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P8 SOLUTIONS TO PROBLEMS RAPIDLY
TIME-VARYING ELECTROMAGNETIC FIELD
Section 8.1 Displacement Current
PROBLEM 8.1 Displacement current in a capacitor with two dielectric layers. (a) As in Fig.8.1, the displacement current in each of the dielectric layers of the capacitor represents a continuation of the conduction current i in the capacitor terminals (through the wire conductors attached to the capacitor plates). With reference to Fig.P8.1, and having also in mind Eqs.(8.4) and (8.5), we thus write
i = Jd1S = Jd2S , (P8.1)
from which the amplitudes (peak-values) of the displacement current densities in the layers come out to be Jd01 = Jd02 =
I 0
S
. (P8.2)
i
S
e 1 e 2
d 1 d 2
E 1
E 2
J d
J d
D 2
® D 1
Figure P8.1 Displacement current densities and electric field intensities in a parallel-plate capacitor with a two-layer perfect dielectric, connected to a low- frequency time-harmonic generator.
(b) Expressing the displacement current densities in terms of the electric field in- tensities in the layers,
Jd1 =
∂D 1
∂t
= ε 1
∂E 1
∂t
and Jd2 =
∂D 2
∂t
= ε 2
∂E 2
∂t
, (P8.3)
Eq.(8.15) then tells us that the amplitudes of these field intensities are
E 01 =
Jd ωε 1
I 0
2 πε 1 f S
and E 02 =
Jd ωε 2
I 0
2 πε 2 f S
. (P8.4)
(c) From Eq.(2.149), the amplitude of the voltage across the capacitor equals
V 0 = E 01 d 1 + E 02 d 2 =
(ε 2 d 1 + ε 1 d 2 ) I 0 2 πε 1 ε 2 f S
. (P8.5)
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P8. Solutions to Problems: Rapidly Time-Varying Field 213
Note that this result for V 0 can as well be obtained using Eq.(3.45) and the capacitance of the capacitor in Fig.P8.1, given in Eq.(2.150), as follows:
i = C dv dt
−→ V 0 =
I 0
ωC
, where C = ε 1 ε 2 S ε 2 d 1 + ε 1 d 2
. (P8.6)
PROBLEM 8.2 Magnetic field due to the displacement current. For the given (instantaneous) time-harmonic voltage of the capacitor v(t) = V 0 sin ωt, the displacement current density in the capacitor dielectric (both layers) is
Jd(t) =
i(t) S
C
S
dv dt
ε 1 ε 2 ωV 0 ε 2 d 1 + ε 1 d 2
cos ωt
C =
ε 1 ε 2 S ε 2 d 1 + ε 1 d 2
. (P8.7)
Due to symmetry, magnetic-field lines in the dielectric are circles centered at the capacitor axis perpendicular to the plates, and an application of the corrected gen- eralized Amp`ere’s law in integral form, Eq.(8.7), to a circular contour (C) of radius r (r ≤ a) centered at the capacitor axis and the flat surface (SC ) spanned over the contour, as in Fig.8.2 and Eq.(8.19), results in the following expression for the magnetic field intensity at an arbitrary point in the dielectric:
H 2 πr = Jdπr^2 −→ H(r, t) =
Jd(t) r 2
ε 1 ε 2 ωV 0 r 2(ε 2 d 1 + ε 1 d 2 )
cos ωt (0 ≤ r ≤ a). (P8.8) In particular, the field intensity at the dielectric-air interface equals
H(a, t) =
ε 1 ε 2 ωV 0 a 2(ε 2 d 1 + ε 1 d 2 )
cos ωt (r = a). (P8.9)
PROBLEM 8.3 Displacement current in an ideal spherical capacitor. This is a spherical version of the displacement-current computation in the parallel- plate capacitor in Example 8.1. The voltage of the capacitor, v(t) = V 0 cos ωt, is a low-frequency one, so that the electric field in the capacitor dielectric can be considered as quasistatic. Due to spherical symmetry, the vector E in the dielectric is radial with respect to the capacitor center, E = E ˆr, as in Fig.2.16, and, using Eqs.(2.117)-(2.119),
E(r, t) =
Q(t) 4 πεr^2
Cv(t) 4 πεr^2
abv(t) (b − a)r^2
C =
4 πεab b − a
. (P8.10)
From Eq.(8.5), the displacement current density vector in the dielectric is also radial, Jd = Jd ˆr, and its magnitude is given by
Jd(r, t) =
dD dt = ε
dE dt
εab (b − a)r^2
dv dt
ωεabV 0 (b − a)r^2 sin ωt (a < r < b). (P8.11) Alternatively, we can first find the conduction current intensity in the capacitor terminals, by means of Eqs.(3.45) and (2.119),
i(t) = C dv dt
4 πωεabV 0 b − a
sin ωt , (P8.12)
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P8. Solutions to Problems: Rapidly Time-Varying Field 215
with ˆr being the radial unit vector in the cylindrical coordinate system whose z-axis coincides with the cable axis.
