm (k0 a)J 0 (ka) − m

and Bm

η0 0 ZT E Jm (ka)Jm (k0 a) (2)0 η0 ZT E Hm (k0 a)Jm (ka)

(2)0 (2) 0 ˜0 E −m Jm (k0 a)Hm (k0 a) − Jm (k0 a)Hm (k0 a) = m j . (2)0 (2) η0 Jm (ka)Hm (k0 a) − ZT E J 0 (ka)Hm (k0 a) η0

m

For a PEC cylinder, use ZT E →0 η0 Then ˜s = − H z

∞ ˜ X E0 n=0

Also,

=⇒

η0

Dm = −

n j −n

0 (k a) ˜0 E 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.

∞ ˜ X ˜0 E E0 i −jk0 ρ cos φ ˜ Hz = 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) ˜0 X J E 0 n ˜ tot = H n j −n Jn (k0 ρ) − (2)0 Hn(2) (k0 ρ) cos nφ z η0 Hn (k0 a) n=0

∞ i ˜0 X n j −n h E (2)0 0 (2) = J (k ρ)H (k a) − J (k a)H (k ρ) cos nφ. n 0 0 0 0 n n n (2)0 η0 n=0 Hn (k0 a)

106

(2)

4.18. The incident field is given by ˜ ˆ cos φ) E0 e−jk0 ρ cos φ . ˜ i = −(ˆ H ρ sin φ + φ η0

˜i = z ˜0 e−jk0 ρ cos φ , ˆE E In region 1 (a ≤ ρ ≤ b), ˜z1 (ρ, φ) = E ˜ 1 (ρ, φ) = H 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), ˜z2 (ρ, φ) = E ˜ 2 (ρ, φ) H φ

=

∞ X

Cn Hn(2) (k0 ρ) cos nφ,

n=0 ∞  X n=0

j − η0



Cn Hn(2)0 (k0 ρ) cos nφ.

˜ 1 = 0 at ρ = a: Boundary condition 1: E z An Jn (ka) + Bn Hn(2) (ka) = 0 =⇒ Bn = −An

Jn (ka) (2)

.

Hn (ka)

So ˜1 E z

=

˜ z1 = H

∞ X

" An Jn (kρ) −

n=0 ∞  X n=0

j − η



Hn(2) (kρ)

#

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

˜0 e E

+

∞ X

Cn Hn(2) (k0 b) cos nφ

n=0

=

∞ X

" An Jn (kb) −

Hn(2) (kb)

n=0

Apply orthogonality to obtain ˜0 2πn j −n Jn (k0 b) + 2πCn H (2) (k0 b) = 2πAn Fn E n or ˜0 An Fn − Cn Hn(2) (k0 b) = n j −n Jn (k0 b)E 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 X ˜0 Jn (ka) j j E 0 (2)0 −jk0 b cos φ (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     ˜0 j E 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 ˜0 n j −n J 0 (k0 b) Gn − Cn Hn(2)0 (k0 b) = E n η

where

" Jn0 (kb)

Gn =

−

Hn(2)0 (kb)

Jn (ka)

#

(2)

cos nφ.

Hn (ka)

Solving the equations ˜0 n j −n Jn (k0 b), An Fn − Cn Hn(2) (k0 b) = E η0 ˜0 n j −n Jn0 (k0 b), An Gn − Cn Hn(2)0 (k0 b) = E η simultaneously, we obtain (2)0

Cn = −n j and An =

−n

˜0 E

Fn Hn (k0 b) − Fn Jn0 (k0 b) −

(2) η0 η Gn Hn (k0 b) η0 η Gn Jn (k0 b)

2j ˜ −n πk0 b E0 n j − . (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 ˜z2 = E ˜2 = H φ

∞ 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 ˜z1 = E ˜1 = H φ

∞ 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 ˜z = 0 at ρ = a. By orthogonality we get Boundary condition 1: E Bn = −An

Jn (k0 a) (2)

,

Hn (k0 a)

so ˜1 E z

= =

∞ 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, ˜1 = H φ

 ∞  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 ).

˜z continuous at ρ = ρ0 . By orthogonality Boundary condition 2: E 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. ˜ φ is discontinuous at the surface current Boundary condition 3: H δ(φ − φ0 ) ˜s = z ˆI˜ . J ρ0 We have

˜ φ (ρ+ , φ) − H ˜ φ (ρ− , φ) = I˜δ(φ − φ0 ) , H 0 0 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

˜ y) = k ψ(x, 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

√

k2 −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> . ˜z = 0. Let 4.21. For a magnetic line source we have TEz fields so that E  Z ∞+j∆ 1   A˜+ (kx )e−jkx x e−jky y dkx , y > 0,   2π −∞+j∆ ˜ z (x, y) = H Z ∞+j∆  1    A˜− (kx )e−jkx x e+jky y dkx , y < 0. 2π −∞+j∆ From (4.225) we have ˜x = E

˜z 1 ∂H , jω˜ c ∂y 111

so

  Z ∞+j∆  −ky ˜+ 1   A (kx )e−jkx x e−jky y dkx ,   2π ωc −∞+j∆ ˜x = E  Z ∞+j∆   +ky ˜− 1    A (kx )e−jkx x e+jky y dkx , 2π −∞+j∆ ω˜ c

y > 0, y < 0.

˜ is continuous at y = 0. This implies BC 1) Tangential H A˜+ (kx ) − A˜− (kx ) = 0

(A)

˜1 − E ˜ 2 ) = −J ˜ ms = −I˜m δ(x). This implies ˆ 12 × (E BC 2) n −A˜+ (kx ) − A˜− (kx ) = I˜m Adding (A) and (B) we obtain

ω˜ c ky

(B)

c

ω˜  −2A˜− (kx ) = I˜m ky so

c

ω˜  A˜+ (kx ) = A˜− (kx ) = −I˜m . 2ky

Finally, Z ω˜ c I˜m ∞+j∆ e−jky |y| −jkx x ˜ Hz (x, y) = − e dkx . 2π 2ky −∞+j∆ 4.22. See Figure 30. In region 1 we need both upward and downward traveling waves:

Figure 30 ˜+ = 1 E z1 2π

Z

˜− = 1 E z1 2π

Z

∞+j∆

−∞+j∆ ∞+j∆

A˜+ (kx )e−jky y e−jkx x dkx ,

ky =

p k 2 − kx2

A˜− (kx )ejky y e−jkx x dkx ,

−∞+j∆

˜+ = 1 H x1 2π ˜− = 1 H x1 2π

Z

∞+j∆



−∞+j∆ Z ∞+j∆  −∞+j∆

ky ωµ ˜



ky − ωµ ˜

A˜+ (kx )e−jky y e−jkx x dkx , 

112

A˜− (kx )ejky y e−jkx x dkx .

In region 2 we need only upward traveling waves: ˜z2 = 1 E 2π

Z

1 = 2π

Z

˜ x2 H

∞+j∆

˜ x )e−jky y e−jkx x dkx , B(k

−∞+j∆ ∞+j∆  −∞+j∆

ωµ ˜ ky



˜ x )e−jky y e−jkx x dkx . B(k

˜z = 0 at y = 0. Boundary condition 1: E A˜− (kx ) = −A˜+ (kx ), Z ∞+j∆ h i 1 + − ˜ ˜ ˜ Ez1 = Ez1 + Ez1 = A˜+ (kx ) e−jky y − ejky y e−jkx x dkx , 2π −∞+j∆  h Z ∞+j∆ i kx 1 + ˜ ˜ A (kx ) e−jky y + ejky y e−jkx x dkx . Hx1 = 2π −∞+j∆ ωµ ˜ ˜z continuous at y = h. Boundary condition 2: E h i + −jky h jky h ˜ ˜ −jky h , A e −e = Be i h ˜ = A˜+ 1 − e2jky h . B Boundary condition 3: 1 ˜ x1 − H ˜ x2 = J˜zs = Iδ(x) ˜ H = I˜ 2π

Z

∞+j∆

e−jkx x dkx ,

−∞+j∆

i h i h ωµ ˜ A˜+ e−jky h + ejky h − A˜+ 1 − e2jky h e−jky h = I˜ , ky ωµ ˜ −jky h e . A˜+ = I˜ 2ky Thus ˜z1 = 1 E 2π

Z

∞+j∆

˜z2 = 1 E 2π

Z

∞+j∆

i ωµ ˜ h −jky (y+h) I˜ e − ejky (y−h) e−jkx x dkx , −∞+j∆ 2ky

h i ωµ ˜ I˜ e−jky y e−jky h − ejky h e−jkx x dkx −∞+j∆ ky Z ∞+j∆ i 1 ωµ ˜ 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, Z ˜I˜ ∞+j∆ e−jky |y−h| − e−jky (y+h) −jkx x ˜z (x, y) = − ω µ E e dkx , 2π −∞+j∆ 2ky 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 , Hz (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 ˜ Hz (x, y) = − e dkx . 2π 2ky1 −∞+j∆ Use Ampere’s law in the form ˜ ˜x = 1 ∂ Hz E jω˜  ∂y to obtain

Z I˜m ∞+j∆ ejky1 (y−h) −jkx x i ˜ Ex (x, y) = − e dkx , 2π −∞+j∆ 2

0 ≤ y < h.

The scattered field in region 1 is ˜ s (x, y) = 1 H z1 2π ˜ s (x, y) = 1 E x1 2π

∞+j∆

Z

A˜1 (kx )e−jky1 y e−jkx x dkx ,

−∞+j∆ Z ∞+j∆  −∞+j∆

ky1 − c ω˜ 1



A˜1 (kx )e−jky1 y e−jkx x dkx .

The scattered field in region 2 is ˜ s (x, y) = 1 H z2 2π ˜ s (x, y) = 1 E x2 2π

Z

∞+j∆

A˜2 (kx )ejky2 y e−jkx x dkx ,

−∞+j∆ Z ∞+j∆  −∞+j∆

ky2 ωc2



A˜2 (kx )ejky2 y e−jkx x dkx .

˜tan continuous at y = 0: Boundary condition 1: E Z Z ∞+j∆ Z ∞+j∆ ky1 ˜ −jkx x ky2 ˜ −jkx x I˜m ∞+j∆ e−jky1 h −jkx x 1 1 − e dkx − A1 e dkx = A2 e dkx , 2π −∞+j∆ 2 2π −∞+j∆ ω˜ c1 2π −∞+j∆ ω˜ c2 or −I˜m

ky1 ky2 e−jky1 h − c A˜1 = c A˜2 . 2 ω˜ 1 ω˜ 2 114

˜ tan continuous at y = 0: Boundary condition 2: H −ω˜ c1 I˜m

e−jky1 h + A˜1 = A˜2 . 2ky1

Solving simultaneously, we obtain ˜c ky2 − ˜c2 ky1 F, A˜1 = 1c ˜1 ky2 + ˜c2 ky1 where

A˜2 = 2F c

˜c2 ky1 ˜c2 ky1 + ˜c1 ky2

−jky1 h

ω˜  e F = −I˜m 1 ky1 2

.

Substitution yields Z I˜m ω˜ c1 ∞+j∆ e−jky1 (y+h) s ˜ Hz1 (x, y) = − RT E (kx )e−jkx x dkx , 2π 2k y1 −∞+j∆ Z ∞+j∆ −jky2 (y−hky1 /ky2 ) c ˜ e 2 ˜ s (x, y) = − Im ω˜ TT E (kx )e−jkx x dkx , H z2 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 = z ˜ 0 e−jkxi x e−jkyi y , ˆH H

kxi = −k cos φ0 ,

kyi = −k sin φ0 ,

i i ˜ i = ηH ˜ 0 (ˆ ˆ cos φ0 )e−jkx x e−jky y E x sin φ0 − y ˜0 ηH i i ˆ + kxi y ˆ )e−jkx x e−jky y . (−kyi x = k

115

q kxi = − k 2 − (kyi )2 ,

Represent the scattered magnetic field in terms of an inverse transform: ˜ zs = 1 H 2π

Z

∞+j∆

˜ y )e−jkx x e−jky y dky , A(k

kx =

−∞+j∆

˜ s use To find E

q k 2 − ky2 .

˜z ˜z ∂H ∂H ˜ = jω˜ ˜ =x ˆ ˆ ∇×H E −y ∂y ∂x

to obtain ˜s = 1 E z 2π

∞+j∆

Z



˜ y) A(k

−∞+j∆

 kx e−jkx x e−jky y dky . ω˜ 

Next we apply the boundary condition ˜ tot = 0 at x = 0, ˆ·E y We obtain

∞+j∆

i ˜ 0 kx e−jkyi y + 1 ηH k 2π

Z

1 2π

Z

i

˜ y) A(k



−∞+j∆

Next use e−jky y =

˜ tot = E ˜i + E ˜ s. where E

∞+j∆

 kx e−jky y dky = 0. ω˜ 

1 e−jky y dky j(ky − kyi )

−∞+j∆

to get i

˜ y ) = −η H ˜ 0 kx A(k k



ω˜  kx



1 . j(ky − kyi )

Substitution into (1) yields ˜s = − 1 H z 2π

Z

∞+j∆

˜0 ηH

−∞+j∆

kxi k



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 ˜ ˜ Hz = 2πj lim (ky − ky ) −η H0 e e . 2π k kx j(ky − kyi ) ky →kyi Use lim kx =

ky →kyi

q

k 2 − (kyi )2 = −kxi

to get i

˜ i = −η H ˜ 0 kx H z k Finally then,



ω˜  −kxi



i

i

e−jky y ejkx x =

ω˜ η ˜ −jkyi y jkxi x H0 e e . k

˜s = z ˜ 0 e−jkyi y ejkxi x . ˆH H

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 40 = 2k0 ,

Zn =

p µ0 /0 = η0 ,

Zn−1 =

p µ0 /40 = 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

κ∆

5

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 T11 T22 − T12 T21

T11 T12 T (n) T (n) T (n+1) T (n+1) T (n+2) T (n+2) 12 11 12 11 12 = 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) T11 T12 Zn−1 , (n) (n) = T21 T22 Zn

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, σ ˜= 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, δ˜ = 0 = 2 2 ω(ω − ωc ) 2.26 × 10−5 0 at 100 MHz. 4.28. (a) Re ˜ − 0 = 0 ωp2

(ω02

ω02 − ω 2 . − 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 ˜ = 50 = 0 1 + 2 , ω0 and

1 ωp2 ω0 [Im ˜]max = −160 = − 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) ˆ E0 e−jβr ejξE . ˇ ω E(r, ˇ) = θ r sin θ (b) ˆ E0 /η0 e−jβr ejξE = φ ˆ E0 /η0 e−jβr ejξE . ˇ H(r, ω ˇ ) = ˆr × θ r sin θ r sin θ 120

(c)    E0 /η0 −jβr jξE ˆ E0 −jβr jξE ˆ ˇ ˇ lim r η0 ˆr × H + E = lim r η0 ˆr × φ e e +θ e e r→∞ r→∞ r sin θ r sin θ   ˆ E0 e−jβr ejξE + θ ˆ E0 e−jβr ejξE = lim −θ r→∞ sin θ sin θ = 0. 

    E0 −jβr jξE E0 /η0 −jβr jξE ˆ ˆ ˇ ˇ lim r ˆr × E − η0 H = lim r ˆr × θ e e − η0 φ e e r→∞ r→∞ r sin θ r sin θ   E0 −jβr jξE ˆ E0 −jβr jξE ˆ = lim φ e e −φ 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 ˇ ˇ ∗ (r) . Re E(r) ·D 4

4.31. Since k2 = ω2 µ ˜(Re ˜ + j Im ˜) = (β − jα)2 = β 2 − 2jαβ − α2 , we have β 2 − α2 = ω 2 µ ˜ Re ˜, Therefore α=

−2αβ = ω 2 µ Im ˜.

ω2µ ˜ Im ˜ 2β 121

0

and we can back-substitute: 2

β −



ω2µ ˜ Re ˜ 2β

2

= ω2µ ˜ Re ˜.

