electromagnetic Fast help
strength
practiceSolution.pdf
EELE 334 ELECTROMAGNETIC THEORY I Fall 2017—Practice Final Exam Solutions 1. [40 points] Short answer (a) [8] For the electric field, we have
€
D1n = D2n with no surface charge density, so
€
ε1E1n = ε2E2n, where
€
ε1 and
€
ε2 are the dielectric permittivity of the two materials. For the magnetic field, we have that
€
B1n = B2n , or
€
µ1H1n = µ2H2n where
€
µ1 and
€
µ2 are the magnetic permeability of the two materials. (b) [8] The displacement current takes into account the fact that a time varying electric field can induce a current elsewhere, such as inside a capacitor. Thus, for consistency, we must consider both the “true” conduction current and the induced current due to the varying electric field. This is important in electromagnetics as both the conduction current and the displacement current are needed in order to maintain conservation of charge, as seen for example in Ampère’s law:
€
H ⋅ d l
C ∫ = IC +
∂ D ∂t
⋅ d s S ∫ , where the second term on the right side is the displacement current.
(c) [8] Ferromagnetic materials have the unique property that they can retain a net magnetization even without an external applied magnetic field. This is due to hysteresis: the magnetization of a ferromagnetic material depends not only on the current externally applied magnetic field, but also on the past (in other words, the material has a “memory”). Applying an external magnetic field to a ferromagnetic material can reorient the magnetic domains in the material, producing a net macroscopic magnetization. Even after this external field is removed, the material may retain a net magnetization.
(d) [8] The current-charge continuity relation is
€
∇⋅ J = −
∂ρv ∂t
. This expresses the relationship
between charge and current, requiring a charge source or sink in order to produce a spatially varying current and therefore conserving net charge. (e) [8] The Brewster angle is the angle at which the reflection coefficient for a plane wave impinging on a planar interface is zero. This can occur only for the parallel polarization at oblique incidence (with non-magnetic materials). 2. [55 points] Transmission Lines
(a) [10]
€
Γ = ZL − Z0 ZL + Z0
= 50 + 25 j( ) −50 50 + 25 j + 50
= 25 j
100 + 25 j = 0.059 + 0.235 j = 0.243e j0.422π .
€
SWR = 1+ Γ 1− Γ
= 1+ 0.243 1−0.243
=1.64
(b) [15]
€
Zin = Z0 1+ Γl 1−Γl
where
€
Γl = Γe − j2βl . Here,
€
Γl = 0.243e j0.422π( )e− j2.4π = 0.243e− j1.978π = 0.243e j0.022π , so
€
Zin = 50( ) 1+ 0.243e j0.022π
1−0.243e j0.022π = 81.906e j0.011π = 81.854 + 2.916 j.
(c) [30] At the midpoint of the line,
€
d = 0.3λ , and
€
Γ0.3λ = Γe − j2β 0.3λ = 0.243e j0.422πe− j1.2π = 0.243e− j0.778π . So the input impedance here is
€
Z0.3λ = Z0 1+ Γd 1−Γd
= 50( ) 1+ 0.243e− j0.778π
1−0.243e− j0.778π = 32.894 − j10.889 or
€
Z0.3λ = 34.650e − j0.102π . The shunt
impedance is
€
Rs = 30, and since it is connected in parallel, the total impedance at the midpoint of
the line is
€
1 Ztot
= 1
Z0.3λ + 1 Rs
= 1
34.650e− j0.102π + 1 30
, so
€
Ztot =16.107−2.405 j =16.285e − j0.047π .
Finally, we must now find the input impedance at the beginning of the line. The reflection
coefficient here is
€
Γtot = Ztot − Z0 Ztot + Z0
= 16.285e− j0.047π −50 16.285e− j0.047π + 50
= −0.511−0.055 j = 0.514e− j0.966π ,
€
Γin2 = Γtote − j2βl = 0.514e− j0.966πe− j1.2π = 0.446−0.256 j = 0.514e− j0.166π and
€
Zin2 = Z0 1+ Γin2 1−Γin2
= 50 1+ 0.514e− j0.166π
1−0.514e− j0.166π = 98.712−68.578 j =120.20e− j0.193π .
3. [45 points] Electrostatics & Magnetostatics (a) [15] The integral form of Gauss’s law is
€
D ⋅ d s
S ∫ = Q. Due to symmetry properties, we can
deduce that the field must only have a radial component, and must depend only on the radial distance from the charge:
€
D = DR R( ) ˆ R . Choosing the Gaussian integration surface to be a sphere
centered at the position of the charge, the surface integral becomes
€
DRR 2sinθdθdφ
0
π
∫ 0
2π
∫ = 2πDRR2 sinθdθ 0
π
∫ = 2πDRR2 −cosθ( ) 0 π
= 4πR2DR . And this equals the total
charge
€
4πR2DR = −Q0, so
€
DR = −Q0 4πR2
Finally, the electric field amplitude is
€
ER = DR ε0
= −Q0 4πε0R
2 ,
and
€
E = E R ˆ R = − ˆ R
Q0 4πε0R
2 .
(b) [15] Ampère’s law can be stated as
€
H ⋅ d l
C ∫ = Ienc . We assume that the wire is coincident
with the z-axis, with the current flowing in the
€
+ ˆ z direction. By symmetry, the magnetic field must have only a
€
ˆ φ component, and can only depend on the radial coordinate
€
r:
€
H = Hφ r( ) ˆ φ .