(b) For a lossy dielectric of the cable, Jd is the same as in Eq.(P8.20), whereas the total (conduction plus displacement) current density vector comes out to be
Jtot = J + Jd = σE + Jd =
r ln(b/a)
[
σv(t) + ε
dv dt
]
ˆr. (P8.21)
PROBLEM 8.6 Conduction to displacement current ratio for water. (a) For a sample of fresh water with εr = 80 and σ = 10−^3 S/m, the ratio of the amplitude of the density of the conduction current to that of the displacement current, Eq.(8.21), is given by
n = |J|max |Jd|max
σ ωε
σ 2 πf εrε 0
224. 7 × 103
f
(f in Hz) , (P8.22)
so that frequencies at which n = 0.001, 1, and 1000 turn out to be f 1 = 224.7 MHz, f 2 = 224.7 kHz, and f 3 = 224.7 Hz, respectively.
(b) In the case of seawater with εr = 80 and σ = 4 S/m, on the other side, we have
n =
898. 8 × 106
f
(f in Hz) , (P8.23)
which results in f 1 = 898.8 GHz, f 2 = 898.8 MHz, and f 3 = 898.8 kHz, respectively.
Section 8.2 Maxwell’s Equations for the Rapidly
Time-Varying Electromagnetic Field
PROBLEM 8.7 Divergence Maxwell’s equations from curl equations. Maxwell’s fourth equation (law of conservation of magnetic flux) for the rapidly time-varying electromagnetic field in differential form, in Eqs.(8.24), is derived from the differential form of the first equation (Faraday’s law of electromagnetic induc- tion) in Example 8.5. Similarly, by taking the divergence of both sides of the differential form of Maxwell’s second differential equation (corrected generalized differential Amp`ere’s law), Eqs.(8.24), and combining this with the differential form of the continuity equation (for rapidly time-varying currents), Eq.(3.39), we obtain [also see Eqs.(8.10), (8.11), and (5.130)] the third equation (generalized differential Gauss’ law), as follows:
∇ · (∇ × H)
≡ 0
= ∇·J+∇·
∂D
∂t
∂ρ ∂t
∂t
(∇ · D) −→ ∇·D = ρ. (P8.24)
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216 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
PROBLEM 8.8 Flux Maxwell’s equations from circulation equations. Exactly as in Example 8.4, we consider an arbitrary closed surface S and a contour C that splits it into two parts, S 1 and S 2 , apply Maxwell’s second equation (corrected generalized Amp`ere’s law) in integral form, Eqs.(8.23), to C and S 1 , on one side, and C and S 2 , on the other, and obtain Eqs.(8.25) and (8.26), ∮
S
J +
∂D
∂t
S
J · dS = −
S
∂D
∂t
· dS = − d dt
S
D · dS. (P8.25)
We then combine this result with the general (high-frequency) continuity equation in integral form, Eq.(3.38), which leads to Maxwell’s third equation (generalized Gauss’ law) in integral form [Eqs.(8.23)]:
d dt
S
D · dS = −
S
J · dS =
v
∂ρ ∂t
dv =
d dt
v
ρ dv −→
S
D · dS =
v
ρ dv. (P8.26) Similarly, the procedure in Eqs.(8.25) and (8.26) now with the integral form of Maxwell’s first equation (Faraday’s law of electromagnetic induction) applied to the contour C and surfaces S 1 and S 2 , respectively, gives the fourth equation (law of conservation of magnetic flux),
S
∂B
∂t · dS = 0 −→
d dt
S
B · dS = 0 −→
S
B · dS = 0 , (P8.27)
where the reason why we have zero rather than an arbitrary constant on the right- hand side of the equation is explained in Example 8.5.