Rearrange to obtain 1 β 4 − β 2ω2µ ˜ Re ˜ − ω 4 µ ˜(Im ˜)2 = 0. 4 Use the quadratic formula to solve for β 2 , picking the positive sign so that β 2 > 0: s " # 2µ (Im ˜)2 ω ˜ Re  ˜ 2 1+ 1+ . β = 2 (Re ˜)2 So

v "s # u u1 p (Im ˜)2 t β=ω µ ˜ Re ˜ 1+ +1 . 2 (Re ˜)2

Then s s " # " # 2µ 2µ 2 2 ω ˜ Re  ˜ (Im  ˜ ) ω ˜ Re  ˜ (Im  ˜ ) α2 = β 2 − ω 2 µ ˜ Re ˜ = 1+ 1+ − ω2µ ˜ Re ˜ = −1 + 1 + 2 (Re ˜)2 2 (Re ˜)2 so

v "s # u u1 p (Im ˜)2 t α=ω µ ˜ Re ˜ 1+ −1 . 2 (Re ˜)2

4.32. Phasor fields:

E0 ˇ ˆ e−jβz . H(r) =y η0

ˇ ˆ E0 e−jβz , E(r) =x

Complex Poynting theorem:  Z I Z  1 1ˇ ˇ∗ 1 ˇ ˇ∗ 1 ∗ ∗ ˇ ˇ ˇ ˇ E · J dV = (E × H ) · dS − 2j ω ˇ E · D − B · H dV − 2 V 2 S 4 V 4 ˇ = 0. Next, The first term is zero because J I Z Z E02 1 1 E02 1 E02 ∗ ∗ ˇ ˇ ˇ ˇ ˆ E×H =z =⇒ (E × H ) · dS = dS − dS = 0. η0 2 S 2 top η0 2 bottom η0 Finally, 1ˇ ˇ∗ 1 ˇ ˇ∗ 1 1 E2 1 E2 1 E · D − B · H = 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 µ ˜˜ = 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 ω vp = = β

dω c = . dβ 2 ω ωc = . 2ω + 10c + 10

∴ 2ω c

122

vg =

4.34.   jωt ˇ ˆ E0 cos(ωt − βz) − y ˆ E0 sin(ωt − βz), E(r, t) = Re E(r)e =x   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 ˇ∗ 1ˇ 1 ˇ ˇ∗ ∗ ˇ ˇ0 e−jβr E0 ejβr 2 sin2 θr2 sin θ dθ dφ = Eθ Hφ dS = E E × H · dS = η0 (βr)2 S 2 0 0 S 2 Z Z π 2π 2 2 ˇ0 | 16π ˇ0 | 2 |E |E = . sin3 θ dθ = dφ 2 η0 β 0 η0 3β 2 0 Z

Pav

4.36. (a) ˇit =

I C

ˇ · dl = H

Z

2π

0

ˇ0 a ˇ E ˆ · φρ ˆ dφ = 2π E0 a . φ ηρ η

(b) Z L Z 2π Z b ˇ0 |2 a2 σ |E σ ˇ ˇ∗ E · E dV = e−2αz ρ dρ dφ dz 2 2 2 ρ 0 0 a V ˇ0 |2 a2 1 − e−2αL σ|E = 2π ln(b/a). 2 2α Z

Pd =

(c) Z Pcs (z) =

ˇ ×H ˇ ∗ ) dS = ˆ · (E n

Z

S

0

2π

Z a

b

ˇ0 |2 a2 |E e−2αz ρ dρ dφ η ∗ ρ2

ˇ0 |2 a2 |E = 2πe−2αz ln(b/a). η∗ (d) Ps = 0 = Pd + j2ˇ ω (Wm − We )Pf =⇒ Pd = − Re[Pf ]. ˇ 0 |2 a2 |E Pf = Pcs z=L − Pcs z=0 = − 2π(1 − e−2αL ) ln(b/a), η∗ ˇ0 |2 a2 1 − e−2αL σ|E ˇ0 |2 a2 Re[1/η ∗ ]2π(1 − e−2αL ) ln(b/a), 2π ln(b/a) = |E 2 2α σ α= . 4 Re[1/η ∗ ] 4.37. From (4.454) we have the structure reflection coefficients for each case as ˜ ˜2 ˜2 ˜ ˜2 ˜2 ˜2 ˜ A = Γ + Γ PA − ΓP − P PA , R ˜ P˜ 2 − Γ ˜ 2 P˜ 2 − Γ ˜ P˜ 2 P˜ 2 1+Γ A B

˜ ˜2 ˜2 ˜ ˜2 ˜2 ˜2 ˜ B = Γ + Γ PB − ΓP − P PB . R ˜ P˜ 2 − Γ ˜ 2 P˜ 2 − Γ ˜ P˜ 2 P˜ 2 1+Γ B B

123

Solving each equation for P˜ 2 gives, respectively, ˜A − Γ ˜ ˜+Γ ˜ 2 P˜ 2 − R ˜ P˜ 2 R Γ A A A P˜ 2 = ˜A − Γ ˜ ˜ P˜ 2 R ˜ + P˜ 2 − Γ ˜ 2R Γ A A A ˜B − Γ ˜ ˜+Γ ˜ 2 P˜ 2 − R ˜ P˜ 2 R Γ B B B P˜ 2 = . ˜B − Γ ˜B ˜ P˜ 2 R ˜ + P˜ 2 − Γ ˜ 2R Γ

(*) (**)

B

B

Equating and clearing the denominators gives    ˜A − Γ ˜A Γ ˜B − Γ ˜B ˜+Γ ˜ 2 P˜ 2 − R ˜ P˜ 2 R ˜ + P˜ 2 − Γ ˜ 2R ˜ P˜ 2 R Γ A A B B    ˜A − Γ ˜A . ˜B − Γ ˜B Γ ˜ + P˜A2 − Γ ˜ 2R ˜ P˜A2 R ˜ P˜B2 R ˜+Γ ˜ 2 P˜B2 − R = Γ Multiplying an collecting terms we find ˜4 − CΓ ˜3 + CΓ ˜−1=0 Γ where

 C=

˜A − R ˜B R

     ˜ AR ˜B 1+R 1 + P˜A2 P˜B2 + P˜A2 − P˜B2 ˜ B P˜ 2 − R ˜ A P˜ 2 R A B

.

This takes the form of (4.461) with the solution given by s  C C 2 ˜ Γ= ± − 1. 2 2 ˜ is found, P˜ may be computed using either (*) or (**). Once Γ 4.38. As in the two thickness method, let ∆B = κ∆A . Then P˜A2 = e−2jkz ∆ = Q and P˜B2 = e−2jkz (κ∆) = Qκ . The transmission coefficients are then ˜2 ˜ ˜2 ˜ A = (1 − Γ )PA = (1 − Γ )Q , T ˜ 2 P˜ 2 ˜ 2 Q2 1−Γ 1−Γ A ˜2 ˜ ˜2 κ ˜ B = (1 − Γ )PB = (1 − Γ )Q . T ˜ 2 P˜ 2 ˜ 2 Q2κ 1−Γ 1−Γ B

˜ 2 gives Solving each equation for Γ ˜ ˜ 2 = Q − TA , Γ ˜ A Q2 Q−T κ ˜ ˜ 2 = Q − TB . Γ ˜ B Q2κ Q−T Equating the expressions and clearing the denominators we find ˜ ˜ A Q2κ−1 − TA Qκ+1 + Q2κ − T ˜B T 124

˜A T ˜ A Q − 1 = 0. Qκ−1 + T ˜B T

(*) (**)

For the special case of κ = 2 this reduces to Q4 − CQ3 + CQ − 1 = 0 where

˜ ˜ A + TA . C=T ˜B T

This takes the form of (4.461) with the solution given by s  C C 2 − 1. Q= ± 2 2 ˜ may be computed using either (*) or (**). Once Q and thus P˜A is found, Γ

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 Jxi , Jzi ,

symmetry about z = 0

i Jyi , Jmz i , Ji 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 ˆ ˜ E = ˆr 2 + cos θ + θ jωµ0 + + sin θ. 4π r r jω0 r2 4π r r jω0 r2 ˆ ˜ = f1 (r) cos θˆr + f2 (r) sin θθ. Write E Case 1: horizontal dipole. See Figure 35.

Figure 35 ˜ and assume ˜Ii and ˜I are directed oppositely. Then ri = r, θi = θ, and Choose di = d, |I˜i | = |I|, ˜ri )tan = −(E ˜r )tan , (E

˜θi )tan = −(E ˜θ )tan . (E

So ˜ +E ˜ i )tan = 0 (E 126

Figure 36 and the boundary condition is satisfied. Case 2: vertical dipole. See Figure 36. ˜ and assume ˜Ii and ˜I 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 ˜ri )tan = −(E ˜r )tan , (E

˜θi )tan = −(E ˜θ )tan . (E

So ˜ +E ˜ i )tan = 0 (E 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 ωµ ˜I˜ ∞+j∆ e−jky |y−h| −jkx x i ˜ Ez (x, y) = − e dkx . 2π −∞+j∆ 2ky The scattered field consists of both upward and downward traveling waves: Z ∞+j∆ h i 1 s ˜ Ez (x, y) = A˜+ (kx )e−jky y + A˜− (kx )ejky y e−jkx x dkx . 2π −∞+j∆ ˜i + E ˜ s = 0 at y = ±d to get the equations To find A˜+ and A˜− , apply the boundary condition E z z ωµ ˜I˜ −jky (d−h) e = A˜+ e−jky d + A˜− ejky d , 2ky ωµ ˜I˜ −jky (d+h) e = A˜+ ejky d + A˜− e−jky d . 2ky Solving simultaneously, we obtain ωµ ˜I˜ −jky d sin ky (d + h) A˜− = e , 2ky sin 2ky d

ωµ ˜I˜ −jky d sin ky (d − h) A˜+ = e . 2ky sin 2ky d 127

Therefore ˜I˜ ˜zs = ω µ E 2π

∞+j∆

Z

−∞+j∆

  e−jky d sin ky (d − h) −jky y sin ky (d + h) jky y −jkx x e + e e dkx . 2ky sin 2ky d sin 2ky d

The term in brackets can be rewritten algebraically to put this into the form of (5.8). ˜ i = 0, J ˜ i 6= 0. From (5.71) we have 5.4. Assume J m ˜ h (r) = A

Z V

˜c ˜ i 0 e−jkR J (r ) dV 0 = ˜c 4π m R

Z

˜ i (r0 )G(r|r0 ) dV 0 . J m

V

Examine (5.53): ˜ h ) − jω A ˜h ˜ = −j ω ∇(∇ · A H k2 Z Z ω˜ c i 0 0 0 c ˜ ˜ i (r0 )G(r|r0 ) dV 0 Jm (r )G(r|r ) dV − jω˜ J  = −j 2 ∇∇ · m k V V Z  h i 1 i 0 0 i 0 0 c ˜ (r )G(r|r ) + ∇∇ · J ˜ (r )G(r|r ) J dV 0 . = −jω˜  m m 2 k V Use

h i ˜ i (r0 )G(r|r0 ) = J ˜ i (r0 ) · ∇G(r|r0 ) + G(r|r0 ) ∇ · J ˜ i (r0 ) ∇· J m m | {zm } =0

to write ˜ = −jω˜ H c c

Z  ZV

= −jω˜ 

˜ i (r0 )G(r|r0 ) + 1 ∇∇G(r|r0 ) · J ˜ i (r0 ) J m m k2



dV 0

¯ e (r|r0 ) · J ˜ i (r0 ) dV 0 G m

V

where

  ∇∇ 0 ¯ ¯ Ge (r|r ) = I + 2 G(r|r0 ). k

Next, examine (5.63): 1 ˜ ˜h E(r) =− ∇×A ˜c Z ˜ i (r0 )G(r|r0 ) dV 0 = −∇ × J m V Z ˜ i (r0 )G(r|r0 )] dV 0 . =− ∇ × [J m V

Use h i ˜ i (r0 ) × ∇G(r|r0 ) ˜ i (r0 )G(r|r0 ) = ∇G(r|r0 ) × J ˜ i (r0 ) + G(r|r0 ) ∇ × J ˜ i (r0 ) = −J ∇× J m m m {zm } | =0

128

and (B.15): h i ˜ i (r0 ) × ∇G(r|r0 ) = J ˜ i (r0 ) × ∇G(r|r0 ) · ¯I J m m   i 0 ˜ (r ) = ∇G(r|r0 ) × ¯I · J m

¯ m (r|r0 ) · J ˜ i (r0 ). =G m So ˜ E(r) =

Z

¯ m (r|r0 ) · J ˜ i (r0 ) dV 0 . G m

V

5.5. Start with V = lim ∇ ∆V →0

I ˜ 0 ˆ J·n dS 0 . 4πR S

˜=z ˆJ˜z with J˜z constant. Then Assume J   I I R 1 0 0 ˜·n ˜ ˆ0 ˆ∇ dS = − lim J dS 0 . V = lim J·n ∆V →0 S ∆V →0 S 4πR 4πR3 We have

 ˜  Jz , 0 ˜ ˆ = −J˜z , J·n   0,

on the top surface of the cube, on the bottom surface, on the other surfaces.

Assume r = 0 (center of cube) so that ( ˆ + y0y ˆ, aˆ z + x0 x −R = −(r − r ) = r = 0 ˆ + y0y ˆ, −aˆ z+xx 0

0

on top, on bottom.

Then Z

R V = − lim 2 J˜z dS 0 ∆V →0 4πR3 top Z aZ a ˆ ˆ + y0y aˆ z + x0 x 0 0 = lim 2 J˜z 2 2 3/2 dx dy 2 0 0 a→0 4π(a + x + y ) −a −a Z Z 2a ˜ a a dx0 dy 0 ˆ lim =z Jz 2 02 0 2 3/2 a→0 4π −a −a (a + x + y ) Z a 2a ˜ dy 0 p ˆ lim Jz · 2a =z 2 2 a→0 4π −a (a2 + y 0 ) 2a2 + y 0 2a ˜ 4π ˆ lim =z Jz · a→0 4π 6a 1˜ ˆ Jz . =z 3 ¯ ·J ˜ =L ¯ · (ˆ ˆ) · n ˆ = 0 on the side Setting this equal to L zJ˜z ), we find that Lzz = 1/3. Since (J˜z z ˜ = x ˜ = y ˆ J˜x and J ˆ J˜y , we see immediately that surfaces, we have Lxz = Lyz = 0. By letting J Lxx = Lyy = Lzz = 1/3 and Lyx = Lyz = Lzx = Lzz = 0. Therefore   1/3 0 0 ¯ =  0 1/3 0  = 1 ¯I. L 3 0 0 1/3 129

5.6. Start with

I ˜ 0 ˆ J·n dS 0 . V = lim ∇ ∆V →0 S 4πR

˜=z ˆJ˜z with J˜z constant. Then Assume J   I I R 1 0 0 ˜ ˜·n ˆ∇ ˆ0 dS = − lim J dS 0 . V = lim J·n ∆V →0 S ∆V →0 S 4πR 4πR3 We have

 ˜  on the top surface of the cylinder, Jz , 0 ˜ ˜ ˆ = −Jz , on the bottom surface, J·n   0, on the curved surface.

