Jackson 6.12. We start with Jackson (6.136): 1 ... Use phasors for this problem, so
that I = I0 cosωt = Re(I0e−iωt) and ... The solution is given by a Bessel function,.
Jackson 6.12 We start with Jackson (6.136): Z Z I 1 ∗ 1 ∗ ~ 3 3 ~ ~ ·n I Vi = J · E d x + 2iω (wl − wm ) d x + S ˆ da 2 i 2 V V (S−Si ) Using Ii = Y Vi (Ii∗ = Y ∗ Vi∗ ), Z Z I 1 ∗ 2 ∗ ~ 3 3 ~ ~ ·n Y |Vi | = J · E d x + 4iω (we − wm ) d x + 2 S ˆ da. 2 V V (S−Si ) Finally, since G = Re Y ∗ and B = Im Y ∗ , Z I Z 1 ∗ ~ 3 3 ~ ~ J · E d x − 4ω Im (we − wm ) d x + 2 Re S ·n ˆ da G= Re |Vi |2 V V (S−Si ) and I Z Z 1 ∗ 3 3 ~ ·n ~ d x − 4ω Re B= S ˆ da (we − wm ) d x + 2 Im J~ · E Im |Vi |2 (S−Si ) V V At low frequencies, ignore the surface integral, assume Ohm’s law holds ~ ~ σ real), and we , wm are real. Thus, (J = σ E, Z 1 ~ 2 d3 x G= σ|E| |Vi |2 V Z 4ω (we − wm )d3 x B= |Vi |2 V
Jackson 6.14 Use phasors for this problem, so that I = I0 cos ωt = Re (I0 e−iωt ) and Maxwell’s equations become ~ = ρ/, ∇ × E ~ = iω B ~ ∇·E ~ = 0, ∇ × B ~ = µ0 J~ − iω/c2 E ~ ∇·B We can rearrange these equations to get the differential equations for the ~ B} ~ = 0. Additionally, by ignoring electric and magnetic fields, (∇2 + k 2 ){E, ~ = Ez (ρ)ˆ edge effects and using the symmetry of the problem, we note that E z ˆ ~ and B = Bφ (ρ)φ. We thus solve our differential equation for Ez : 1 ∂ ∂ 2 2 ρ Ez (ρ) + k 2 Ez (ρ) = 0 (∇ + k )Ez (ρ) = ρ ∂ρ ∂ρ 1
The solution is given by a Bessel function, Ez (ρ) = AJ0 (kρ) To find the amplitude, we use the boundary condition Ez = σ/0 . The alterR nating current I induces a charge Q = I dt = iI0 /ωe−iωt . To lowest order, though, this is a constant. Since the charge distribution is not necessarily constant over the surface, we integrate over the entire plate and set this equal to the total charge: Z a Q iI0 Ez ρ dρ = − = − 2π 0 ω0 0 The integral of J0 is J1 , so, a iI0 2π J1 (ka)A = − k ω0 which yields A=− Thus
iI0 k iI0 =− 2πaω0 J1 (ka) 2πac0 J1 (ka) ~ = − iI0 J0 (kρ) zˆ. E 2πac0 J1 (ka)
Using similar reasoning, ~ =− B
iI0 J1 (kρ) ˆ φ. 2πac2 0 J1 (ka)
6.14(a) To second order in k, J0 (kρ) ≈ 1 − J1 (ka) ≈
k 2 ρ2 + ... 4
ka + ... 2
Plugging this in, iI0 2 k 2 a2 k 2 ρ 2 ~ E = − 1+ + zˆ 2πac0 ka 8 4 2 2 2 2 iI ρ k a k ρ 0 ~ = − B 1+ − φˆ 2πac2 0 a 8 8 2
6.14(b) 0 ~ 2 |E| 4 Z Z a 0 I02 d 4 3 wE d x = + ... 4 4π 2 a2 c2 20 0 k 2 a2 I02 d = 4πa2 ω 2 0 wE =
Similarly, 1 ~ 2 |B| 4µ0 Z µ0 I02 d ω 2 a2 3 wB d x = 1+ 4π 8 12c2 wM =