PROBLEM 8.9 Magnetic from electric field of an antenna using Maxwell’s equations. Starting from the electric field intensity vector radiated by the antenna, given by E(r, θ, t) = E 0 sin θ cos(ωt − βr) ˆθ/r = Eθ ˆθ (β = ω
ε 0 μ 0 ), and using Maxwell’s first equation in differential form, Eqs.(8.24), and the formula for the curl in spherical coordinates, Eq.(4.85), which retains here only one (the fifth) term (out of six terms), we can write
∇ × E = −μ 0
∂H
∂t
∂H
∂t
μ 0
∇ × E = −
μ 0 r
∂r
(rEθ ) φˆ
E 0 sin θ μ 0 r
∂r
cos(ωt − βr) φˆ = − βE 0 sin θ μ 0 r
sin(ωt − βr) φˆ , (P8.28)
so that, by integration in time, the magnetic field intensity vector of the antenna, at the same far point, amounts to
H = −
βE 0 sin θ μ 0 r
sin(ωt − βr) dt φˆ = E 0 sin θ r
ε 0 μ 0
cos(ωt − βr) φˆ ( β ωμ 0
ω
ε 0 μ 0 ωμ 0
ε 0 μ 0
. (P8.29)
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218 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
108 t − z −
π 2
ˆy
]
A/m −→ H = 0. 5
[
e−jz^ xˆ + e−jz^ e−jπ/^2 ˆy
]
A/m
= 0. 5
2 e−jz^ ( ˆx − j ˆy) A/m (z in m). (P8.33)
(c) In the last case, similarly,
E(t) = − 0 .5 sin 0. 01 y sin(3× 106 t) ˆz V/m = − 0 .5 sin 0. 01 y cos
3 × 106 t − π 2
ˆz V/m
= 0.5 sin 0. 01 y cos
3 × 106 t +
π 2
ˆz V/m −→ E = 0. 25
2 sin 0. 01 y ejπ/^2 ˆz V/m
= j0. 25
2 sin 0. 01 y ˆz V/m (y in m) , (P8.34) where the use is also made of the trigonometric identity − cos α = cos(α + π).
PROBLEM 8.12 Converting complex vectors to instantaneous expres- sions. (a) We first recast the two components of the vector E in polar (exponential) form, as in Eq.(8.64),
E = jE 0 ejθ^0 sin βz e−jβx^ ˆx + E 0 ejθ^0 cos βz e−jβx^ ˆz = E 0 sin βz ej(−βx+θ^0 +π/2)^ ˆx
+E 0 cos βz ej(−βx+θ^0 )^ ˆz
j = ejπ/^2
, (P8.35)
and then invoke Eq.(8.66) to convert them to the instantaneous (time-domain) ex- pressions,
E(t) = E 0
2 sin βz cos(ωt − βx + θ 0 + π/2) ˆx + E 0
2 cos βz cos(ωt − βx + θ 0 ) ˆz
= −E 0 sin βz sin(ωt − βx + θ 0 ) ˆx + E 0 cos βz cos(ωt − βx + θ 0 ) ˆz (P8.36) [cos(α + π/2) = − sin α].
(b) In a similar way,
H = jhH 0 ejψ^0 sin
( (^) π a
x
e−jβz^ xˆ + H 0 ejψ^0 cos
( (^) π a
x
e−jβz^ ˆz
= hH 0 sin
( (^) π a
x
ej(−βz+ψ^0 +π/2)^ ˆx + H 0 cos
( (^) π a
x
ej(−βz+ψ^0 )^ ˆz −→
H(t) = −hH 0
2 sin
( (^) π a
x
sin(ωt−βz +ψ 0 ) ˆx+H 0
2 cos
( (^) π a
x
cos(ωt−βz +ψ 0 ) ˆz. (P8.37)
(c) Here, using the facts that j^2 = −1, j^3 = −j, and 1/j = −j, we have
E = bI e−jβr
[
(jβr)^2
(jβr)^3
]
ˆr +
[
jβr
(jβr)^2
(jβr)^3
]
θ^ ˆ
= bI ejψ^ e−jβr
[
(βr)^2
j (βr)^3
]
ˆr +
[
j βr
(βr)^2
j (βr)^3
]
θ^ ˆ
, (P8.38)
so that Eq.(8.66) gives [also see Eqs.(P8.35) and (P8.36)]
E(t) = −bI
[
cos(ωt − βr + ψ) (βr)^2
sin(ωt − βr + ψ) (βr)^3
]
ˆr
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P8. Solutions to Problems: Rapidly Time-Varying Field 219
[
sin(ωt − βr + ψ) βr
cos(ωt − βr + ψ) (βr)^2
sin(ωt − βr + ψ) (βr)^3
]
θ^ ˆ
. (P8.39)
Section 8.8 Maxwell’s Equations in Complex
Domain
PROBLEM 8.13 Divergence-free equivalent electric displacement vec- tor. (a) Using the differential Maxwell’s third equation and continuity equation in the complex domain, Eqs.(8.81) and (8.82), we have
∇ · (εeE) = ∇ ·
[(
ε − j
σ ω
E
]
= ∇ · (εE) −
j ω ∇ · (σE) = ∇ · D −
j ω
∇ · J
= ρ − j ω
(−jωρ) = ρ − ρ = 0. (P8.40)
(b) Similarly, from the corresponding integral equations, namely, the third equation in Eqs.(8.80) and the complex version of the first equation in Eqs.(8.34), ∮
S
(εeE) · dS =
S
[(
ε − j
σ ω
E
]
· dS =
S
(εE) · dS −
j ω
S
(σE) · dS
S
D · dS −
j ω
S
J · dS =
v
ρ dv −
j ω
−jω
v
ρ dv
= 0. (P8.41)
Of course, Eqs.(P8.40) and (P8.41), namely, the facts that εeE is a divergence- free vector and that the flux of εeE through any closed surface is zero, are equivalent to each other, like the differential and integral forms of the law of conservation of magnetic flux (Maxwell’s fourth equation) in Eqs.(8.81) and (8.80).