Assume r = 0 (center of cylinder) so that ( ˆ + y0y ˆ = aˆ ˆ + ρ0 sin φ0 y ˆ, aˆ z + x0 x z + ρ0 cos φ0 x on top, 0 0 −R = −(r − r ) = r = ˆ + y0y ˆ = −aˆ ˆ + ρ0 sin φ0 y ˆ , on bottom. −aˆ z + x0 x z + ρ0 cos φ0 x Then Z

R dS 0 J˜z 3 ∆V →0 4πR top Z 2π Z a ˆ + ρ0 sin φ0 y ˆ 0 0 0 aˆ z + ρ0 cos φ0 x = lim 2 J˜z ρ dρ dφ 2 + ρ0 2 )3/2 a→0 4π(a 0 0 " # a 1 ˜ 1 ˆ lim 2(2π) Jz a − p =z a→0 4π ρ0 2 + a2 0   1 1 = −ˆ z lim J˜z a √ − a→0 2a2 a   1 ˆJ˜z 1 − √ =z 2 ˜ ˆ. = 0.293Jz z

V = − lim 2

¯ · (ˆ Setting this equal to L zJ˜z ), we find that Lzz = 0.293. 5.7. (a) −jkr ˜ ω) = lim e = lim 1 = 0. lim ψ(r, r→∞ r→∞ 4πr r→∞ 4πr (b)   ∂ ψ˜ ∂ e−jkr 1 e−jkr = = (−jkr − 1) 2 , ∂r ∂r 4πr 4π r ! −jkr   −jkr −jkr ˜ e ∂ ψ 1 e = lim − e = 0. + (−jkr − 1) 2 lim r jk ψ˜ + = lim r jk r→∞ r→∞ r→∞ ∂r 4πr 4π r 4πr

130

5.8. Begin with ∇2 A = ∇(∇ · A) − ∇ × (∇ × A) Use ∇·A=

ˆ and ∇ = ∇t + u

∂ . ∂u

  ∂Au ∂ ˆ Au ) = ∇ t · A t + ˆ · (At + u ∇t + u ∂u ∂u

to write    ∂Au ∂ ˆ ∇t · A t + ∇(∇ · A) = ∇t + u ∂u ∂u ∂Au ∂ 2 Au ∂At ˆ ˆ ∇t · +u + ∇t . = ∇t (∇t · At ) + u ∂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): ˆ · At = u ˆ · [A − u ˆ Au ] = u ˆ · A − Au = Au − Au = 0. u (B.89):  (ˆ u · ∇t )φ =



∂ ˆ · ∇−u ˆ u ∂u

 ˆ · ∇φ − φ=u

∂φ = 0. ∂u

(B.90):   ∂ ∂φ ˆ ˆ . φ = ∇φ − u ∇t φ = ∇ − u ∂u ∂u (B.91): 

 ∂φ ∂φ ∂φ ˆ · (∇φ) = u ˆ · ∇t φ + u ˆ ˆ · (∇t φ) + u =u = , ∂u ∂u ∂u    ∂ ∂φ ∂φ ˆ · ∇t + u ˆ ˆ ·u ˆ (ˆ u · ∇)φ = u φ=u = . ∂u ∂u ∂u (B.92):   ∂φ ∂φ ∂φ ˆ · ∇φ − u ˆ ˆ · (∇t φ) = u u = − = 0. ∂u ∂u ∂u 131

(B.93):   ∂ ∂φ ∂φ ∂φ ∂φ ˆ ˆ +u ˆ · ∇φ − ∇t · (ˆ uφ) = ∇ − u · (ˆ uφ) = ∇ · (ˆ uφ) − = φ∇ · u = − = 0. ∂u ∂u ∂u ∂u ∂u (B.94):   ∂φ ∂ ˆ) ˆ ˆ −u ˆ × ∇φ × (ˆ uφ) = ∇ × (ˆ uφ) − (ˆ u×u ∇t × (ˆ uφ) = ∇ − u = φ∇ × u ∂u ∂u   ∂φ ∂ ˆ ×u ˆ ˆ φ = −ˆ u × ∇t φ − u = −ˆ u × ∇t φ. = −ˆ u × ∇t + u ∂u ∂u (B.95): 

 ∂ ˆ ˆ Au )] ∇t × (ˆ u × A) = ∇ − u × [ˆ u × (At + u ∂u   ∂ ˆ = ∇−u × (ˆ u × At ) ∂u ∂At ˆ ×u ˆ× = ∇ × (ˆ u × At ) − u ∂u  ∂At ∂At ˆ u ˆ· + = ∇ × (ˆ u × At ) − u ∂u ∂u ∂At = ∇ × (ˆ u × At ) + ∂u ∂At ˆ ∇ · At − (ˆ =u u · ∇)At + ∂u   ∂At ∂At ∂ ˆ ∇t + u ˆ · At − + =u ∂u ∂u ∂u ˆ ∇t · A t . =u (B.96): ˆ × (∇t × A) = u ˆ × [∇t × (ˆ u uAu + At )] ˆ × [−ˆ =u u × ∇t Au + (∇t × At )] ˆ × (∇t × At ) = −ˆ u × (ˆ u × ∇ t Au ) + u ˆ) = −ˆ u(ˆ u · ∇t Au ) + ∇t Au (ˆ u·u = ∇ t Au .

132

(B.97):    ∂ ˆ × (∇t × At ) = u ˆ × ∇−u ˆ × At u ∂u   ∂At ˆ × (∇ × At ) − u ˆ× u ˆ× =u ∂u   ∂At ∂At ˆ) ˆ u ˆ· + (ˆ u·u = ∇(ˆ u · At ) − (ˆ u · ∇)At − u ∂u ∂u ∂At ∂At =− + ∂t ∂t = 0. (B.98): ˆ · (ˆ ˆ ) = 0. u u × A) = A · (ˆ u×u (B.99): ˆ × (ˆ ˆ (ˆ ˆ ) = −[A − u ˆ (ˆ u u × A) = u u · A) − A(ˆ u·u u · A)] = −At . 5.10. (B.100):   ∂φ ∂ ˆ . ˆ φ = ∇t φ + u ∇φ = ∇t + u ∂u ∂u (B.101): 

 ∂ ˆ ˆ Au ) ∇ · A = ∇t + u · (At + u ∂u ∂ ∂Au = ∇t · A t + (ˆ u · At ) + ∇t · (ˆ uAu ) + ∂u ∂u ∂Au = ∇t · A t + . ∂u (B.102):   ∂ ˆ ˆ Au ) ∇ × A = ∇t + u × (At + u ∂u ∂At ∂Au ˆ× ˆ) = ∇t × A t + u + ∇t × (ˆ uAu ) + (ˆ u×u ∂u ∂u  ∂At ˆ× − ∇ t Au . = ∇t × A t + u ∂u (B.103): 

∂φ ˆ ∇ φ = ∇ · (∇φ) = ∇ · ∇t φ + u ∂u 2



∂ = ∇t · (∇t φ) + ∂u

133



∂φ ∂u



= ∇2t φ +

∂2φ . ∂u2

(B.104): 

 ∂At ˆ× ∇ × (∇ × A) = × ∇t × A t + u − ∇ t Au ∂u      ∂At ∂At ˆ × ∇t × ˆ× + ∇t × u = ∇t × (∇t × At ) + u − ∇ t Au ∂u ∂u    2 ∂Au ∂ At ˆ× u ˆ× +u − ∇t . 2 ∂u ∂u Use

∂ ˆ ∇t + u ∂u





ˆ× ∇t × u and







∂At − ∇ t Au ∂u



 ˆ ∇t · =u



∂At − ∇ t Au ∂u



  2    2 ∂Au ∂ At ∂Au ∂ At ˆ× u ˆ× − ∇ = − − ∇ u t t ∂u2 ∂u ∂u2 ∂u

to get, finally, ∇ × (∇ × A) = ∇t × (∇t × At ) −

  ∂ 2 At ∂Au ∂ 2 ˆ − ∇ − u (∇ · A ) − ∇ A t t t t u . ∂u2 ∂u ∂u

ˆ × ∂/∂u of (5.90): 5.11. Take u −ˆ u×

˜u ∂E ˆ × ∇t u ∂u

!

Use −ˆ u×

˜u ∂E ˆ × ∇t u ∂u

ˆ× +u

!

˜t ∂2E ˆ× u ∂u2

˜u ∂E = ∇t ∂u

! ˆ× = −jω µ ˜u

ˆ× and u

˜t ˜i ∂H ∂J mt ˆ× −u . ∂u ∂u

˜t ∂2E ˆ× u ∂u2

! =−

˜t ∂2E ∂u2

to get ˜t ˜t ˜i ˜u ∂ 2 E ∂H ∂J ∂E mt ˆ ˆ − = −jω µ ˜ u × − u × . ∂u ∂u2 ∂u ∂u Now multiply (5.92) by −jω µ ˜: ∇t

˜t ∂H ˜i . ˜ u − jω µ ˆ × jω µ ˆ× = k 2 Et − jω µ ˜J u ˜ ∇t H ˜u t ∂u Add (*) and (**): ∇t or 

˜t ˜i ˜u ∂ 2 E ∂E ∂J mt ˜ t − jω µ ˜i − u ˜ u = k2 E ˆ ˆ − + u × jω µ ˜ ∇ H ˜ J × t t ∂u ∂u2 ∂u

 ˜i ˜u ∂2 ∂E ∂J mt 2 ˜ ˜i . ˜u + u ˆ ˆ + jω µ ˜ u × ∇ H × + jω µ ˜J + k E = ∇ t t t t ∂u2 ∂u ∂u

134

(*)

(**)

5.12. Begin with ∇ × E = −Jm − µ

∂H , ∂t

∇×H=J+

Write ˆ Jmu − µ ∇ × E = −Jmt − u

∂E . ∂t

∂Ht ∂Hu − µˆ u . ∂t ∂t

Use (B.102) to get ˆ× ∇ t × Et + u

∂Et ∂Ht ∂Hu ˆ × ∇t Eu = −Jmt − u ˆ Jmu − µ −u − µˆ u . ∂u ∂t ∂t

Equate transverse components: ˆ× u

∂Ht ∂Et ˆ × ∇t Eu = −Jmt − µ −u . ∂u ∂t

Equate longitudinal components: ∇t × Et = −ˆ uJmu − µˆ u

∂Hu . ∂t

Do the same with Ampere’s law: ˆ× ∇t × Ht + u

∂Et ∂Eu ∂Ht ˆ × ∇t Hu = Jt + u ˆ Ju +  −u + ˆ u , ∂u ∂t ∂u

ˆ× u

So

∂Ht ∂Et ˆ × ∇ t Hu = J t +  −u , ∂u ∂t ∂Eu ˆ Ju + ˆ . ∇t × Ht = u u ∂t

ˆ× u

∂Et ∂Ht ˆ × ∇t Eu = −Jmt − µ −u ∂u ∂t

(1)

ˆ× u

∂Ht ∂Et ˆ × ∇t Hu = −Jt +  −u . ∂u ∂t

(2)

and

ˆ × ∂/∂u to (1): Apply u −

∂ 2 Et ∂Eu ∂Jmt ∂ 2 Ht + ∇ = −ˆ u × − µˆ u × . t ∂u2 ∂u ∂u ∂t∂u

(3)

∂ 2 Ht ∂Hu ∂Jt ∂ 2 Et − µˆ u × ∇t =µ + µ 2 . ∂t∂u ∂u ∂u ∂t

(4)

Apply µ ∂/∂t to (2): µˆ u×

Subtract (4) from (3) to get  2  ∂ 1 ∂2 ∂Eu ∂Hu ∂Jmt ∂Jt ˆ× − 2 2 Et = ∇ t + µˆ u × ∇t +u +µ . 2 ∂u v ∂t ∂u ∂u ∂u ∂u ˆ × ∂/∂u to (2): Next, apply u −

∂ 2 Ht ∂Jt ∂ 2 Et ˆ + ∇ H = u × + ˆ u × . t u ∂u2 ∂u ∂t∂u 135

(5)

Apply  ∂/∂t to (1): ˆ u×

∂ 2 Et ∂Eu ∂Jmt ∂ 2 Ht − ˆ u × ∇t = − − µ 2 . ∂t∂u ∂t ∂t ∂t

Add (5) and (6) to get   2 ∂Eu ∂ 1 ∂2 ∂Jmt ∂Jt ˆ× u × ∇t − 2 2 Ht = ∇t Hu − ˆ + −u . 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 

ˆ · to both sides to obtain Apply u ∇2 Eu − µ or



∂ 2 Eu ∂Ju 1 ∂ρ ˆ · (∇t × Jmt ) + µ =u + , 2 ∂t ∂t  ∂u

1 ∂2 ∇ − 2 2 v ∂t 2

 Eu =

1 ∂ρ ∂Ju ˆ · (∇t × Jmt ). +µ +u  ∂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 ∂ ˆ Hy ) = x ˆ ˆ ˆ ˆ − (ˆ xHx + y +y − ˆ z× x +y . ∂z 2 v 2 ∂t2 ∂x∂z ∂y∂z ∂x∂t ∂y∂t Equate x-components to obtain: 

∂2 1 ∂2 − ∂z 2 v 2 ∂t2

Equate y-components to obtain: 

∂2 1 ∂2 − ∂z 2 v 2 ∂t2

Hx =

∂ 2 Hz ∂ 2 Ez + . ∂x∂z ∂y∂t

Hy =

∂ 2 Hz ∂ 2 Ez − . ∂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 ˆ Ey ) = x ˆ ˆ ˆ ˆ − (ˆ xEx + y +y + µˆ z× x +y . ∂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 − ∂z 2 v 2 ∂t2

 Ey =

∂ 2 Ez ∂ 2 Hz +µ . ∂y∂z ∂x∂t

The third and fourth equations from Problem 5.12 with u = z yield the wave equations. 136

(6)

ˆ Πe and Πh = 0. Then 5.14. Let Πe = u   ∂2 2 ˆ Πe = 0. ∇ − µ 2 u ∂t Use (B.105): ˆ ∇ 2 Πe , ∇2 (ˆ uΠe ) = u so

  1 ∂2 2 ∇ − 2 2 Πe = 0. v ∂t

(*)

We also have ˆ Πe ). E = ∇ × (∇ × Πe ) = ∇ × (∇ × u Use (B.104): E = ∇t

∂Πe ˆ ∇2t Πe . −u ∂u

By (*) we have ∂ 2 Πe 1 ∂ 2 Πe = , ∂u2 v 2 ∂t2  2  ∂Πe ∂ Πe 1 ∂ 2 Πe ˆ E = ∇t +u − 2 . ∂u ∂u2 v ∂t2 ∇2 Πe = ∇2t Πe +

so

Finally, H = ∇ ×

∂ ∂ (ˆ uΠe ) =  ∇ × (ˆ uΠe ). ∂t ∂t

Use (B.102): H=

∂ ∂Πe (−ˆ u × ∇t Πe ) = −ˆ u × ∇t . ∂t ∂t

ˆ Πh and Πe = 0. Then 5.15. Let Πh = u   ∂2 ˆ Πh = 0. ∇2 − µ 2 u ∂t Use (B.105): ˆ ∇2 Πh , ∇2 (ˆ uΠh ) = u so

  1 ∂2 2 ∇ − 2 2 Πh = 0. v ∂t

We also have ˆ Πh ). H = ∇ × (∇ × Πh ) = ∇ × (∇ × u Use (B.104): E = ∇t

∂Πh ˆ ∇2t Πh . −u ∂u

By (*) we have ∇2 Πh = ∇2t Πh +

∂ 2 Πh 1 ∂ 2 Πh = , ∂u2 v 2 ∂t2

137

(*)

so ∂Πh ˆ H = ∇t +u ∂u



∂ 2 Πh 1 ∂ 2 Πh − ∂u2 v 2 ∂t2

 .

Finally, from (5.54), ∂ ∂ (ˆ uΠh ) = −µ ∇ × (ˆ uΠh ). ∂t ∂t

E = −µ∇ × Use (B.102): E = −µ

∂ ∂Πh (−ˆ u × ∇t Πh ) = µˆ u × ∇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 = −ˆ u × ∇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 = µˆ u × ∇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

˜0 E ˜ =y ˆ e−jkz . H η

˜ =x ˜0 e−jkz , ˆE E

138

(a) TEy , use (5.96): ˜y ∂E ˜y ˆ × ∇t H + jω µ ˜y ∂y ! ˜0 ∂ E −jkz ˆ×z ˆ = jω µ ˜y e ∂z η

˜ t = ∇t k2 E

ˆ (−jk) = jω µ ˜x = k 2 ηˆ x

˜0 E e−jkz η

˜0 E e−jkz . η

So ˜t = x ˜0 e−jkz . ˆE E (b) TMx , use (5.95): ˜x ˜x ∂H ∂E ˆ×z ˆ − jω˜ c x ∂x ∂z   c ∂ −jkz ˜0 e ˆ = jω˜ y E ∂z ˜0 e−jkz = ω˜ c kˆ yE

˜ t = ∇t k2 H

= So

k 2 ˜ −jkz ˆ E0 e y . η ˜0 E ˜t = y ˆ e−jkz . H η

(c) Use (5.124): ˜ = jω µ ˜h ˆ × ∇t Π E ˜y ∂ ˆ×z ˆ = jω µ ˜y ∂z

˜0 E e−jkz ηk 2

!