6.14(c) Z 4ω χ = (wM − wE ) d3 x 2 I µ0 dω d ω 2 a2 = − 1+ 8π 12c2 0 πωa2 −1 + ωL = ωC Thus µ0 d 8π π0 a2 C = d √ √ The √ resonance is given by ωres = 1/ LC ≈ 2 2(c/a). From this, ka ≈ 2 2 = 2.828. J0 has a root near 2.405, which is close to the estimate of ka above. L =
Jackson 7.16 7.16(a) Using the substitutions ∂t → −iω and ∇ → i~k, Maxwell’s relevant equations ~ = −iω D ~ and i~k × E ~ = iω B. ~ Using B ~ = µ0 H, ~ we get become i~k × H ~k × (~k × E) ~ = iω(~k × µ0 H) ~ = −ω 2 µD ~ 3
7.16(b) Using vector idententies and ~k = kˆ n, we write the above result as ~ n − k2E ~ = −µ0 ↔ ~ E k 2 (ˆ n · E)ˆ We can write this equation component-wise in a basis that is aligned with the princial axes of the medium: 2 X v ni nj + − 1 δij Ej = 0. vi2 j We can define two matrices, N = ni nj and W = (1/vi2 − 1)δij and write this as a eigenvalue problem: ~ =0 (N − v 2 W)E In order for this to hold, the det(N − v 2 W) = 0. Many lines of algebra later, it is shown that the Fresnel equations correspond to this determinant. Also, since it is quadratic in v, there are two solutions that are in general different.
7.16(c) ~a · D ~b = D
X
2i Eai Ebi
i
Using the Fresnel equations, Ei = Di =
~ n · E) ni vi2 (ˆ vi2 − v 2 ~ ni (ˆ n · E) µ0 (vi2 − v 2 ) ~b ) X (ˆ n · E~a )(ˆ n·E
n2i µ20 (vi2 − va2 )(vi2 − vb2 ) i ~ b ) X n2 (ˆ n · E~a )(ˆ n·E n2i i = − µ20 (va2 − vb2 ) vi2 − va2 ) vi2 − va2 ) i
~a · D ~b = D
= 0
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Extra Problem Start with the general form of the energy density: 1 ∂ ∂ 2 2 ~ + ~ u = (ω)|E| (ωµ)|B| 4 ∂ω ∂ω 1 ∂ ~ 2 ∂µ ~ 2 = +ω |E| + µ + ω |H| 4 ∂ω ∂ω p √ ~ = /µ(~k × E), ~ so In the case of plane waves, k = ω µ and H 1 ~ ~ ∗| = 1 S = |E ×H 2 2 and
r
~ 2 |E| µ
1 ∂ ∂µ ~ 2 u = ( + ω ) + (µ + ω ) |E| 4 ∂ω µ ∂ω
Calculate the group velocity as −1 −1 ∂ω ω ∂µ ∂µ ∂ √ vg = µ + √ = + = µ ∂k ∂ω 2 µ ∂ω ∂ω −1 ω ∂ ω ∂µ 1 1+ + = √ µ 2 ∂ω 2µ ∂ω Calculating the requested ratio, p ~ 2 1/2 /µ|E| |S| = ∂µ ∂ ~ 2 u 1/4 ( + ω ∂ω ) + µ (µ + ω ∂ω ) |E| p 2 /µ = ∂µ ∂ ) + ω ∂ω 1/4 (2 + ω ∂ω −1 1 ω ∂ ω ∂µ 1+ + = √ µ 2 ∂ω 2µ ∂ω = vg . Thus the energy flows through the material at the group velocity. We can ∂µ ∂k ∂ ∂ ∂k simplify further by noting that ∂ω = ∂k = 0 and ∂ω = ∂µ = 0 such ∂ω ∂k ∂ω the group velocity and energy density reduce to the expressions given in the book.
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