PROBLEM 8.14 Boundary condition for the equivalent displacement vector. With the use of the boundary condition for normal components of the complex electric flux density vector, D, in Eq.(8.85), and that for normal compo- nents of the complex current density vector, J, in Eqs.(8.86), we obtain
n ˆ · (εe1E 1 ) − ˆn · (εe2E 2 ) = ˆn ·
[(
ε 1 − j
σ 1 ω
E 1
]
− ˆn ·
[(
ε 2 − j
σ 2 ω
E 2
]
= nˆ · (ε 1 E 1 ) − ˆn · (ε 2 E 2 ) −
j ω
[ˆn · (σ 1 E 1 ) − ˆn · (σ 2 E 2 )]
= ˆn · D 1 − ˆn · D 2 −
j ω
[nˆ · J 1 − nˆ · J 2 ] = ρs −
j ω
−jωρs
= 0. (P8.42)
This result, namely, the boundary condition telling us that the normal compo- nent of the vector εeE (εe = ε−jσ/ω) must be continuous across a boundary surface between two media in a high-frequency time-harmonic electromagnetic field, can, of course, be derived, as in Fig.2.10(b), from the corresponding integral equation for the flux of the vector εeE (previous problem).
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P8. Solutions to Problems: Rapidly Time-Varying Field 221
in Eqs.(8.92) and (8.93) ought to be replaced by jω × jω = −ω^2 in the complex domain, which gives the complex forms of wave equations for Lorenz potentials:
∇^2 V + ω^2 εμV = −
ρ ε
and ∇^2 A + ω^2 εμA = −μJ. (P8.48)
(b) To derive complex wave equations in Eqs.(P8.48) from complex Maxwell’s equa- tions in differential form, paralleling the time-domain derivation in Eqs.(8.89)-(8.93), we first substitute the complex version of Eq.(6.43),
E = −jωA − ∇V , (P8.49)
into the complex generalized Gauss’ law in differential form, Eqs.(8.81), where D is previously expressed as εE. This, combined with Eq.(2.92), results in
∇ · (∇V ) = ∇^2 V = −
ρ ε
− jω∇ · A. (P8.50)
Similarly, substituting Eqs.(6.28) and (P8.49) into the corrected generalized Amp`ere’s law in Eqs.(8.81) with H replaced by B/μ and employing Eq.(4.124), we can write
∇ × (∇ × A) = ∇(∇ · A) − ∇^2 A = μJ + ω^2 εμA − jωεμ∇V. (P8.51)
Finally, with ∇ · A expressed as
∇ · A = −jωεμV , (P8.52)
which is the complex-domain Lorenz condition [Eq.(8.115)], Eqs.(P8.50) and (P8.51) simplify to the respective wave equations in Eqs.(P8.48).
PROBLEM 8.18 One-dimensional source-free wave equation in complex form. The complex-domain equivalent of the one-dimensional wave equation (for U ) in Eq.(8.97) is given by
d^2 U dR^2
ω^2 c^2
U = 0 −→
d^2 U dR^2
β = ω c
, (P8.53)
where the use is made of Eq.(8.111) defining the phase coefficient β. We take the second derivative of U = e−jβR,
d^2 U dR^2
d^2 dR^2
e−jβR^ = (−jβ)^2 e−jβR^ = −β^2 U , (P8.54)
and what we obtain is the wave equation in Eq.(P8.53), which proves that this expression for U is a solution of the 1-D wave equation. From Eq.(8.96), we then have that
V =
U
R
e−jβR R
(P8.55)
is a solution of the three-dimensional wave equation in complex form for the electric potential V (from the previous problem), namely, the complex version of Eq.(8.92).