˜0 E e−jkz ηk 2 ˜0 e−jkz . ˆE =x

= ωµ ˜kˆ x

Use (5.125):  2  ˜h ˜0 ∂Π ∂ E 2 ˜ ˜ ˜h = y ˆ ˆ k2 Π ˆ e−jkz . +y + k Πh = y H = ∇t 2 ∂y ∂y η (d) Use (5.122):  2  ˜e ∂Π ∂ 2 ˜ ˜ ˜e = x ˜0 e−jkz . ˆ ˆ k2 Π ˆE E = ∇t +x + k Πe = x ∂x ∂x2

139

Use (5.123): ˜ = −jω˜ ˜e ˆ × ∇t Π H c x ˜0 E e−jkz k2

∂ ˆ×z ˆ = −jω˜ c x ∂z = ω˜ c kˆ y ˆ =y

!

˜0 E e−jkz k2

˜0 E e−jkz . η

(e), (f) Similar. 5.19. Using the given identity, we have Z

(ψ˜m ∇ ψ˜n − ψ˜n ∇ ψ˜m ) dS = 2

2

I Γ

S

∂ ψ˜n ∂ ψ˜m ψ˜m + ψ˜n ∂n ∂n

! dl.

2 ψ ˜n : Use ∇2 ψ˜n = ∇2t ψ˜n = −kc,n

Z

2 ˜ ˜ (−kc,n ψn ψm

+

2 ˜ ˜ kc,m ψm ψn ) dS

=

2 (kc,m

−

2 kc,n )

S

Z

ψ˜n ψ˜m dS =

S

I Γ

∂ ψ˜m ∂ ψ˜n − ψ˜n ψ˜m ∂n ∂n

! dl.

The boundary conditions are that ψ˜e = 0 on Γ for TM modes and ∂ ψ˜h /∂n = 0 on Γ for TE modes. Since kc,m 6= kc,n , we must have Z ψ˜n ψ˜m dS = 0. S

So

Z

2 ˜ 2 ˜ (kc,n ψn )(kc,m ψm ) dS = 0.

S

For TM modes,

2 ψ ˜n kc,n

= ez,n . In this case Z e˜z,n e˜z,m dS = 0 S

so that

Z 

e˜z,n e±jkz,n z

  e˜z,m e±jkz,m z dS = 0

S

or

Z

˜z,n E ˜z,m dS = 0. E

S

TE modes are treated similarly. 5.20. (a) For TM modes we write ˜ =E ˜t + z ˜z , ˆE E

˜m · E ˜n = E ˜ tm · E ˜ tn + E ˜zm E ˜zn . E

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 ˜ ˜ Etm · Etn = −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 a Z b βπx δπx απy γπy ∓jkzn z ∓jkzm z βπ δπ ˜ ˜ cos Etm · Etn 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

But Z 0

so

a

Z b απy βπx δπx γπy sin cos cos dx = sin dy = 0, a a b b 0 Z ˜ tm · E ˜ tn dS = 0. E CS

Next, ˜zm E ˜zn = k 2 k 2 Aαβ Aγδ e∓jkzn z e∓kzm z sin βπx sin απy sin δπx sin γπy E cn cm a b a b Z Z a Z b δπx απy γπy βπx 2 2 ˜zm E ˜zn dS = kcn sin dx sin dy = 0. E kcm Aαβ Aγδ e∓jkzn z e∓kzm z sin sin a a b b CS 0 0 Thus

Z

˜m · E ˜ n dS = 0. E

CS

(b) Write ˜m ·H ˜n = H ˜ tm · H ˜ tn H 

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

e

 Z a Z b βπx δπx απy γπy ∓jkzn z ∓jkzm z απ γπ ˜ ˜ Hm · Hn dS = kzn kzm Yem Yen Aαβ Aγδ e e sin sin dx cos cos dy b b 0 a a b b 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

Z

The treatment for TE modes is similar.

141

5.21. ˇ ∗) = z ˇ t ) × (−jk ∗ )∇t ψˇ∗ ] ˆ · (ˇ ˆ · [Zh (ˆ z et × h z×h t z h ∗ ˇ ˆ · [(ˆ = −kz k Zh z z × ∇t ψh ) × ∇t ψˇ∗ ] z

h

ˆ · [∇t ψˇh∗ × (ˆ = |kz | Zh z z × ∇t ψˇh )] ˆ · [ˆ = |kz |2 Zh z z(∇t ψˇ∗ ) · (∇t ψˇh ) − ∇t ψˇh (ˆ z · ∇t ψˇ∗ )] 2

h

h

= |kz | Zh (∇t ψˇh ) · (∇t ψˇh∗ ). 2

Pav =

1 Re 2

Z

ˇ ∗ ) dS = ˆ · (ˇ z et × h t

CS

1 Re{Zh }|kz |2 2

Z

(∇t ψˇh ) · (∇t ψˇh∗ ) dS.

CS

Use (B.35): Z (∇t a · ∇t b +

a∇2t b) dS

I a

= Γ

S

∂b dl. ∂n

Let a = ψˇh and b = ψˇh∗ . Then Z I ∂ ψˇ∗ 2 ˇ∗ ∗ ˇ ˇ ˇ (∇t ψh · ∇t ψh + ψh ∇t ψh ) dS = ψˇh h dl. ∂n CS Γ Since ∂ ψˇh /∂n = 0 on Γ and ∇2t ψˇh = −kc2 ψˇh , we have Z Z Z ∗ 2 ˇ∗ 2 ˇ ˇ ˇ (∇t ψh ) · (∇t ψh ) dS = − ψh ∇t ψh dS = kc CS

CS

ψˇh ψˇh∗ dS.

CS

Therefore

Z 1 2 2 Pav = Re{Zh }|kz | kc ψˇh ψˇh∗ dS. 2 CS For a lossless guide, kz = β and Zh = ω ˇ µ/β. In this case, Z Z 1ω 1 ˇµ 2 2 ∗ 2 ˇ ˇ Pav = β kc ψh ψh dS = ω ˇ µβkc ψˇh ψˇh∗ dS. 2 β 2 CS CS

5.22. We have  hwe i = 4 so

Z lZ 0

hwe i  = l 4

ˇ ·E ˇ ∗ dS dz, E

CS

Z

ˇ ·E ˇ ∗ dS. E

CS

Substitute from (5.199) and (5.200): Z hwe i  = |Zh |2 |kz |2 (ˆ z × ∇t ψˇh ) · (ˆ z × ∇t ψˇh∗ ) dS l 4 CS  2 Z  ω ˇµ β2 (ˆ z × ∇t ψˇh ) · (ˆ z × ∇t ψˇh∗ ) dS = 4 β CS Z  2 ˆ · [∇t ψˇh∗ × (ˆ = (ˇ ω µ) z z × ∇t ψˇh )] dS 4 ZCS  2 = (ˇ ω µ) ∇t ψˇh∗ · ∇t ψˇh dS 4 CS Z  2 2 = (ˇ ω µ) kc ψˇh ψˇh∗ dS 4 CS 142

(cf., Problem 5.21). 5.23. For TM modes, 1 Pav = ω ˇ βkc2 2 Use

Z

ψˇe ψˇe∗ dS.

CS

 mπy   nπx  sin ψˇe (x, y) = Anm sin a b

to get (Pav )nm

Z a Z b    mπy  1 2 nπx 2 2 sin = ω ˇ βnm kc,nm |Anm | dx sin2 dy 2 a b 0 0 1 a b 2 = ω ˇ βnm kc,nm |Anm |2 · · . 2 2 2

5.24. For TE modes, Pav Use

1 ˇ µβkc2 = ω 2

Z

ψˇh ψˇh∗ dS.

CS

 mπy   nπx  cos ψˇh (x, y) = Bnm cos a b

to get Z b Z a  nπx   mπy  1 2 cos2 dx dy ˇ µβnm kc,nm |Bnm |2 cos2 (Pav )nm = ω 2 a b 0 0 1 a b 2 = ω ˇ µβnm 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 ˜ s exists along the direction θ = π. This is given by x-direction, only the x-component of E s s ˜ ˜ Ex = −Eθ . θ=π θ=π

φ=0

In the far zone we have ˜s = E ˜0 e E θ

−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 lim

θ→π

d 1 P (cos θ) = lim (− sin θ)Pn10 (cos θ) θ→π dθ n

where

Pn10 (x) =

dPn1 (x) . dx

So ˜ Ex

˜0 e = −E θ=π

=

−jkr

kr

cos(0)

∞ X

j

n+1



n=1

(−1)n (−1)n − n(n + 1)bn + n(n + 1)cn 2 2



∞ ˜0 e−jkr X E j n+1 n(n + 1)(cn − bn ). 2 kr n=1

Use cn = −

Jˆn (ka) an , ˆ n(2) (ka) H

bn = −

Jˆn0 (ka) an ˆ n(2)0 (ka) H

to get "

# " # ˆ n(2) (ka) − H ˆ n(2)0 (ka)Jˆn (ka) Jˆn0 (ka) Jˆn (ka) Jˆn0 (ka)H − (2) an = an . cn − bn = ˆ n(2)0 (ka) H ˆ n (ka) ˆ n(2)0 (ka)H ˆ n(2) (ka) H H Use d Jˆn0 (x) = [xjn (x)] = jn (x) + xjn0 (x), dx (2) (2)0 ˆ n(2)0 (x) = d [xh(2) H n (x)] = hn (x) + xhn (x), dx and (E.99) to get ˆ (2) (x) − H ˆ (2)0 (x)Jˆn (x) Jˆ0 (x)H n

n

= [jn (x) +

n 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 ˆ n (ka)H ˆ n(2) (ka) H

an = j −n

where

2n + 1 . n(n + 1)

Therefore ˜ Ex

θ=π

∞ ˜0 e−jkr X E j −n (2n + 1) 1 = j n+1 (−1)n n(n + 1)j (2)0 2 kr n(n + 1) H ˆ n (ka)H ˆ n(2) (ka) n=1

∞ ˜0 e−jkr X E 2n + 1 . =− (−1)n (2)0 2 kr ˆ n (ka)H ˆ n(2) (ka) H n=1

Now use (5.167):   s 2 2 |E | σ = lim 4πr , ˜ i |2 r→∞ |E

σ = lim 4πr2

˜i = x ˜0 e−jkz . ˆE E

2 2 ∞ e−jkr 2 X E ˜ 2n + 1 0 (−1)n (2)0 (2) 2 kr ˆ ˆ Hn (ka)Hn (ka) n=1

˜0 |2 |e−jkz |2 |E 2   ∞ 1 λ 2 X (−1)n (2n + 1) = 4π 4 2π ˆ (2)0 ˆ (2) n=1 Hn (ka)Hn (ka) 2 ∞ λ2 X (−1)n (2n + 1) = . 4π ˆ (2)0 ˆ (2) n=1 Hn (ka)Hn (ka) r→∞

5.27. Start with λ2 σ= 4π

2 ∞ X (−1)n (2n + 1) . ˆ (2)0 ˆ (2) n=1 Hn (ka)Hn (ka)

Use ˆ n(2) (x) = Jˆn (x) − j N ˆn (x) = xjn (x) − jxnn (x). H For x  1, (E.59) and (E.60) give jn (x) ≈ So

Therefore

2n n! xn , (2n + 1)!

nn (x) ≈ −

(2n)! −(n+1) x . 2n n!

ˆ (2) (x) ≈ −j N ˆn (x) = j (2n)! x−n , H n 2n n!

ˆ (2)0 (x) ≈ −nj (2n)! x−n−1 , H n 2n n!  2 ˆ (2) (x)H ˆ (2)0 (x) ≈ n (2n)! x−2n−1 . H n n 2n n!

2 ∞ X n (−1) (2n + 1) 2n+1 σ≈ . h i2 (ka) 4π (2n)! n=1 n 2n n! λ2

146

Since ka  1, the first term in the series dominates:   2  6 2+1 1 2 1 2 σ∼λ ∼ λ−4 . ∼λ λ λ 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 = −µ0 , ∂t   1 1 ∂2A − ∇2 A − (∇ × A) × ∇ = Ji − 0 2 , µ µ ∂t   2 ∂ A 1 ∇2 A − µ0 2 + µ∇ × A + ∇ = Ji . ∂t µ Because

  ∇µ 1 =− 2, ∇ µ µ

we can also write ∇2 A − µ0

∂2A ∇µ − (∇ × A) × 2 = Ji . 2 ∂t µ

5.29. A(r, t) = F

 Z µ0 e−jkR 0 0 ˜ J(r , ω) dV 4π V R Z i 1 −1 h ˜ 0 F J(r , ω)e−jω(R/c) dV 0 R ZV 1 J(r0 , t − R/c) dV 0 . V R

−1

µ0 4π µ0 = 4π =



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 ˆ λx + y ˆ λy , λ=x

|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

G(λ|ρ0 ) = −

e−jλ·ρ . k 2 − λ2

Therefore G(ρ|ρ0 ) = −

1 (2π)2

1 =− (2π)2

∞

0

e−jλ·ρ jλ·ρ 2 e d λ 2 2 −∞ k − λ "Z Z ∞ 0 dλx ejλx (x−x )

ZZ

Write k 2 − λ2x − λ2y =

−∞

∞

−∞

# 0 ejλy (y−y ) dλy . 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) (2) H0 (k|ρ

1 − ρ0 |) = − π

Z

∞

−∞

0

√

2

ejα(x−x ) e−j k −α √ k 2 − α2

to finish the problem. 149

2 |y−y 0 |

dα

Figure 39 5.31. See Figure 40.

Figure 40

˜ ˜y = − ∂ψ , H ∂x

˜ ˜x = − kz ∂ ψ , E ω˜  ∂x

2 ˜ ˜z = kx ψ. E jω˜ 

Region 1: ψ˜1 = C1 sin kx1 xe−jkz z . Region 0: ψ˜0 = (C2 sin kx0 x + C3 cos kx0 x)e−jkz z . ˜z = 0 at x = d =⇒ ψ˜ = 0 at x = d : Boundary condition: E 2 2 C2 sin kx0 d2 + C3 cos kx0 d2 = 0 " ψ˜0 = C2 sin kx0 x − C2

sin kx0 d2

=⇒

C3 = −C2

sin kx0 d2 cos kx0 d2

.

#

cos kx0 x = C¯2 sin kx0 (x − d2 ). cos kx0 d2 ( ˜ −kx0 C¯2 cos kx0 (x − d2 ), region 0, ∂ ψ ˜y = − H = e−jkz z ∂x −kx1 C1 cos kx1 x, region 1. 150

˜ y continuous at x = BC: H

a 2 a kx0 C¯2 cos kx0 ( a−d 2 ) = kx1 C1 cos kx1 2 .

˜z continuous at x = BC: E

a 2 2 2 kx1 kx0 ) = C1 sin kx1 a2 . C¯2 sin kx0 ( a−d 2 jω0 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 H(r) for all r ∈ V by placing a magnetic dipole at location r and using the Lorentz reciprocity theorem Z I ˜ m ) dV 0 . ˜−H ˜ pm · J ˜ pm · J ˜ dS 0 = ˜ ×H ˜ pm − E ˜ pm × H) ˆ 0 · (E (1) (E − n V0

S

˜ pm are the fields due to a magnetic ˜ pm and H ˆ is the normal directed into the region V , and E Here n ˜ ˜ dipole. E and H are the desired unknowns, and we have excluded a small volume element Vδ around the point determined by r. Let us use

˜ ˜ pm = − jω ∇ × (˜ E pm ψ), 4π and calculate each term in (1).