We choose the integration contour to be a circle perpendicular to and centered on the wire. Thus
the integral becomes
€
Hφrdφ 0
2π
∫ = 2πrHφ . Since this equals the total current, we find that
€
2πrHφ = Ienc , or
€
Hφ = I0 2πr
, and thus
€
H = ˆ φ
I0 2πr
.
(c) [15]
€
H =
1 µ0 ∇ × A =
1 µ0
ˆ r 1 r ∂Az ∂φ
− ˆ φ ∂Az ∂r
'
( )
*
+ , . Since
€
∂Az ∂φ
= 0, we find that
€
∂Az ∂r
= ∂ ∂r
− µ0J0 4
r2 $
% &
'
( ) = −
µ0J0r 2
. Thus,
€
H =
1 µ0 ∇ × A = − ˆ φ
1 µ0
− µ0J0r
2 &
' (
)
* + = ˆ φ
J0r 2
.
4. [60 points] Plane wave propagation (a) [10] In order to have a circularly polarized wave, it is necessary that
€
ay = ax . Since the wave propagates in the
€
−ˆ z direction, it is RHCP if
€
δ = + π2 . (b) [15] With
€
ay = 2ax , the wave is elliptically polarized unless
€
δ = 0 or
€
δ = π , in which case the wave is linearly polarized. Since a value of
€
δ is not specified, the state of polarization of this wave is indeterminate. (c) [10] To determine the conductive properties of the material, we estimate using
€
" " ε " ε
= σ ωε
= 4
2π( )103 80( ) 8.85x10−12( ) = 8.99x105. Since
€
" " ε " ε
>>1, this is a good conductor.
(d) [10] Using the good conductor approximation, we know that:
€
α ≈ πfµσ = π 103( ) 4πx10−7( )4 = 0.126 Np/m
€
β = α = 0.126 rad/m
€
ηc ≈ 1+ j( ) α σ
= 1+ j( ) 0.126 4
= 0.0445e j π 4
(e) [15] The average power density decays as
€
e+2αz , since the wave propagates in the
€
−ˆ z direction. Thus, we wish to find the depth
€
z = −d at which the power density decays to the given threshold:
€
1⋅e−2αd =10−4 . Taking the natural log of both sides and inserting the above value for
€
α : −2 0.126( )d = −9.21. Thus, we find that d =36.55 m. 5. [50 points] Plane waves & Oblique incidence
(a) [10] The magnetic field phasor for the incident wave is given by
€
˜ H = 1
− jωµ ∇ × ˜ E . First we
evaluate ∇× Ei = ŷ ∂Ax ∂z
− ∂Az ∂x
$
% &
'
( )
€
= E 0 i ˆ y
∂ ∂z
cosθi exp − jk1 x sinθi + z cosθi( )[ ]( ) + ∂ ∂x
sinθi exp − jk1 x sinθi + z cosθi( )[ ]( ) %
& ' (
) *
€
= − jk1E 0 i ˆ y cos2 θi + sin
2 θi( ) exp − jk1 x sinθi + z cosθi( )[ ][ ] = − jk1E 0i ˆ y exp − jk1 x sinθi + z cosθi( )[ ] .
Thus,
€
˜ H i = 1
− jωµ − jk1E 0
i ˆ y exp − jk1 x sinθi + z cosθi( )[ ]( ) = k1E 0
i
ωµ ˆ y exp − jk1 x sinθi + z cosθi( )[ ].
Alternatively,
€
˜ H i = E 0
i
η1 ˆ y exp − jk1 x sinθi + z cosθi( )[ ], where
€
η1 = µ1 ε1
= µ0 ε0
= 377Ω.
(b) [15] Note that this is the parallel polarization, since the electric field vector lies in the plane
of incidence. The reflected magnetic field phasor is
€
˜ H r = − ˆ y E 0
r
η1 exp − jk1 x sinθi − z cosθi( )[ ] ,
where
€
E0 r = Γ||E0
i and
€
Γ|| = η2cosθt −η1cosθi η2cosθt + η1cosθi
. The transmitted angle
€
θt is given by Snell’s law:
€
sinθi = εr sinθt . Thus
€
˜ H r = − ˆ y Γ|| E 0
i
η1 exp − jk1 x sinθi − z cosθi( )[ ]. To find the time-dependent
expression for the magnetic field, we use
€
H r x, y, z, t( ) = Re ˜ H e jωt{ }, so
€
H r = − ˆ y
Γ|| E 0 i
η1 cos ωt − k1 x sinθi − z cosθi( )[ ] .
(c) [15] The phasor of the transmitted electric field is
€
˜ E t = ˆ x cosθt − ˆ z sinθt( ) E 0 t exp − jk2 x sinθt + z cosθt( )[ ] , where
€
E0 t = τ ||E0
i , τ || = 2η2 cosθi
η2 cosθt +η1cosθi ,
and
€
η2 = µ2 ε2
= 1 εr
µ0 ε0
= 377Ω εr
, and
€
k2 = ω µ2ε2 = ω µ0εrε0 = εr ω c
. Thus,
€
˜ E t = ˆ x cosθt − ˆ z sinθt( )τ || E 0 i exp − jk2 x sinθt + z cosθt( )[ ].
(d) [10] The transmissivity for the parallel polarization is defined to be
€
T|| = τ || 2 η1cosθt η2cosθi
.
Plugging in:
€
T|| = 4η1η2cosθi cosθt η2cosθt + η1cosθi( )
2 .
