Problem 6.12 The electromagnetic generator shown in Fig. 6-12 is connected to an
electric bulb with a resistance of 150 Ω. If the loop area is 0.1 m2 and it rotates
at 3,600 revolutions per minute in a uniform magnetic flux density B0 = 0.4 T,
determine the amplitude of the current generated in the light bulb.
Solution: From Eq. (6.38), the sinusoidal voltage generated by the a-c generator is
Vemf = Aω B0 sin(ω t +C0 ) = V0 sin(ω t +C0 ). Hence,
V0 = Aω B0 = 0.1 ×
I=
2π × 3, 600
× 0.4 = 15.08 (V),
60
V0 15.08
=
= 0.1 (A).
R
150
Problem 6.18 An electromagnetic wave propagating in seawater has an electric
field with a time variation given by E = ẑE0 cos ω t. If the permittivity of water is
81ε0 and its conductivity is 4 (S/m), find the ratio of the magnitudes of the conduction
current density to displacement current density at each of the following frequencies:
(a) 1 kHz
(b) 1 MHz
(c) 1 GHz
(d) 100 GHz
Solution: From Eq. (6.44), the displacement current density is given by
∂
∂
D=ε E
∂t
∂t
and, from Eq. (4.67), the conduction current is J = σ E. Converting to phasors and
taking the ratio of the magnitudes,
¯
¯ ¯ ¯
¯ ¯
¯e
e ¯¯
σ
¯ J ¯ ¯ σE
.
¯=
¯ ¯=¯
e ¯ ωεr ε0
¯e
Jd ¯ ¯ jωεr ε0 E
Jd =
(a) At f = 1 kHz, ω = 2π × 103 rad/s, and
¯ ¯
¯e
¯
4
¯J¯
= 888 × 103 .
¯ ¯=
3
e
¯ Jd ¯ 2π × 10 × 81 × 8.854 × 10−12
The displacement current is negligible.
(b) At f = 1 MHz, ω = 2π × 106 rad/s, and
¯ ¯
¯
¯e
4
¯J¯
= 888.
¯ ¯=
6 × 81 × 8.854 × 10−12
¯
¯e
2
π
×
10
Jd
The displacement current is practically negligible.
(c) At f = 1 GHz, ω = 2π × 109 rad/s, and
¯ ¯
¯e
¯
4
¯J¯
= 0.888.
¯ ¯=
9 × 81 × 8.854 × 10−12
¯e
¯
2
π
×
10
Jd
Neither the displacement current nor the conduction current are negligible.
(d) At f = 100 GHz, ω = 2π × 1011 rad/s, and
¯ ¯
¯e
¯
4
¯J¯
= 8.88 × 10−3 .
¯ ¯=
11
−12
¯e
¯
2
π
×
10
×
81
×
8.854
×
10
Jd
The conduction current is practically negligible.
Problem 6.25 Given an electric field
E = x̂E0 sin ay cos(ω t − kz),
where E0 , a, ω , and k are constants, find H.
Solution:
E = x̂ E0 sin ay cos(ω t − kz),
e = x̂ E0 sin ay e− jkz ,
E
e
e = − 1 ∇×
×E
H
jω µ
·
¸
1
∂
∂
− jkz
− jkz
ŷ (E0 sin ay e
=−
) − ẑ (E0 sin ay e
)
jω µ
∂z
∂y
E0
[ŷ k sin ay − ẑ ja cos ay]e− jkz ,
=
ωµ
e jω t ]
H = Re[He
½
¾
E0
− jπ /2 − jkz jω t
]e
e
= Re
[ŷ k sin ay + ẑ a cos ay e
ωµ
³
E0 h
π ´i
=
ŷ k sin ay cos(ω t − kz) + ẑ a cos ay cos ω t − kz −
ωµ
2
E0
[ŷ k sin ay cos(ω t − kz) + ẑ a cos ay sin(ω t − kz)] .
=
ωµ
Problem 6.29 The magnetic field in a given dielectric medium is given by
H = ŷ 6 cos 2z sin(2 × 107t − 0.1x) (A/m),
where x and z are in meters. Determine:
(a) E,
(b) the displacement current density Jd , and
(c) the charge density ρv .
Solution:
(a)
H = ŷ 6 cos 2z sin(2 × 107t − 0.1 x) = ŷ 6 cos 2z cos(2 × 107t − 0.1 x − π /2),
e = ŷ 6 cos 2z e− j0.1 x e− jπ /2 = −ŷ j6 cos 2z e− j0.1 x ,
H
e = 1 ∇×
e
×H
E
jωε
¯
¯
¯ x̂
ŷ
ẑ ¯¯
1 ¯¯
∂ /∂ x
∂ /∂ y
∂ /∂ z¯¯
=
jωε ¯¯
0 − j6 cos 2z e− j0.1 x 0 ¯
½ ·
¸
·
¸¾
1
∂
∂
=
x̂ − (− j6 cos 2z e− j0.1 x ) + ẑ
(− j6 cos 2z e− j0.1 x )
jωε
∂z
∂x
¶
µ
¶
µ
j0.6
12
− j0.1 x
− j0.1 x
+ ẑ
.
sin 2z e
cos 2z e
= x̂ −
ωε
ωε
From the given expression for H,
ω = 2 × 107 (rad/s),
β = 0.1 (rad/m).
Hence,
up =
and
εr =
µ
c
up
ω
= 2 × 108
β
¶2
=
Using the values for ω and ε , we have
µ
3 × 108
2 × 108
(m/s),
¶2
= 2.25.
e = (−x̂ 30 sin 2z + ẑ j1.5 cos 2z) × 103 e− j0.1 x (V/m),
E
£
¤
E = −x̂ 30 sin 2z cos(2 × 107t − 0.1 x) − ẑ 1.5 cos 2z sin(2 × 107t − 0.1 x)
(kV/m).
(b)
e = εE
e = εr ε0 E
e = (−x̂ 0.6 sin 2z + ẑ j0.03 cos 2z) × 10−6 e− j0.1 x
D
∂D
,
Jd =
∂t
(C/m2 ),
or
e
e = (−x̂ j12 sin 2z − ẑ 0.6 cos 2z)e− j0.1 x ,
Jd = jω D
Jd = Re[e
Jd e jω t ]
¤
£
= x̂ 12 sin 2z sin(2 × 107t − 0.1 x) − ẑ 0.6 cos 2z cos(2 × 107t − 0.1 x)
(c) We can find ρv from
(A/m2 ).
∇ · D = ρv
or from
∇·J = −
∂ ρv
.
∂t
Applying Maxwell’s equation,
ρv = ∇ · D = ε ∇ · E = εr ε0
µ
∂ Ex ∂ Ez
+
∂x
∂z
¶
yields
½
¤
∂ £
−30 sin 2z cos(2 × 107t − 0.1 x)
∂x
¾
¤
∂ £
−1.5 cos 2z sin(2 × 107t − 0.1 x)
+
∂z
£
¤
= εr ε0 −3 sin 2z sin(2 × 107t − 0.1 x) + 3 sin 2z sin(2 × 107t − 0.1 x) = 0.
ρv = εr ε0