˜ ˜ pm = 1 ∇ × ∇ × (˜ H pm ψ). 4π µ ˜

(a) I

  1 0 ˜ ˜ ˆ · E× ∇ × ∇ × (˜ pm ψ) dS 0 n 4π µ ˜ S I 1 ˜ − ∇0 2 [˜ ˜ dS 0 ˜ × [∇0 (∇0 · [˜ ˆ 0 · {E = n pm ψ]) pm ψ]} 4π µ ˜ S I 1 2 ˜ dS 0 . ˜ 0·p ˜ × [∇0 (˜ ˜ m ) − ∇0 [˜ ˆ 0 · {E pm ψ]} n pm · ∇0 ψ˜ + ψ∇ = 4π µ ˜ S

0

˜ ×H ˜ pm ) dS 0 = ˆ · (E n

S

I



0

Note that 2 2 2 2 ˜ ˜ +z ˜ +y ˜ =x ˆ∇0 [˜ ˆ ∇0 [˜ ˆ ∇0 [˜ pmz ψ] pmy ψ] pmx ψ] pm ψ] ∇0 [˜ 2˜ 2˜ 2˜ = (ˆ xp˜mx ∇0 ψ) + (ˆ yp˜my ∇0 ψ) + (ˆ zp˜mz ∇0 ψ) 2 ˆ p˜my + z ˆp˜mz )∇0 ψ˜ = (ˆ xp˜mx + y 2˜ ˜ m ∇0 ψ. =p

So RHS = But

∇0 2 ψ˜

=

−k 2 ψ˜

for r 6=

r0 ,

1 4π µ ˜

I

2˜ ˜ −p ˜ × [∇0 (˜ ˆ 0 · {E ˜ m ∇0 ψ]} n pm · ∇0 ψ) dS 0 .

S

so

1 RHS = 4π µ ˜

I

˜ +p ˜ dS 0 . ˜ × [∇0 (˜ ˆ 0 · {E ˜ m k 2 ψ]} n pm · ∇0 ψ)

S

Next use A × ∇φ = φ∇ × A − ∇ × (φA) to write I 1 0 ˜ ˜ E]} ˜ ×p ˜ − ∇0 × [(˜ ˜ dS 0 ˆ 0 · {(E ˜ m )k 2 ψ˜ + (˜ RHS = n pm · ∇0 ψ)(∇ × E) pm · ∇0 ψ) 4π µ ˜ S I 1 ˜ n0 · (−J ˜ ×p ˜ m − jω µ ˜ dS 0 − third term ˜ m )k 2 ψ˜ + (˜ = {ˆ n0 · (E pm · ∇0 ψ)ˆ ˜H)} 4π µ ˜ S I 1 ˜ n0 · (J ˜ 2 ψ˜ − (˜ ˜ m + jω µ ˜ dS 0 − third term = {˜ pm · (ˆ n0 × E)k pm · ∇0 ψ)ˆ ˜H)} 4π µ ˜ S 152

where

X 1 I ˜ E] ˜ dS 0 . ˆ 0 · ∇0 × [(˜ n pm · ∇0 ψ) third term = 4π µ ˜ Sn n

˜ By Stokes’s theorem, Split each surface along lines of discontinuity of E. I I 0 0˜ ˜ 0 ˜ ˜ pm · ∇0 ψ). ˆ · ∇ × [(˜ (dl0 · E)(˜ n pm · ∇ ψ)E] dS = Γna +Γnb

Sn =Sna +Snb

˜ is continuous the two line integrals cancel and the third term is zero. Note that if E So I

) (I XI 1 0 0 0 2 0 0 0 ˜ dS − ˜ ˜ ˜ m + jω µ ˜ ˜ H ˜ pm ) dS = ˆ ·(E× ˜ m· ˆ · (J n (dl · E)∇ ψ˜ . p [(ˆ n × E)k ψ˜ − n ˜H)∇ ψ] 4π µ ˜ Γ +Γ S S na nb n 0

0

(b) I

I jω ˜ × H} ˜ ˜ ˜ dS 0 ˆ · (Epm × H) dS = − ˆ 0 · {[∇0 × (˜ n n pm ψ)] 4π S I S jω ˜ dS 0 ˜ × (∇0 × [˜ ˆ 0 · [H n pm ψ])] = 4π S I jω ˜ dS 0 ˜ · (∇0 × [˜ (ˆ n0 × H) pm ψ]) = 4π S I jω ˜ 0×p ˜ · {∇0 ψ˜ × p ˜ m + ψ∇ ˜ m } dS 0 = (ˆ n0 × H) 4π S I jω ˜ dS 0 ˜ × ∇0 ψ] ˜ m · [(ˆ p n0 × H) = 4π S I jω ˜ × ∇0 ψ˜ dS. ˜ m · (ˆ = n0 × H) p 4π S 0

Combining terms (a) and (b) gives the LHS of (1) as I ˜ ×H ˜ pm − E ˜ pm × H) ˜ dS 0 ˆ 0 · (E LHS = − n S (I ) XI 1 0 2 0 0 0 0 0 ˜ dS − ˜ ˜ m + jω µ ˜ ˜ ˜m · ˆ · (J =− p [(ˆ n × E)k ψ˜ − n ˜H)∇ ψ] (dl · E)∇ ψ˜ 4π µ ˜ Γna +Γnb S n I jω ˜ × ∇0 ψ˜ dS ˜ m · (ˆ + p n0 × H) 4π S  I 1 ˜ dS 0 ˜ × ∇0 ψ˜ + k 2 (ˆ ˜ ψ˜ − n ˜ m + jω µ ˜ 0 ψ] ˆ 0 · (J ˜m · =− p [−jω µ ˜(ˆ n0 × H) n0 × E) ˜H)∇ 4π µ ˜ S  XI 0 ˜ 0˜ − (dl · E)∇ ψ . n

Γna +Γnb

Now calculate the RHS of (1). 153

(c) Z jω 0 ˜ ·J ˜ dV 0 ˜ ˜ [∇0 × (˜ pm ψ)] Epm · J dV = − 4π V 0 V0 Z jω ˜ 0×p ˜ · [∇0 ψ˜ × p ˜ m + ψ∇ ˜ m ] dV 0 =− J 4π V 0 Z jω ˜ dV 0 ˜ × ∇0 ψ] [˜ pm · (J =− 4π V 0 Z jω ˜ dV 0 . ˜ × ∇0 ψ) ˜m · =− p (J 4π 0 V

Z

(d) Z

Z ˜ ·J ˜ m dV 0 ˜ m dV 0 = 1 ˜ pm · J H {∇0 × [∇0 × (˜ pm ψ)]} 4π µ ˜ 0 0 V ZV 1 ˜ − ∇0 2 (˜ ˜ ·J ˜ m dV 0 = [∇0 (∇0 · [˜ pm ψ)] pm ψ]) 4π µ ˜ V0 Z 1 2˜ ˜ 0·p ˜ m · [∇0 (˜ ˜ m) − p ˜ m ∇0 ψ] J pm · ∇0 ψ˜ + ψ∇ dV 0 = 4π µ ˜ V0 Z 1 ˜ +p ˜ dV 0 ˜ m · [∇0 (˜ ˜ m k 2 ψ] = J pm · ∇0 ψ) 4π µ ˜ V0 Z 1 ˜ − (˜ ˜ 0·J ˜ dV 0 ˜ m [˜ ˜ m + k 2 (˜ ˜ m )ψ] = [∇0 · (J pm · ∇0 ψ]) pm · ∇0 ψ)∇ pm · J 4π µ ˜ V0 Z I 1 ˜ dV 0 − ˜ ·n ˜ m )ψ˜ + jω ρ˜m (˜ ˜ m [˜ ˆ 0 dS 0 [k 2 (˜ pm · J pm · ∇0 ψ)] J pm · ∇0 ψ] = 4π µ ˜ V0 S Z  I 1 2˜ ˜ 0˜ 0 0 ˜ 0˜ 0 ˜m · = p [k Jm ψ + jω ρ˜m ∇ ψ] dV − (ˆ n · Jm )∇ ψ dS . 4π µ ˜ V0 S

Combining, we get the RHS of (1): Z  I 1 ˜ − k 2 ψ˜J ˜ dV 0 + (ˆ ˜ × ∇0 ψ) ˜ m − jω ρ˜m ∇0 ψ] ˜ m )∇0 ψ˜ dS 0 . ˜m · RHS = p [−jω µ ˜(J n0 · J 4π µ ˜ V0 S Equating both sides of (1) gives  Z I 1 0 ˜ 0˜ 0 0˜ 2 ˜˜ 0˜ 0 ˜ ˜m · − p [jω µ ˜(J × ∇ ψ) + k ψ Jm + jω ρ˜m ∇ ψ] dV − (ˆ n · Jm )∇ ψ dS 4π µ ˜ V0 S I 1 ˜ dS 0 ˜ × ∇0 ψ˜ + k 2 (ˆ ˜ ψ˜ − n ˜ m + jω µ ˜ 0 ψ] ˜m · ˆ 0 · (J =− p [−jω µ ˜(ˆ n0 × H) n0 × E) ˜H)∇ 4π µ ˜ S  XI 0 ˜ 0˜ − (dl · E)∇ ψ , n

hence Z

Γna +Γnb

˜ dV 0 ˜ × ∇0 ψ˜ + k 2 ψ˜J ˜ m + jω ρ˜m ∇0 ψ] [jω µ ˜J I I 0 0 2 0 0 0 0 ˜ ˜ ˜ ˜ × ∇ ψ + k (ˆ ˜ ψ − jω µ ˜ = [−jω µ ˜(ˆ n × H) n × E) ˜(ˆ n · H)∇ ψ] dS −

V0

S

Γna +Γnb

154

˜ ˜ 0 ψ, (dl0 · E)∇

Z

˜ dV 0 ˜ × ∇0 ψ˜ − jω˜ ˜ m + ρ˜m ∇0 ψ] [J ψ˜J µ ˜ 0 V I I ˜ dS 0 − ˜ × ∇0 ψ˜ − jω˜ ˜ ψ˜ − (ˆ ˜ 0 ψ] = [−(ˆ n0 × H) (ˆ n0 × E) n0 · H)∇ S

˜ 0 ψ˜ (dl0 · E)∇

Γna +Γnb

1 . jω µ ˜

(2)

˜ is continuous Now single out the contributions due to the surface Σ and the volume Vδ . Since E on Σ we have I ˜ 0 ψ˜ = 0. (dl0 · E)∇ ΓΣa +ΓΣb

Also, I lim

˜ dS 0 ˜ × ∇0 ψ˜ − jω˜ ˜ ψ˜ − (ˆ ˜ 0 ψ] [−(ˆ n0 × H) (ˆ n0 × E) n0 · H)∇ # I " ˆ ˆ ×E ˜ ˆ R R R ˆ × H) ˜ × ˆ · H) ˜ (R = lim + jω˜  + (R dS 0 δ→0 Σ δ2 δ δ2 # Z 2π Z π " ˆ ×E ˜ ˆ 1 1 R R ˜ R ˆ · R) ˆ ˆ R ˆ · H) ˜ ˆ · H) ˜ = lim H( − R( + jω˜  + (R δ 2 sin θ0 dθ0 dφ0 δ→0 0 δ2 δ2 δ δ2 0 Z 2π Z π ˜ ˆ × E)] ˜ sin θ0 dθ0 dφ0 = lim [H(r) + jω˜ δ(R

δ→0 Σ

δ→0 0

0

˜ = 4π H(r), # ˆ ˆ 1 ρ ˜ R R m ˜m + ˜× lim − jω˜ J δ 2 sin θ dδ dθ dφ J δ→0 Vδ δ2 δ µ ˜ δ2   Z δ ρ˜m ˆ ˜ ˆ ˜ = lim 4π J × R − jω˜ δ Jm + R dδ δ→0 0 µ ˜ = 0. Z

"

˜ Substitution back into (1) gives the desired result for H(r). ˜ from E ˜ using Faraday’s law 6.2. Derive H ˜ = −J ˜ m − jω µ ˜ ∇×E ˜H. ˜ term by term: Calculate ∇ × E Z ˜ dV 0 ˜ = 1 ∇ × [−J ˜ m × ∇0 ψ˜ + ρ˜ ∇0 ψ˜ − jω µ ˜ ψ] ∇×E ˜J 4π  ˜ V Z 1 X ˜ dS 0 ˜ × ∇0 ψ˜ + (ˆ ˜ 0 ψ˜ − jω µ ˜ ψ] + ∇× [(ˆ n0 × E) n0 · E)∇ ˜(ˆ n0 × H) 4π n Sn I X 1 1 ˜ ˜ 0 ψ. − ∇× (dl0 · H)∇ 4π n jω˜  Γna +Γnb

155

(a) Z

˜ dV 0 = ˜ m × ∇0 ψ] [−J

∇×

Z

˜ dV 0 ˜ m × ∇ψ) ∇ × (J

ZV

V

=

˜ ×J ˜ m − ∇ × (ψ˜J ˜ m )] dV 0 ∇ × [ψ∇

VZ

=− ZV =− ZV =− ZV =−

˜ m ) dV 0 ∇ × ∇ × (ψ˜J ˜ m ]) − ∇2 (ψ˜J ˜ m )] dV 0 [∇(∇ · [ψ˜J ˜ dV 0 ˜ m ]) − J ˜ m ∇2 ψ] [∇(∇ · [ψ˜J ˜ ·J ˜ dV 0 . ˜ m · ∇ψ˜ + ψ∇ ˜m) − J ˜ m ∇2 ψ] [∇(J

V

Continuing, we use the fact that ∇2 ψ˜ + k 2 ψ˜ = −4πδ(r − r0 ): Z Z 0˜ 0 ˜ −J ˜ ˜ m · ∇ψ) ˜ m (−k 2 − 4πδ(r − r0 ))]] dV 0 ∇ × [−Jm × ∇ ψ] dV = − [∇(J V V Z ˜ + ψk ˜ 2J ˜ ˜ m · ∇ψ) ˜ m ] dV 0 = −4π Jm (r) − [∇(J ZV ˜ + ψk ˜ 2J ˜ m (r) − [−∇(J ˜ m · ∇0 ψ) ˜ m ] dV 0 = −4π J ZV ˜ 0·J ˜ − ψk ˜ 2J ˜ m (r) − [∇(ψ∇ ˜ m − ∇0 · [J ˜ m ψ]) ˜ m ] dV 0 = −4π J ZV

˜ 2J ˜ m ] dV 0 + ∇ [−jω∇(ψ˜ρ˜m ) − ψk

Z

˜ dV 0 ˜ m ψ] ∇0 · [J I 2˜ 0 ˜ ˜ ˜ ˜ dS 0 ˜ ˜ m ψ) ˆ 0 · (J = −4π Jm (r) − [−jω(˜ ρm ∇ψ + ψ∇˜ ρm ) − ψk Jm ] dV − ∇ n V S Z I 0˜ 2˜ 0 ˜ ˜ dS 0 ˜ ˜ m ψ)} = −4π Jm (r) − [jω ρ˜m ∇ ψ + ψk Jm ] dV − ∇{ˆ n0 · ( J V S Z I 0˜ 2˜ 0 ˜ ˜ ˜ m ∇ψ˜ dS 0 ˆ0 · J = −4π Jm (r) − [jω ρ˜m ∇ ψ + ψk Jm ] dV − n ZV IS 0 2 0 ˜ J ˜ m (r) − [jω ρ˜m ∇ ψ˜ + ψk ˜ m ] dV + (ˆ ˜ m )∇0 ψ˜ dS 0 . = −4π J n0 · J ˜ m (r) − = −4π J

ZV

V

V

S

(b) Z ∇× V

ρ˜ 0 ˜ 0 ∇ ψ dV = ˜

Z V

  Z ρ˜ ˜ ρ˜ ∇ × − ∇ψ dV 0 = − ∇ × ∇ψ˜ dV 0 = 0. ˜  ˜ V

156

(c) Z

˜ dV 0 = −jω µ ˜ ψ] [−jω µ ˜J ˜

∇×

Z ZV

V

= −jω µ ˜

˜ dV 0 ˜ ψ] ∇ × [J ˜ × J] ˜ + ψ∇ ˜ dV 0 [∇ψ˜ × J

ZV = −jω µ ˜

˜ × ∇0 ψ˜ dV 0 . J

V

(d) Z ∇×

˜ × ∇0 ψ˜ dS 0 = −∇ × (ˆ n0 × E)

Z

˜ × ∇ψ˜ dS 0 (ˆ n0 × E)

ZSn

Sn

˜ × (ˆ ˜ n0 × E)}] ˜ − ∇ × {ψ(ˆ ˜ dS 0 [ψ∇ n0 × E) Z ˜ n0 × E) ˜ dS 0 . =∇×∇× ψ(ˆ = −∇ ×

Sn

Sn

(e) Z ∇×

Z

0

˜ 0 ψ˜ dS 0 = (ˆ n · E)∇

Sn

˜ dS 0 ˜ (ˆ n0 · E)(∇ × ∇0 ψ)

Sn

Z =−

˜ dS 0 ˜ (ˆ n0 · E)(∇ × ∇ψ)

Sn

= 0. (f) Z ∇×

˜ dS 0 = −jω µ ˜ ψ] [−jω µ ˜(ˆ n0 × H) ˜

Sn

Z ZSn

= −jω µ ˜

˜ dS 0 ˜ ψ] ∇ × [(ˆ n0 × H) ˜ × (ˆ ˜ + ∇ψ˜ × (ˆ ˜ dS 0 [ψ∇ n0 × H) n0 × H)]

Sn

Z = −jω µ ˜

˜ × ∇0 ψ˜ dS 0 . (ˆ n0 × H)

Sn

(g) Z ∇× Γna +Γnb

Z 1 1 0 ˜ 0˜ ˜ ˜ × ∇0 ψ˜ + (dl0 · H)∇ ˜ (dl · H)∇ ψ = [∇(dl0 · H) × ∇0 ψ] jω˜  jω˜  Γna +Γnb Z 1 ˜ =− (dl0 · H)∇ × ∇ψ˜ jω˜  Γna +Γnb = 0.