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222 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
Section 8.10 Computation of High-Frequency
Potentials and Fields in Complex Domain
PROBLEM 8.19 Gradient of the source-to-field distance. The operator ∇ in the gradient of the source-to-field distance (R) in Fig.8.7 acts on the coordi- nates of the field point (P), and an application of the formula for gradient in the Cartesian coordinate system, Eq.(1.102), with R given by Eq.(8.124) and x, y, and z as independent variables (x′, y′, and z′^ are fixed in this operation) leads to
∇R =
∂R
∂x
xˆ +
∂R
∂y
ˆy +
∂R
∂z
ˆz , where R =
(x − x′)^2 + (y − y′)^2 + (z − z′)^2. (P8.56) Since the individual components of ∇R can be computed as
∂R ∂x
(x − x′)^2 + (y − y′)^2 + (z − z′)^2
2(x − x′) =
(x − x′) R
∂R
∂y
(y − y′) R
∂R
∂z
(z − z′) R
, (P8.57)
we obtain, having in mind Eqs.(8.123),
∇R =
(x − x′) ˆx + (y − y′) ˆy + (z − z′) ˆz R
r − r′ R
= Rˆ (R = |r − r′|). (P8.58) Of course, this result matches the one in Eq.(8.122).
PROBLEM 8.20 High-frequency current in a semicircular wire conduc- tor. (a) As in Eq.(8.137), the complex rms line charge density accompanying the current I(φ) = I 0 cos φ (0 ≤ φ ≤ π) in Fig.8.16 is found to be
Q′^ =
j ω
dI dl
j ωa
dI dφ
jI 0 ωa
sin φ. (P8.59)
(b) Similarly to the computation in Eq.(8.138), the electric scalar potential at an arbitrary point (defined by a coordinate z) along the z-axis in Fig.8.16 is computed as V =
4 πε 0
l
Q′(φ) e−jβR^ dl R
jI 0 e−jβR 4 πε 0 ωR
∫ (^) π
φ=
sin φ dφ = − jI 0 e−jβR 2 πε 0 ωR ( β = ω
ε 0 μ 0 , R =
a^2 + z^2 , dl = a dφ
. (P8.60)
(c) Since the angle φ in Fig.8.10(b) is for 90◦^ smaller than that (angle φ) in Fig.8.16, the decomposition of the elementary line vector dl along the wire in Eq.(8.139) can be written, for the situation in in Fig.8.16, as
dl = − dl sin (φ − 90 ◦) ˆx + dl cos (φ − 90 ◦) ˆy = dl cos φ ˆx + dl sin φ yˆ , (P8.61)
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224 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
and the complex H-vector is H = −I 0 az(1 + jβR) e−jβR^ yˆ/(8R^3 ).
(f) Using Eq.(8.66), the instantaneous (time-domain) equivalent of A in Eq.(P8.62) is A(t) =
2 μ 0 I 0 a 8 R
cos(ωt − βR) ˆx , (P8.68)
and that of E in Eqs.(P8.64) and (P8.65) [see also Eq.(8.136)]
E(t) =
2 ωμ 0 I 0 a 8 R
[
sin(ωt − βR) −
1 + β^2 R^2 β^2 R^2
sin(ωt − βR + arctan βR)
]
ˆx
2 I 0 z
1 + β^2 R^2 2 πωε 0 R^3
sin(ωt − βR + arctan βR) ˆz. (P8.69)
PROBLEM 8.21 High-frequency line current along 3/4 of a circle. Com- putations in this problem parallel those in the previous one, as well as those in Example 8.13.
(a) The complex line charge density along the wire in Fig.8.17 is
I(φ) = I 0 sin φ −→ Q′^ =
j ω
dI dl
j ωa
dI dφ
jI 0 ωa cos φ (0 ≤ φ ≤ 3 π/2). (P8.70)
(b) The electric scalar potential at the z-axis in Fig.8.17 amounts to
V =
4 πε 0
l
Q′(φ) e−jβR^ dl R
jI 0 e−jβR 4 πε 0 ωR
∫ (^3) π/ 2
φ=
cos φ dφ = −
jI 0 e−jβR 4 πε 0 ωR
, (P8.71)
where β = ω
ε 0 μ 0 and R =
a^2 + z^2.