157

˜ Now, let’s calculate H. ˜ ˜ ˜ = − ∇ × E − Jm H jω µ ˜ jω µ ˜  Z  ˜ ˜ Jm ρ˜m 0 ˜ Jm 1 ˜ ˜ =− + + ∇ ψ − jω˜ ψ J dV 0 jω µ ˜ jω µ ˜ 4π V µ ˜ Z X 1 Z 1 0 ˜ 0˜ 0 ˜ × ∇0 ψ˜ dV 0 (ˆ − J n · Jm )∇ ψ dS + jω µ ˜ 4π 4π Sn V n X 1 Z ˜ n0 × E)] ˜ dS 0 − ∇ × ∇ × [ψ(ˆ ω µ ˜ 4π Sn n Z X 1 ˜ × ∇0 ψ˜ dS 0 + (ˆ n0 × H) 4π n Sn  Z  ρ˜m 0 ˜ 1 0˜ ˜ ˜ ˜ = ∇ ψ − jω˜ ψ Jm + J × ∇ ψ dV 0 4π V µ ˜ Z 1 X ˜ × ∇0 ψ˜ dS 0 + (ˆ n0 × H) 4π n Sn X 1 Z ˜ n0 × E)] ˜ dS 0 . ˜ + (ˆ ˜ m )∇0 ψ} − {∇ × ∇ × [ψ(ˆ n0 · J jω µ ˜ 4π Sn n The volume integral is exactly what we need, but the surface integral still requires manipulation. Z Z 0 0 ˜ 0˜ ˜ − jω µ ˜ 0 ψ˜ dS 0 ˆ 0 · (−∇0 × E n ˜H)∇ (ˆ n · Jm )∇ ψ dS = Sn Sn Z ˜ + jω µ ˜ ψ˜ dS 0 ˆ 0 · (∇0 × E n ˜H) =∇ ZSn ˜ − ∇0 ψ˜ × E ˜ dS 0 . ˜ ψ] ˜ + jω µ ˜ ψ) ˆ 0 · (∇0 × [E =∇ n ˜H Sn

Use Stokes’s theorem on the first term: I Z Z 0 ˜ ˜ 0˜ 0 ˜ ˜ ψ˜ dS 0 RHS = ∇ (dl · E)ψ − ∇ ∇ ψ × E dS + ∇ jω µ ˜(ˆ n0 · H) Γna +Γnb Sn Sn I Z Z ˜ 0 ψ˜ − ∇ ˜ · ∇ψ˜ dS 0 − jω µ ˜ 0 ψ˜ dS 0 =− (dl0 · E)∇ (ˆ n0 × E) ˜ (ˆ n0 · H)∇ Sn IΓna +Γnb Z Z Sn ˜ 0 ψ˜ − ∇∇ · ˜ ψ˜ dS 0 − jω µ ˜ 0 ψ˜ dS 0 . =− (dl0 · E)∇ (ˆ n0 × E) ˜ (ˆ n0 · H)∇ Γna +Γnb

Sn

158

Sn

Thus Z Z 0 0 ˜ 0˜ 0 ˜ ˜ ψ˜ dS 0 ˜ (ˆ n0 × E) {∇ × ∇ × [ψ(ˆ n × E)] + (ˆ n · Jm )∇ ψ} dS = [∇ × ∇ × −∇∇·] Sn Sn I Z ˜ 0 ψ˜ − jω µ ˜ 0 ψ˜ dS 0 − (dl0 · E)∇ ˜ (ˆ n0 · H)∇ Γna +Γnb

Sn

Z

˜ ψ˜ dS 0 + remaining terms (ˆ n0 × E) = −∇2 Sn Z ˜ 2 ψ˜ dS 0 + remaining terms (ˆ n0 × E)∇ =− Z Sn ˜ 2 ψ˜ dS 0 + remaining terms. = (ˆ n0 × E)k Sn

˜ Back-substitution gives the desired result for H. 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 i ˜ ˜ i × ∇0 G ˜i G ˜ + ρ˜m ∇0 G ˜ − jω˜ ˜ H(r, ω) = J c J dV 0 m µ ˜ V +Vm i XZ h ˜ × ∇0 G ˜ 0G ˜ G ˜ + (ˆ ˜ + jω˜ ˜ dS 0 (ˆ n0 × H) n0 · H)∇ c (ˆ n0 × E) + n6=m Sn

+

X n6=m

1 jω µ ˜

I

˜ 0 G, ˜ (dl0 · E)∇

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  ρ˜im 0 ˜ i 0˜ c ˜i ˜ ˜ ˜ H(r, ω) = J ×∇G+ ∇ G − jω˜  Jm G dV 0 µ ˜ Vm Z h i ˜ × ∇0 G ˜ 0G ˜ G ˜ + (ˆ ˜ + jω˜ ˜ dS 0 (ˆ n0 × H) n0 · H)∇ c (ˆ n0 × E) − Sm I 1 ˜ 0 G, ˜ − r ∈ Vm . (dl0 · E)∇ jω µ ˜ Γma +Γmb Note the minus signs introduced by reversal of the normal vector. Now subtract the two equations for points within Vm :  Z  i ρ ˜ m i 0 0 c i ˜ ×∇G ˜ G ˜+ ˜ − jω˜ ˜ dV 0 0= J ∇G J m µ ˜ V N Z h i X ˜ × ∇0 G ˜ 0G ˜ G ˜ + (ˆ ˜ + jω˜ ˜ dS 0 + (ˆ n0 × H) n0 · H)∇ c (ˆ n0 × E)

+

n=1 Sn N X n=1

1 jω µ ˜

I

˜ 0 G, ˜ (dl0 · E)∇

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 i ˜ h ˜ φ = j I e−jkR1 + e−jkR2 − (2 cos kl)e−jkr . H 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 ˜ h ˜ φ = j I e−jkr ejkl cos θ + e−jkr e−jkl cos θ − (2 cos kl)e−jkr H 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 Next, by (6.21),  −jkR2 −jkr ˜  z − l e−jkR1 η I z + l e z e ˜ρ = j E + − (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  −jkr ejkl cos θ ˜ cos θ e−jkr e−jkl cos θ cos θ e−jkr ˜ρ ≈ j η I cos θ e E + − (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 By (6.22),   η I˜ e−jkR1 e−jkR2 e−jkr ˜ Ez = −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

˜ −jkr ˜ =ρ ˜ρ + z ˜z = j η I e ˆE ˆE ˆ sin θ]. E F (θ, kl)[ˆ ρ cos θ − z 2π r

Let ˆ cos θ − z ˆ sin θ = [ˆ ˆ sin φ] cos θ − z ˆ sin θ. V=ρ x cos φ + y 161

Then ˆ·V Vθ = θ ˆ·x ˆ·y ˆ·z ˆ ) cos φ cos θ + (θ ˆ ) sin φ cos θ − (θ ˆ) sin θ = (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 = ˆr · V ˆ ) cos φ cos θ + (ˆr · y ˆ ) sin φ cos θ − (ˆr · z ˆ) sin θ = (ˆr · x = (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, ˆ ·V Vφ = φ ˆ ·x ˆ ·y ˆ ·z ˆ ) cos φ cos θ + (φ ˆ ) sin φ cos θ − (φ ˆ) sin θ = (φ = (− sin φ) cos φ cos θ + (cos φ) sin φ cos θ − (0) sin θ = 0. Finally then, ˜ −jkr ˆ ηI e ˜ = θj F (θ, kl). E 2π r 6.6. We have ˜i = z ˜ −jk|z| δ(x)δ(y), ˆIe J and R = (ρ2 + (z − z 0 )2 )1/2 , Now

and

  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 .

Z µ ˜I˜ l −jk|z 0 | e−jkR 0 ˜ ˆ Ae = z e dz 4π −l R ˜ ˆ 1 ∂ Aez . ˜ = 1∇ × A ˜ e = −φ H µ ˜ µ ˜ ∂ρ

Therefore Z 0 Z l −jkR −jkR ˜ I˜ ∂ 0e 0e ˜φ = − I ∂ H 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 162

Use

0

0

e−jk(R±z ) ∂ ∂ e−jk(R±z ) = ±ρ 0 ∂ρ R ∂z R[R ∓ (z − z 0 )]

to get " # " # −jk(R−z 0 ) −jk(R+z 0 ) ˜ ˜ 0 l e e I I ˜φ = − − −ρ ρ H 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 ˜ use To find E, " # ˜φ ∂ H 1 1 1 1 ∂ ˆH ˜ = ˜ = ˜ φ) = ˜ φ) , ˆ E ∇×H ∇ × (φ −ˆ ρ +z (ρH 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 ˜ Hφ ≈ − − 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 ˜ cross ˆr into both sides of To find E, ˜ ˜ ≈ ˆr × E H η to obtain ˜ = ˆr × (ˆr × E) ˜ = ˆr(ˆr · E) ˜ − E(ˆ ˜ r · ˆr) = −E ˜ ηˆr × H or ˜ = −ηˆr × H. ˜ E ˆ = −θ, ˆ we compute Since ˆr × φ 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) . Eθ ≈ η 2π sin θ r 4π sin θ r 163

Figure 41 6.7. Refer to Figure 41. By (6.31) we have Z ˇe = a

ˇ i (r0 , ω)ejk0 ˆr·r0 dV 0 J

ZVπ =

ˆ 0 I(φ ˇ 0 )ejk0 ˆr·r0 a dφ0 where I(φ) ˇ φ = Iˇ0 e−jk0 a|φ| .

−π

But ˆr · r0 = (ˆ ˆ sin θ sin φ + z ˆ cos θ) · (ˆ ˆ sin φ0 )a x sin θ cos φ + y x cos φ0 + y = a(sin θ cos φ cos φ0 + sin θ sin φ sin φ0 ) = a sin θ cos(φ − φ0 ) and ˆ 0 = −ˆ ˆ cos φ0 φ x sin φ0 + y ˆ cos θ cos φ − φ ˆ sin φ) sin φ0 + (ˆr sin θ sin φ + θ ˆ cos θ sin φ + φ ˆ cos φ) cos φ0 = −(ˆr sin θ cos φ + θ = ˆr(− sin θ cos φ sin φ0 + sin θ sin φ cos φ0 ) ˆ cos θ cos φ sin φ0 + cos θ sin φ cos φ0 ) + θ(− ˆ + φ(sin φ sin φ0 + cos φ cos φ0 ) ˆ cos θ sin(φ − φ0 ) + φ ˆ cos(φ − φ0 ). = ˆr sin θ sin(φ − φ0 ) + θ Therefore

Z

π

a ˇθ = a cos θ

ˇ 0 ) sin(φ − φ0 )ejka sin θ cos(φ−φ0 ) dφ0 I(φ

−π

and

Z

π

a ˇφ = 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

and

a ˇφ = ˇ I0 a

0

0

sin(φ − φ0 )e−jk0 a[|φ |−sin θ cos(φ−φ )] dφ0

−π

π

Z

π

Z

a ˇθ = cos θ ˇ I0 a

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 = a ˇθ a ˇθ + a ˇφ a ˇ∗φ sin θ dθ dφ. 2 0 0 32π Thus, 

2PR = η0 Rr = |Iˇ0 |2

k0 a 4π

2 Z

2π

Z

0

π

0

" 2 2 # a ˇ a ˇ θ + θ sin θ dθ dφ. Iˇ a Iˇ a 0

0

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 π

a ˇφ ≈ Iˇ0 a

Z

a ˇθ ≈ Iˇ0 a sin θ

Z



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 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 a ˇθ 2 a ˇθ 2 Rr = = η0 Iˇ a + Iˇ a sin θ dθ dφ. 4π |Iˇ0 |2 0 0 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 ˜a = E ˜0 x ˜0 x ˆ e−jkz z=0 = E ˆ. E We have ˜ eq = −2ˆ ˜ a = −2ˆ ˜0 x ˜0 , ˆ ) = −2ˆ J n×E z × (E yE ms Z ˜ ms (r0 )ejkˆr·r0 dS 0 , ˜h = a J 0

0

S 0

ˆr · r = ˆr · (ˆ ˆ y ) = x0 sin θ cos φ + y 0 sin θ sin φ. xx + y

166

Therefore ˜0 ˜h = −2ˆ a yE

Z

a/2

Z

−a/2

a/2

ejkx

0

sin θ cos φ jky 0 sin θ sin φ

e

dx0 dy 0

−a/2

jkx0 sin θ cos φ a/2 ejky 0 sin θ sin φ a/2 e ˜0 = −2ˆ yE 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) ˜ = −2ˆ yE0 a · j k a2 sin θ cos φ j k a2 sin θ sin φ ˜0 a2 sin πX · sin πY = −8ˆ yE πX πY

where X=

a sin θ cos φ, λ

Y =

a sin θ sin φ. λ

Then, by (6.55), ˜h ˜ = jk ˆr × A E 0  −jkr  e jk 2 sin πX sin πY ˜ ˆ )E0 a · 0 . = − 8(ˆr × y 0 πX πY 4πr Here ˆ cos θ sin φ − φ ˆ cos φ) = φ ˆ cos θ sin φ + θ ˆ cos φ, ˆr × y ˆ = ˆr × (ˆr sin θ sin φ + θ so

−jkr

˜ = − 2jk0 E ˜0 a2 e E π r

ˆ cos θ sin φ + θ ˆ cos φ) sin πX · sin πY . (φ πX πY

Finally, −jkr ˜ ˆ cos φ − θ ˆ cos θ sin φ) sin πX · sin πY . ˜ = ˆr × E = − 2jk0 E ˜ 0 a2 e (φ H η ηπ r πX πY

6.10. Refer to Figure 44. The field in the aperture is ˜ a (r) = ρ ˆ E

V˜0 V˜0 −jkz ˆ e =ρ . ρ ln(b/a) ρ ln(b/a) z=0

We have

Z ˜h = a

˜ eq (r0 )ejkˆr·r0 dS 0 J ms

S

where ˜ eq = −2ˆ ˜ a = −2ˆ ˆ J n×E z×ρ ms

V˜0 V˜0 ˆ = −2φ . ρ ln(b/a) ρ ln(b/a)

Here ˆ 0 = ρ0 x ˆ cos φ0 + ρ0 y ˆ sin φ0 r0 = ρ0 ρ

167

Figure 44 so ˆr · r0 = (ˆr · x ˆ )ρ0 cos φ0 + (ˆr · y ˆ )ρ0 sin φ0 = ρ0 cos φ0 sin θ cos φ + ρ0 sin φ0 sin θ sin φ = ρ0 sin θ cos(φ − φ0 ). Also, ˆ 0 = −ˆ ˆ cos φ0 φ x sin φ0 + y ˆ cos θ cos φ − φ ˆ sin φ) sin φ0 + (ˆr sin θ sin φ + θ ˆ cos θ sin φ + φ ˆ cos φ) cos φ0 = −(ˆr sin θ cos φ + θ = ˆr(− sin θ cos φ sin φ0 + sin θ sin φ cos φ0 ) ˆ cos θ cos φ sin φ0 + cos θ sin φ cos φ0 ) + θ(− ˆ + φ(sin φ sin φ0 + cos φ cos φ0 ) ˆ cos θ sin(φ − φ0 ) + φ ˆ cos(φ − φ0 ). = ˆr sin θ sin(φ − φ0 ) + θ Hence Z

2π

Z

a ˜θ = 0

a

b

−2 cos θ sin(φ − φ0 )

V˜0 0 0 ejkρ sin θ cos(φ−φ ) ρ0 dρ0 dφ0 = 0 ρ0 ln(b/a)

168

(because the integrand is odd in u = φ − φ0 ) and V˜0 0 0 −2 cos(φ − φ0 ) 0 ejkρ sin θ cos(φ−φ ) ρ0 dρ0 dφ0 ρ ln(b/a) 0 a   Z Z b 2π 2V˜0 jkρ0 sin θ cos u =− cos u e du dρ0 ln(b/a) a 0 Z b   2V˜0 =− 2πjJ1 (kρ0 sin θ) dρ0 . . . by (E.83) ln(b/a) a b Z 1 4πj V˜0 0 = J0 (kρ sin θ) . . . since J1 (x) dx = −J0 (x) ln(b/a) k sin θ Z

2π

b

Z

a ˜φ =

a

4πj V˜0 J0 (kb sin θ) − J0 (ka sin θ) = . ln(b/a) k sin θ Then ˜h ˜ = jk ˆr × A E ˜  −jkr  jk e ˆ = r × ˜ ah ˜ 4πr −jkr 4πj V ˜0 J0 (kb sin θ) − J0 (ka sin θ) jk ˆ e (ˆr × φ)˜ ˜ 4πr ln(b/a) k sin θ −jkr V˜0 J0 (kb sin θ) − J0 (ka sin θ) ˆe =θ r ln(b/a) sin θ

=

and

−jkr ˜ V˜0 J0 (kb sin θ) − J0 (ka sin θ) ˆe ˜ = ˆr × E = φ H . η r η ln(b/a) sin θ

6.11. −jkr

e ˜ e (r) = z ˆ A 4πr ˆ =z

e−jkr 4πr

Z

L/2

˜ 0 )ejkz 0 cos θ dz 0 I(z

−L/2

"Z

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

e

0

" # e−jkr ˜ 1 − e−jk(cos θ+1)L/2 1 − ejk(cos θ−1)L/2 ˆ I0 − . =z 4πr jk(cos θ + 1) jk(cos θ − 1) 6.12. Refer to Figure 45. ˜ ms = −2ˆ ˜ A = −2ˆ ˜0 = −ˆ ˜0 , ˆE J z×E z×x y 2E −jkr Z ˜h = e ˜ ms (r0 )ejkˆr·r0 dS 0 A J 4πr S 169

cos θ

dz 0

Figure 45 ˆr · r0 = ˆr · (ˆ ˆ y 0 ) = x0 sin θ cos φ + y 0 sin θ sin φ, xx 0 + y −jkr

˜ h = −ˆ ˜0 e A y 2E 4πr

Z

a

e

jkx0 sin θ cos φ

0

Z

a

dx

ejky

0

sin θ sin φ

dy 0

−a

−a

−jkr sin(ka sin θ cos φ) sin(ka sin θ sin φ) ˜0 e = −ˆ y 2E 2 2 4πr k sin θ cos φ k sin θ sin φ

˜ = jkˆr × J ˜ msT E

=⇒

˜φ = jk J˜msθ E

ˆ cos θ sin φ + φ ˆ cos φ, ˆ = ˆr sin θ sin φ + θ y −jkr

˜φ = jk cos θ sin φ(−2E ˜0 ) e E ˜0 = −2j E

πr

sin(ka sin θ cos φ) sin(ka sin θ sin φ) k 2 sin2 θ sin φ cos φ

cos θ e−jkr sin(ka sin θ cos φ) sin(ka sin θ sin φ). πrk sin2 θ cos φ

170

Chapter 7 7.1. (a) Begin with (7.14) and (7.17): Z z dV˜ (z) 2 V˜ (t) dt + C1 , =γ dz −l Z z 2 ˜ (z − t)V˜ (t) dt + C1 z + C2 . V (z) = γ

(1)

−l

To find C1 and C2 , apply the boundary conditions. At z = −l we have ˜ ˜ ˜ = − 1 dV (z) . + V˜ (−l) where I(z) Vg = Zg I(−l) Z dz This gives us Vg = −

Zg ˜ 0 V (−l) + V˜ (−l) Z

or Vg = −

Zg C1 − C1 l + C2 . Z

(2)

At z = 0 we have ˜ V˜ (0) = ZL I(0) which gives us −γ

2

Z

0

ZL tV˜ (t) dt + C2 = −γ 2 Z −l

Z

0

ZL V˜ (t) dt − C1 . Z −l

(3)

Solving, we obtain C1 =

Z γ2 Zl + ZT

Z

0

tV˜ (t) dt −

−l

γ 2 ZL Zl + ZT

Z

0

V˜ (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

tV˜ (t) dt −

−l

Zg + Zl 2 ZL γ ZT + Zl Z

Z

0

V˜ (t) dt.

−l

Substituting into (1) and simplifying, we get the integral equation for V˜ (z): Z z Z 0 Z 0 Zg (t − ZL /Z) ˜ Zt − ZL ˜ Zz − ZL 2 2 2 ˜ ˜ V (z) = γ (z−t)V (t) dt− Vg +γ (l+z) V (t) dt+γ V (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 V˜ (z) =

N X n=1

171

an Pn (z)

and point match at zm : N X

an Pn (z) =

n=1

N X

zm

 Zzm − ZL (zm − t)Pn (t) dt − Vg 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  Z an γ 2

n=1

Write γ2

zm

Z

(t − zm )Pn (t) dt = Umn −l

Z

2

0

−γ (l + zm ) −l

−γ 2

Z

0

−l

  m < n, 0, 1 = − 8 γ 2 ∆2 , m = n,   2 γ ∆(zn − zm ), m > n,

Zt − ZL Zzn − ZL Pn (t) dt = Wmn = −γ 2 (l + zm )∆ , Zl + ZT Zl + ZT

Zg (t − ZL /Z) Zg (zn − ZL /Z) Pn (t) dt = Zmn = −γ 2 ∆ , Zl + ZT Zl + ZT

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 ˜ V (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
Vg −γl −γz V˜ (z) = e e + Γeγz . 2 Note that the MoM solution matches well with the theoretical voltage. 7.2. Z zZ a

a

u

 Z u  z Z z Z u d F (t) dt du = u F (t)dt − u F (t) dt du 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

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

(z − t)kz2 (t)g(t) dt + C1 z + C2

g(z) = − 0

twice with respect to z. The results are 0

Z

g (z) = −

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

˜ Z z 2 2E0 = −kz (z − t) sin kz (d − t) dt D 0  ˜  sin kz (d − t) z 2 2E0 z − t cos kz (d − t) − = −kz D kz kz2 0 ˜0 2E = [sin kz (d − z) + zkz cos kz d − sin kz d]. D The second term on the RHS of (7.41) is ˜0 Z d a − z 2E k2 (d − t) sin kz (d − z) dt T2 = a−d D z 0  ˜0  d − t a − z 2E sin kz (d − t) d 2 = k cos kz (d − t) − a − d D z kz kz2 0 ˜0 a − z 2E = k 2 [−dkz cos kz d + sin kz d] . a−d D z Substituting these into (7.43), we get ˜0 2E ˜0 z − d sin kz (d − z) = T1 + T2 + 2E D a−d ˜0 2E = [sin kz (d − z) + zkz cos kz d − sin kz d] D ˜0 z − d ˜0 a − z 2E 2E [−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): ˜ = H

" # ˜y ˜y ∂E ∂E j ˆ −ˆ x +z . ωµ ˜(z) ∂z ∂x

Substitute into Ampere’s law: "

˜x ∂H ˜z ∂H ˜ =y ˆ ∇×H − ∂z ∂x We obtain

# ˜y (x, z). = jω˜ (z)E

" # " # ˜y ˜y ∂ j ∂E ∂ j ∂E ˜y , − − = jω˜ (z)E ∂z ωµ ˜(z) ∂z ∂x ω µ ˜(z) ∂x 174

" # " # ˜y ˜y ∂ 1 ∂E ∂ 1 ∂E ˜y , + = −ω 2 ˜(z)E ∂x µ ˜(z) ∂x ∂z µ ˜(z) ∂z ˜y ˜y ˜z 1 ∂2E 1 ∂2E µ ˜0 (z) ∂ E ˜y , + − = −ω 2 ˜(z)E µ ˜(z) ∂x2 µ ˜(z) ∂z 2 µ ˜(z) ∂y so

˜y ˜y ˜y ∂2E ∂2E µ ˜0 (z) ∂ E ˜y = 0, + − + k2 E 2 2 ∂x ∂z µ ˜(z) ∂z

k 2 (z) = ω 2 µ ˜(z)˜ (z).

˜y (x, z) = f (x)g(z), substitute to get Let E g(z)

d2 f (x) d2 g(z) µ ˜0 (z) dg(z) + f (x) − f (x) + k 2 f (x)g(z) = 0, dx2 dz 2 µ ˜(z) dz

and divide by f g: 1 d2 f 1 d2 g 1 µ ˜0 (z) dg + − = −k 2 . f dx2 g dz 2 g µ ˜(z) dz Now separate variables to obtain 1 d2 g 1 µ ˜0 (z) dg − = −kz2 g dz 2 g µ ˜(z) dz

1 d2 f = −kx2 , f dx2 where

kx2 + kz2 = k 2 . So

1 d˜ µ(z) dg d2 g − + kz2 g = 0. dz 2 µ ˜(z) dz dz

Integrate to obtain an integral equation: Z z Z 0 0 g (z) − h(t)g (t) dt + 0

z

f (t)g(t) dt = C¯1 .

0

If µ ˜(z) is twice differentiable, we can integrate the second term by parts: 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) − 0

0

Then

Z

0

g (z) + h(z)g(z) −

z

z

h0 (t)g(t) dt.

0

Z

0

h (t)g(t) dt + 0

z

f (t)g(t) dt = C¯1 + 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 g(z) +

z

Z h(t)g(t) dt +

0

z

(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 ˜0 , g(0) + ag 0 (0) = 2E We find C2 =

˜0 2E a − C1 h0 h0

a=

where

jη0 . ωµ ˜(0) cos φ0

h0 = 1 − ah(0)

and

Z d ˜0 2E 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):     g1 (d) g1 (d) 0 0 ˜ 2E0 = A1 g1 (0) − g2 (0) + aA1 g1 (0) − g (0) . g2 (d) g2 (d) 2 We obtain A1 =

˜0 g2 (d) 2E 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) =

˜0 2E [g1 (z)g2 (d) − g1 (d)g2 (z)]. F + aG

(b) g(0) ˜0 E 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 µ ˜r ˜r0 /κ and ν = 2kx /κ. Use k 2 (z) = ω 2 µ ˜˜(z) = ω 2 µ ˜r µ0 0 ˜r0 eκz = k02 µ ˜r ˜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 µ ˜r ˜r0 eκz − kx2 ]Bν (λeκz/2 ) = 0 4 4

or Bν00 (u) +

1 −κz/2 0 4 e Bν (u) + [k02 µ ˜r ˜r0 − kx2 e−κz ] 2 2 Bν (u) = 0 λ λ κ

where u = λeκz/2 . This can be rewritten as Bν00 (u)

  1 0 ν2 + Bν (u) + 1 − 2 Bν (u) = 0. u 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 g(z) = −

z

(z − t)kz2 (t)g(t) dt +

0

a−z a−d

Z

d

˜0 (d − t)kz2 (t)g(t) dt + 2E

0

z−d a−d

where kz2 (z) = ω 2 µ ˜˜(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

+

an

n=1

a − zm a−d

(zm − t)kz2 (t)Pn (t) dt

0

n=1 N X

zm

an d

Z

˜0 (d − t)kz2 (t)Pn (t) dt + 2E

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

˜ 0 zm − d , bm = 2E a−d Am = δmn + Umn + Vmn , Z zm Umn = (zm − t)kz2 (t)Pn (t) dt, 0

and Vmn

a − zm =− a−d

d

Z

(d − t)kz2 (t)Pn (t) dt.

0

To compute Umn we consider three cases. (1) m > n Z

Umn = = =

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 2 k0 ˜r0 2 [(zm − t)κ + 1] κ zn −∆/2 (

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 Umn = =

zm

(zm − t)kz2 (t) dt zm −∆/2 zm eκt 2 k0 ˜r0 2 [(zm − t)κ + 1] K zm −∆/2 (

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 Vmn

a − zm =− a−d

d

Z

(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 − P˜ 2 where 1 − Γ0 P˜ 2

Γ0 =

179

1 + jakz , 1 − jakz

P˜ = 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 ωµ˜ jη k 2 − k02 sin2 φ0 cos φ0 p ˜ ˜c − sin2 φ0 ωµ ˜r µ0 cos φ0 − η0 k02 µ p r r ωµ ˜r µ0 cos φ0 + η0 k02 µ ˜r ˜cr − sin2 φ0 q p √ ωµ ˜r µ0 cos φ0 − µ00 ω µ0 0 µ ˜r ˜cr − sin2 φ0 q p √ ωµ ˜r µ0 cos φ0 + µ00 ω µ0 0 µ ˜r ˜cr − sin2 φ0 p ˜ ˜c − sin2 φ0 µ ˜r cos φ0 − µ p r r . µ ˜r cos φ0 + µ ˜r ˜cr − sin2 φ0 0 1 + j ωµ˜ jη cos φ0

180

0.03

7.10. In region 0, ˜0 E ˜ i = −ˆ H x e−jk0 z , η0 ˜ Er ˜r = x ˆ ejk0 z . H η0

˜i = y ˜0 e−jk0 z , ˆE E ˜r = y ˜r ejk0 z , ˆE E In region 2,

˜t E ˜ t = −ˆ H x ejk0 z , η0

˜t = y ˜t e−jk0 z , ˆE E

We can develop an integral equation for the electric field in region 1 by starting with Maxwell’s equations. Faraday’s law: ˜ = jω µ ˜ ∇×E ˜H

=⇒

˜ ˜ x = −j ∂ Ey . H ωµ ˜ ∂z

˜ = jω˜ ˜ ∇×H c E

=⇒

˜y = jω˜ c E

Ampere’s law

Combine to obtain ˜y = − jω˜ c E or

˜x ∂H . ∂z

˜y j ∂2E ωµ ˜ ∂z 2

˜y ∂2E ˜y = 0, + k2 E ∂z 2

k2 = ω2 µ ˜˜c .