(c) The elementary line vector dl along the wire is given in Eq.(8.139), so that the components of the magnetic vector potential at the z-axis come out to be
Ax = − μ 0 4 π
l
I(φ) dl sin φ e−jβR R
μ 0 I 0 a e−jβR 4 πR
∫ (^3) π/ 2
0
sin^2 φ dφ
3 μ 0 I 0 a e−jβR 16 R , Ay =
μ 0 I 0 a e−jβR 4 πR
∫ (^3) π/ 2
0
sin φ cos φ dφ
μ 0 I 0 a e−jβR 8 πR
∫ (^3) π/ 2
0
sin 2φ dφ =
μ 0 I 0 a e−jβR 8 πR
, (P8.72)
and hence A = μ 0 I 0 a e−jβR(− 3 π ˆx + 2 ˆy)/(16πR).
(d) Using the expression for the source-to-field unit vector Rˆ in Eq.(8.142), the x-component of the electric field vector at the z-axis in Fig.8.17 is found to be
Ex = −jωAx +
4 πε 0
l
Q′(φ) dl
(1 + jβR) e−jβR R^2
a R cos φ
= −jωAx
jI 0 a(1 + jβR) e−jβR 4 πωε 0 R^3
∫ (^3) π/ 2
0
cos^2 φ dφ = −jωAx −
3jI 0 a(1 + jβR) e−jβR 16 ωε 0 R^3
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P8. Solutions to Problems: Rapidly Time-Varying Field 225
3jωμ 0 I 0 a e−jβR 16 R
1 + jβR β^2 R^2
. (P8.73)
Similarly, the other two components of E are calculated as
Ey = −jωAy −
jI 0 a(1 + jβR) e−jβR 4 πωε 0 R^3
∫ (^3) π/ 2
0
sin φ cos φ dφ = −jωAy
jI 0 a(1 + jβR) e−jβR 8 πωε 0 R^3
jωμ 0 I 0 a e−jβR 8 πR
1 + jβR β^2 R^2
Ez =
jI 0 z(1 + jβR) e−jβR 4 πωε 0 R^3
∫ (^3) π/ 2
0
cos φ dφ = −
jI 0 z(1 + jβR) e−jβR 4 πωε 0 R^3
. (P8.74)
(e) The vector dl × Rˆ needed for the magnetic field evaluation is that in Eq.(8.145), so that
Hx =
I 0 az(1 + jβR) e−jβR 4 πR^3
∫ (^3) π/ 2
0
sin φ cos φ dφ =
I 0 az(1 + jβR) e−jβR 8 πR^3
Hy = I 0 az(1 + jβR) e−jβR 4 πR^3
∫ (^3) π/ 2
0
sin^2 φ dφ = 3 I 0 az(1 + jβR) e−jβR 16 R^3
Hz = I 0 a^2 (1 + jβR) e−jβR 4 πR^3
∫ (^3) π/ 2
0
sin φ dφ = I 0 a^2 (1 + jβR) e−jβR 4 πR^3
, (P8.75)
and H = I 0 a(1 + jβR) e−jβR(2z ˆx + 3πz ˆy + 4a ˆz)/(16πR^3 ).
(f) By means of Eq.(8.66), the instantaneous magnetic potential equals
A(t) =
2 μ 0 I 0 a 16 πR
(− 3 π ˆx + 2 yˆ) cos(ωt − βR) , (P8.76)
whereas the instantaneous electric field vector is
E(t) =
2 ωμ 0 I 0 a 16 R
[
− sin(ωt − βR) +
1 + β^2 R^2 β^2 R^2
sin(ωt − βR + arctan βR)
]
ˆx
2 ωμ 0 I 0 a 8 πR
[
sin(ωt − βR) +
1 + β^2 R^2 β^2 R^2
sin(ωt − βR + arctan βR)
]
ˆy
2 I 0 z
1 + β^2 R^2 4 πωε 0 R^3
sin(ωt − βR + arctan βR) ˆz. (P8.77)
PROBLEM 8.22 Abrupt change of current intensity in a circular loop. (a) As βa = ω/c 0 = 3.33 (c 0 = 3 × 108 m/s), the quasistatic condition βa ≪ 1 [Eq.(8.130)] is not met, which verifies that this is a high-frequency current.
(b) Based on Eq.(8.66), we convert the loop current to the complex domain,
i(t, φ) = cos
3 φ 2
sin 10^10 t A = cos
3 φ 2
cos
1010 t −
π 2
A (0 < φ < 2 π ; t in s)
−→ I(φ) =
cos
3 φ 2
e−jπ/^2 = −j707.1 cos
3 φ 2
mA , (P8.78)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
P8. Solutions to Problems: Rapidly Time-Varying Field 227
d B
S
z
R
z
P
O
d I P'
d r (^) J s( ) r ®
r a
b
O
Figure P8.2 Computing the magnetic field (at the z-axis) due to a high-frequency circular surface current distribution given by Js = Js0(a/r) φˆ (a ≤ r ≤ b).