˜y (z). Then we have the differential equation Let g(z) = E 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 ˜0 + E ˜r = g(0). E (1) Continuity of tangential H gives η0 0 g (0). ωµ ˜

(2)

η0 0 g (0). ωµ ˜

(3)

˜0 + E ˜r = −j −E Subtracting (2) from (1), we obtain ˜0 = g(0) + j 2E 181

˜ gives At z = d, continuity of tangential E

Continuity of tangential H gives −j

˜t e−jk0 d . g(d) + E

(4)

η0 0 ˜t e−jk0 d . g (d) = E ωµ ˜

(5)

Subtracting (5) from (4), we obtain g(d) + j

η0 0 g (d) = 0. ωµ ˜

(6)

Using (3) and the facts that g(0) = C2 and g 0 (0) = C1 , we get ˜ 0 − j η 0 C1 . C2 = 2E ωµ ˜ 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 0

d

η0 (d − t)g(t) dt − j ωµ ˜

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, " # ˜y ˜y ∂ H ∂ H 1 ˜ = jω˜ ˜ =⇒ E ˜ = ˆ ∇×H c (z)E −ˆ x +z , jω˜ c (z) ∂z ∂x so ˜x = − E Then

˜y ∂H 1 , jω˜ c (z) ∂z "

˜z = E

˜x ∂ E ˜z ∂E ˜ =y ˆ ∇×E − ∂z ∂x

˜y ∂H 1 . jω˜ c (z) ∂x

# ˜ = −jω µ ˜(z)H,

so " # ˜x ∂ E ˜z 1 ∂ E ˜y = − − H jω µ ˜(z) ∂z ∂x ! !#   " ˜y ˜y 1 ∂ 1 ∂H ∂ 1 ∂H 1 = − − − jω µ ˜(z) jω ∂z ˜c (z) ∂z ∂x ˜c (z) ∂x " # ˜y ˜y ˜y 1 ∂˜ c /∂z ∂ H 1 ∂2H 1 ∂2H =− 2 − c 2 + c + c ω µ ˜(z) (˜ ) ∂z ˜ ∂z 2 ˜ ∂x2 " # ˜y ˜y ˜y ∂2H ∂2H 1 ˜c0 (z) ∂ H =− 2 + − . k (z) ∂x2 ∂z 2 ˜c (z) ∂z 182

Thus

˜y ˜y ˜y ∂2H ∂2H ˜c0 (z) ∂ H ˜ y = 0. + − + k 2 (z)H ∂x2 ∂z 2 ˜c (z) ∂z

˜ y (x, z) = f (x)g(z): Seek a product solution. Let H 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 ˜ yi = H ˜ 0 e−jkx0 x e−jkz0 z , H ˜xi = η0 cos φ0 H ˜ 0 e−jkx0 x e−jkz0 z , E

˜ yr = H ˜ r e−jkx0 x ejkz0 z , H ˜xr = −η0 cos φ0 H ˜ 0 e−jkx0 x ejkz0 z , E

where kx0 = k0 sin φ0 ,

kz0 = k0 cos φ0 ,

so that 2 2 kx0 + kz0 = k02 = ω 2 µ ˜0 0 .

To satisfy the boundary conditions we must have kx = kx0 . Thus, f (x) = e−jkx0 x . ˜ continuous at z = 0: Boundary condition 1, tangential E ˜ 0 cos φ0 − η0 H ˜ r cos φ0 = η0 H ˜ continuous at z = 0: Boundary condition 2, tangential H ˜0 + H ˜ r = g(0). H ˜ is zero at z = d: Boundary condition 3, tangential E g 0 (d) = 0. 183

j ω˜ c (0)

g 0 (0).

Thus we have the three equations ˜0 − H ˜r = H

j g 0 (0), ω˜ c (0)η0 cos φ0

(1)

˜0 + H ˜ r = g(0), H

(2)

g 0 (d) = 0.

(3)

and Adding (1) and (2), we get an alternative form of the boundary condition ˜ 0 = g(0) + ag 0 (0), 2H

a=

j ω˜ c (0)η0 cos φ0

.

Now return to the differential equation for g(z): g 00 (z) + h(z)g 0 (z) + kz2 (z)g(z) = 0, Integrate: g 0 (z) +

Z

z

Z

h(t)g 0 (t) dt +

0

h(z) = −

˜c0 (z) . ˜c (z)

z

kz2 (t)g(t) dt = C1 .

0

Integrate again: z

Z

Z

0

(z − t)h(t)g (t) dt =

g(z) + 0

z

(z − t)kz2 (t)g(t) dt = C1 z + C2 .

0

To find C1 and C2 we apply (1’) and (3). We obtain d

Z C1 =

[h(t)g 0 (t) + kz2 (t)g(t)] dt,

C2 = 2H0 − aC1 .

0

Substitution yields Z

z

Z

0

(z − t)h(t)g (t) dt +

g(z) + 0

z

(z − t)kz2 (t)g(t) dt

0

˜ 0 − aC1 = C1 z + 2 H ˜ 0 + C1 (z − a) = 2H Z d ˜ = 2H0 + (z − a) [h(t)g 0 (t) + kz2 (t)g(t)] dt. 0

So the integro-differential equation for g(z) is Z g(z) =

d

˜ 0, [(z − a) − (z − t)U (z − t)][kz2 (t)g(t) + h(t)g 0 (t)] dt + 2H

0

holding for 0 ≤ z ≤ d, where U (x) is the unit step function.

184

(1’)

7.12. Z Amn = where Z G(u) = 0

2π

(m−n+ 12 )∆

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 Amn =

−(m−n+ 21 )∆

Z G(v)(−dv) =

−(m−n− 21 )∆

−(m−n− 12 )∆

Z G(v) dv =

−(m−n+ 12 )∆

(m−n+ 12 )∆

(m−n− 21 )∆

7.13. Refer to Figure 48.

Figure 48 Use the Schelkunoff equivalence principle of § 6.3.4. V˜0 /2 V˜0 ˆ ˜ sm (r) = −2ˆ ˜ = −2ˆ ˆ J n×E z×ρ = −φ , ρ ln(b/a) ρ ln(b/a) Z −jkR ˜ h (r) = ˜ sm (r0 ) e A 0 J dS 0 , 4πR S ˜h ˜ =−1∇×A E 0 Z e−jkR V˜0 ˆ0 =∇× φ dS 0 0 ρ ln(b/a) 4πR S   Z −jkR 1 V˜0 dS 0 0e ˆ = ∇× φ . 4π ln(b/a) S R ρ0 Use (B.49) to write 

−jkR

ˆ0 e ∇× φ

R



ˆ0 × ∇ = −φ 185



e−jkR R



G(u) du = Anm .

where, by (B.74),  ∇ Then

e−jkR R



 = −R

1 + jk R



e−jkR . R2

  −jkR Z V˜0 dS 0 1 e 1 0 ˆ ˜ . E= φ ×R + jk 4π ln(b/a) S R R 2 ρ0

ˆ0 × R = φ ˆ 0 × (ˆ ˆ 0 ρ0 ) = ρ ˆ 0 z + ρ0 z ˆ for points on the z-axis, so But φ zz − ρ  −jkR Z b  Z 2π 1 1 V˜0 e 0 0 ˜ Ez = + jk dρ0 ρ dφ 2 4π ln(b/a) 0 R R a  −jkR Z b 0 V˜0 ρ 1 e = + jk dρ0 . 2 ln(b/a) a R R R Now ∂ ∂ρ0 where



e−jkR R

∂R ∂ = 0 0 ∂ρ ∂ρ



 =−

1 + jk R



e−jkR ∂R R ∂ρ0

q ρ0 ρ0 ρ0 2 + z 2 = p = . R ρ0 2 + z 2

Therefore  −jkR  Z b V˜0 ∂ e dρ0 0 2 ln(b/a) a ∂ρ R  −jkR1  V˜0 e e−jkR2 = − 2 ln(b/a) R1 R2 √ √ where R1 = a2 + z 2 and R2 = b2 + z 2 . ˜z = − E

7.14. For a straight wire aligned along z, use ∂ ∂ = , ∂u ∂z

¯

ˆ ·u ˆ 0 = 1, u

0

e−jk0 R(z−z ) G(u, u0 ) = 2π ¯ , R(z − z 0 )

¯= R

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 E (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 R R(−jk ¯ ∂R 0 ) − 1 −jk0 R = 2π ¯ = 2π e , ¯2 ∂z ∂z ∂z R R ¯ ∂R ∂ p z − z0 = (z − z 0 )2 + a2 = ¯ ∂z ∂z R ∂G(z − z 0 ) z − z0 ¯ −jk0 R¯ . = G1 (z − z 0 ) = −2π ¯ 3 (1 + jk0 R)e ∂z R 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 E (z). I(z ) 0 ∂z 0 η0 −L −L Use integration by parts in the first integral: L Z L ˜ 0 0 ∂ I(z ) 0 ∂G1 (z − z ) 0 0 0 ˜ ˜ I(z ) dz = I(z )G1 (z − z ) − G1 (z − z 0 ) dz 0 , 0 0 ∂z ∂z −L −L −L

Z

L

so Z

L

−L

"

# ˜ 0) ∂ I(z 8π 2 k0 ˜ i 0 2˜ 0 0 0 E (z), G (z − z ) + k I(z )G(z − z ) dz = −j 1 0 ∂z 0 η0

˜ ˜ 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 N X an G1 (zm − [zn − ∆/2]) − G1 (zm − [zn + ∆/2]) + k02

#

zn +∆/2

0

G(zm − z ) dz

0

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 where

Z

zn +∆/2

0

0

G(zm − z ) dz = zn −∆/2

k02

Z

zm −zn +∆/2

G(u) du zm −zn −∆/2

¯

ejk0 R(u) , G(u) = 2π ¯ R(u)

¯ R(u) =

p u2 + a2 ,

and zm − zn = −L + (m − 1/2)∆ − [−L + (n − 1/2)∆] = (m − n)∆. 187

= −j

8π 2 k0 ˜ i E (zm ) η0

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

u ¯ G1 (u) = −2π ¯ 3 (1 + jk0 R)e−jk0 R , R

and bm = −j where

8π 2 k0 ˜ i E (zm ) η0

" # e−jk0 R1 (z) e−jk0 R2 (z) V0 , E (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 ˜ i (u), ˜ 0 )K(u, u0 ) du0 = j 8π k0 E I(u η0 Γ where

 ∂2 2 0 ˆ ) G(u, u0 ), − k0 (ˆ u·u 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,

ˆ ˆ = φ, u

∂ 1 ∂ = , ∂u b ∂φ

ˆ ·u ˆ 0 = cos(φ − φ0 ), u

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 0

2π

2

˜ 0 )Π(φ, φ0 ) dφ0 = A sin k0 bφ + B cos k0 bφ − j 8π I(φ η0

Z

φ

˜ i (ζ) sin k0 b(φ − ζ) dζ. E

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 I2 = 0

and use

φ

∂G(χ − φ0 ) cos k0 b(χ − φ) dχ ∂φ0

∂G(χ − φ0 ) ∂G(χ − φ0 ) = − ∂φ0 ∂χ

to write φ

φ I2 = −G(χ − φ ) cos k0 b(χ − φ) 0 −

Z

= −G(φ − φ0 ) + G(φ0 ) cos k0 bφ −

Z

0

k0 bG(χ − φ0 ) sin k0 b(χ − φ) dχ

0 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 ˆr · r0 = ˆr · (ˆ zz 0 ) = z 0 cos θ and k0 = ω/c, we get Z L µ0 e−jk0 r ˜ ˜ 0 , ω)ejk0 ˆr·r0 dz 0 Eθ (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

L

  r z cos θ 0 I z ,t − + dz 0 . c c −L

Z

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

 mπ 2L

−L

 [z − L] e−γm z cos θi dz

 mπ  mπ i L = −γm cos θi sin [z − L] − cos [z − L]
mπ 2 2 2L 2L 2L + (γ cos θ ) −L m i 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

h

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

L

Z

L

Z

L

Z

Cm1 =

and L

Cm2 = −L −L L Z L

Z =

−L

−L

2 γm Im (z)Im (z 0 )g(z − z 0 , sm ) dz 0 dz c  mπ   mπ  2 γm sin [z − L] sin [z 0 − L] g(z − z 0 , sm ) dz 0 dz. c 2L 2L

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 ± Cm

1 = 2

Z

L

−L

L

 mπ 2

−L

2L

Z

  mπ  2 γm 0 g(z − z , sm ) cos [z ± z 0 ] dz 0 dz. F (z − z , sm ) ∓ c 2L 0

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 R(ξ,ζ) ¯ e G(ζ, s) = ¯ ζ) dξ. R(ξ, 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) = be the Fourier series coefficients of

1 2π

I(φ0 , s)

Z

π

8π 2 γ i E (φ, s), bη0

−π ≤ φ ≤ π,

0

I(φ0 , s)e−jnφ dφ0

−π

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: 2N X In,m (s) In (s) = . s − sn,m m=1

So

2N X ∞ X

αn (s)

m=1 n=−∞

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 [jk0 cos θi sin k0 (u − zm ) − k0 cos k0 (u − zm )] I=− 2 2 2 k0 − k0 cos 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 bm

  jk0 zm cos θi 8π 2 ˜ j cos θi sin k0 (L + zm ) cos k0 (L + zm ) e = −j − − E0 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 E0 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 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

∇ G2D = = = = So 2

∇ G2D +

k02 G2D

  k02 1 (2)00 (2)0 (2) = H0 (k0 ρ) + H (k0 ρ) + H0 (k0 ρ) . 4j k0 ρ 0

(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 = [ˆ ρ · ∇G2D ] dS + k0 G2D dS

I=

Γ

S

195

by (B.26), where Γ is the contour of the disk. But ˆ ∇G2D = ρ

∂ 1 (2) k0 (2)0 k0 (2) ˆ H0 (k0 ρ) = −ˆ H0 (k0 ρ) = ρ ρ H1 (k0 ρ) ∂ρ 4j 4j 4j

so 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 Z

2π

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), ˜ ˜ (2) ˜zi = − ω µ E IH0 (k0 |ρ − ρ0 |). 4 For ρ on the strip, we use ρ0 = dˆ y and ρ = xˆ x so that p |ρ − ρ0 | = d2 + x2 . Thus bm =

p 4 ˜i ˜ (2) (k0 x2m + d2 ). Ez (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.

An+1,n

Amn

1 = k0

Z

3 k ∆ 2 0

1 = k0

1 k ∆ 2 0

Z

k0 ∆(m−n+ 12 )

k0 ∆(m−n− 12 ) (2) H0 (|u|) du

197

(2)

H0 (|u|) du.

1 = k0

Z

3 k ∆ 2 0 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

An−1,n

1 = k0

Z

− 12 k0 ∆

− 32 k0 ∆

Let An+1,n

1 = k0

Z

(2) H0 (|u|) du

3 k ∆ 2 0 1 k ∆ 2 0

h

(2) H0 (u)

1 = k0

Z

3 k ∆ 2 0 1 k ∆ 2 0

(2)

H0 (|x|) dx = An+1,n .

1 − f0 (u) du + k0 i

198

Z

3 k ∆ 2 0 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 k ∆ 2 0 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 k0 ∆     2 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 k ∆ 2 0 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 ˜xs − J˜x (x)Z i (x) = −E ˜xi (x) E where Z i (x) =

1 . [˜ σ (x) + jω(˜ (x) − 0 )]t

Substitution from (7.156) and (7.155) gives the equation −

jω ∂ 2 A˜sex ˜0 sin φ0 ejk0 x cos φ0 − jω A˜sex − J˜x (x)Z i (x) = −E 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 2 ∂2 2 ˜s = − k0 J˜x (x)Z i (x) + k0 E ˜0 sin φ0 ejk0 x cos φ0 . + k A 0 ex ∂x2 jω jω

The solution to this differential equation is A˜sex (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 Z ˜0 1 E k0 x i jk0 x cos φ0 e +j Z (u)J˜x (u) sin k0 (x − u) du. fp (x) = −j ω sin φ0 ω 0

201

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 A˜sex 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 ˜0 1 E 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 0

x

˜0 1 4E 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

Z xm N 4Z i X − Pn (u) sin k0 (xm − u) du = bm an η0 0 n=1

where bm = Examine

Z

˜0 1 4E ejk0 xm cos φ0 . η0 k0 sin φ0

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

Z an

n=1

or

(m−n+ 21 )∆

(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 Fmn

 4Z i   
 
 1 1   k0 η0 cos k0 m − n + 2 ∆ − cos k0 m − n − 2 ∆ , n < m, i = 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 ¯ = B∆, and C¯ = 4C. Then Let a ¯1 = 4a1 , a ¯2 = 4a2 , a ¯3 = 4a3 , A¯ = A∆2 , B ¯ + C¯ a ¯1 = A¯ + 2B ¯ + C¯ a ¯2 = 9A¯ + 6B ¯ + C. ¯ a ¯3 = 25A¯ + 10B Eliminating A¯ gives the two equations ¯ − 8C¯ a ¯2 − 9¯ a1 = −12B ¯ − 24C. ¯ a ¯3 − 25¯ a1 = −40B ¯ then gives Eliminating B ¯ 60¯ a1 − 40¯ a2 + 12¯ a3 = 32C, and so Fa (0) = C =

1 C¯ = [15a1 − 10a2 + 3a3 ] . 4 8

205

1.0