PROBLEM 8.23 High-frequency circular surface current over a hollow plate. From Eq.(8.66), the complex rms surface current density vector over the plate in Fig.8.5 equals
Js(r) =
Js0a r
φˆ (a ≤ r ≤ b). (P8.85)
Invoking the superposition principle, we subdivide the plate into elemental rings of width dr, as shown in Fig.P8.2. Using Eq.(3.13), the current of a ring of radius r is
dI = Js(r) dr , (P8.86)
and it can be viewed as the current of an equivalent circular loop of radius r with a high-frequency current of constant magnitude ( dI). Based on Eq.(8.135), the magnetic flux density vector of this loop at an arbitrary point P at the z-axis (defined by a coordinate z) in Fig.P8.2 amounts to
dB =
μ 0 dI r^2 (1 + jβR) e−jβR 2 R^3
ˆz , β = ω
ε 0 μ 0 , R =
r^2 + z^2 , (P8.87)
and the total field B at the point P is given by
B =
S
dB = μ 0 Js0a 2
ˆz
∫ (^) b
r=a
(1 + jβR) e−jβR^ r dr R^3
. (P8.88)
With the use of Eq.(1.62) to change variables in integration (see Fig.1.14) and the expression for the derivative in R of e−jβR/R in Eq.(8.121) to carry out the integration (instead of differentiation), we then have
B =
μ 0 Js0a 2
ˆz
∫ (^) b
r=a
(1 + jβR) e−jβR R^2
dR = −
μ 0 Js0a 2
ˆz
∫ (^) b
r=a
d dR
e−jβR R
dR
μ 0 Js0a 2
ˆz
∫ (^) b
r=a
d
e−jβR R
μ 0 Js0a 2
ˆz
e−jβR R
b
r=a
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228 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
μ 0 Js0a 2
e−jβ
√ a^2 +z^2 √ a^2 + z^2
e−jβ
√ b^2 +z^2 √ b^2 + z^2
ˆz. (P8.89)
Note that the magnetic field computed in Problem 4.4 represents the quasistatic version, for βR ≪ 1 [Eq.(8.130)], so for e−jβ
√a (^2) +z 2 ≈ 1 and e−jβ
√b (^2) +z 2 ≈ 1, of that in Eq.(P8.89).
PROBLEM 8.24 Uniform high-frequency plate current. (a) Using the second expression in Eqs.(8.116), the complex rms magnetic vector potential at an arbitrary point along the plate axis normal to its plane, z-axis in Fig.P8.3, can be found as A = μ 0 4 π
S
Js e−jβR^ dS R
μ 0 Js 4 π
∫ (^) a
r=
e−jβR R
(^2) ︸ ︷︷ ︸πr dr dS ( Js = const , β = ω
ε 0 μ 0 , R =
r^2 + z^2
, (P8.90)
where Js can be taken out of the integral because it is constant (the same at every point of the plate) and dS is the area of an elementary ring (adopted for integra- tion) of radius r and width dr, shown in Fig.1.14 (and Fig.P8.3) and computed in Eq.(1.60). By means of Eq.(1.62), we then change integration variables, from r to R in Fig.P8.3, and obtain
A =
μ 0 Js 2
∫ (^) a
r=
e−jβR^ dR =
jμ 0 Js 2 β
e−jβR
∣a r=0 =
jμ 0 Js 2 β
e−jβ
√ a^2 +z^2 − e−jβ|z|
(P8.91)
B
S
z
R
z
P
O
d r P'
J s
r a
x
y
R
a
A
J s
Figure P8.3 Evaluation of the magnetic vector potential and flux density vec- tor (at the z-axis) due to a high-frequency surface current, of density Js, flowing uniformly (Js = const) over a circular plate in free space.
(b) Similarly, starting with Eq.(8.128), the magnetic flux density vector at the plate axis is given by
B =
μ 0 4 π
S
(Js dS) × Rˆ (1 + jβR) e−jβR R^2
μ 0 4 π
Js ×
S
R^ ˆ (1 + jβR) e
−jβR R^2
(^2) ︸ ︷︷ ︸πr dr dS
(P8.92)
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230 Branislav M. Notaroˇs: Electromagnetics (Pearson Prentice Hall)
where [similarly to the computation in Eq.(8.153)] θˆ in the integral is replaced by its z-component, − sin θ ˆz (Fig.P8.4), as it is obvious (before the integration) that A has a z-component only, due to symmetry. In addition, since the function in the integrand depends on the angle θ only, the surface dS′^ is extended to the surface dS in the form of a thin ring of radius a sin θ and width a dθ [see Fig.1.16 and Eq.(1.65)] – our surface-integration strategy is always to adopt as large as possible dS (Section 1.5). Finally, the resulting integral in θ is solved in Eq.(5.46), and it equals 4/3.
J s
O
q
z
d S'
A a
S
d S
Figure P8.4 Finding the magnetic vector potential (at the point O) due to a high- frequency surface current, with density Js = Js(θ) ˆθ (0 ≤ θ ≤ π), over the surface of a sphere in free space.
PROBLEM 8.26 High-frequency surface currents over a sphere, φ- directed. We apply the superposition principle and subdivide the sphere surface with current into thin rings of width dlr = a dθ, as in Fig.1.16. Each such ring can be viewed as an equivalent circular wire loop with the high-frequency current (constant along the loop)
dI = Js dlr = Jsa dθ = Js0a dθ (0 ≤ θ ≤ π) , (P8.98)
where Js = Js0 ej0^ = Js0 is the complex rms current density over the surface (Js = Js φˆ).
(a) Eq.(8.134) then tells us that the magnetic vector potential at the sphere center due to all these rings (equivalent loops) is zero, and so is the total potential, A = 0.
(b) From Eq.(8.135), the magnetic flux density vector at the sphere center due to the current loop whose position is defined by an angle θ, and whose radius is ar = a sin θ (Fig.1.16), is given by
dB =
μ 0 dI (a sin θ)^2 (1 + jβa) e−jβa 2 a^3
ˆz , β = ω
ε 0 μ 0 , R = a , (P8.99)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
P8. Solutions to Problems: Rapidly Time-Varying Field 231
so that the total complex B-vector amounts to
B =
S
dB =
μ 0 Js0(1 + jβa) e−jβa 2
ˆz
∫ (^) π
θ=
sin^2 θ dθ =
πμ 0 Js0(1 + jβa) e−jβa 4
ˆz. (P8.100) Note that the magnetic field computed in Problem 4.8 represents the quasistatic version, for βa ≪ 1 [Eq.(8.130)], i.e., for βa ≈ 0, of that in Eq.(P8.100).
PROBLEM 8.27 High-frequency volume current in a hollow hemi- sphere. (a) Since J = J ˆz = const, Eq.(8.82) tells us that the volume charge density is ρ = 0 throughout the volume of the hemisphere in Fig.8.18. With reference to Fig.P8.5, the surface charge density ρs1 is given by Eq.(8.150), and the charge densities ρs2 and ρs3 are found in a similar fashion,
ρs1 = j ω
(nˆ · J 1 − ˆn · J 2 ) = − j ω
ˆn ·J = − j ω
J cos θ , ρs2 = j ω
J cos θ , ρs3 = j ω
J.
(P8.101)
O
q
q
q
J (^) rs
n
J
J
n n
rs
rs
z
Figure P8.5 Evaluation of the distribution of surface charges accompanying the high-frequency uniform volume current in the hollow hemisphere in Fig.8.18, and of the electric scalar potential (at the point O) due to these charges.
(b) Using the first expression in Eqs.(8.116), the electric scalar potential at the center of the hemisphere (point O) due to the surface charge of density ρs1 in Fig.P8.5 equals
V 1 =
4 πε 0
S 1
ρs1 e−jβR^ dS 1 R
jJ e−jβb 4 πωε 0 b
∫ (^) π/ 2
θ=
cos θ (^2) ︸πb 2 sin︷︷ θ dθ︸ dS 1
jJb e−jβb 2 ωε 0
∫ (^) π/ 2
0
sin θ cos θ dθ = −
jJb e−jβb 4 ωε 0
∫ (^) π/ 2
0
sin 2θ dθ = −
jJb e−jβb 4 ωε 0
(P8.102)
where β = ω
ε 0 μ 0 and dS 1 is the area of an elementary ring (adopted for integra- tion) of radius b sin θ and width b dθ, shown in Fig.1.16 and computed in Eq.(1.65). By the same token, the potential due to the charge of density ρs2 is
V 2 =
4 πε 0
S 2
ρs2 e−jβa^ dS 2 a
jJa e−jβa 4 ωε 0
. (P8.103)
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