HW8 solutions - Itai Cohen Group

Phys 3327 Fall ’11 Solutions 8
Solutions: Homework 8
Ex. 8.1: EM Waves Incident on a Conductor
Consider a conductor lying in the z > 0 half space with conductivity σ m and dielectric constant ǫ m , and a
dielectric in the z > 0 half space with dielectric constant ǫ. A wave E 0 (r,t) = E 0 0e i(kz−ωt) e x is normally
incident on the conductor, and ω ≪ σ m . We also assume ǫ m 4π.
(a) First, allthe resultsfornormalincidenceholdinthiscase, butwiththe refractiveindexoftheconductor
now a complex number. The refractive index of the dielectric
and complex refractive index of the conductor
√
to leading order in ǫ m ω/σ m . Note here α ≪ 1.
n 1 = √ ǫ , (1)
n m = ǫ m +i 4πσ m
ω
√
2πσm
≃ (1+i)
ω
≡ (1+i)α −1 (2)
The complex magnitude of the reflected electric field E 1 (r,t) = E 0 1 ei(kz−ωt) e x is then
E 0 1
E 0 0
= n m −n 1
n m +n 1
= 1−√ ǫα(1−i)/2
1+ √ ǫα(1−i)/2
≃ [ 1− √ ǫα(1−i)/2 ] 2
≃ 1− √ ǫα(1−i) , (3)
to leading order in α. Taking real parts then
[ (1−
E 1 (r,t) = E0
0 √ ) √ ]
ǫα cos[kz −ωt]− ǫαsin[kz −ωt] e x . (4)
Since the reflected field propagates in the −z direction, the magnetic field B 1 = −n 1 e z ×E 1 , so
B 1 (r,t) = − √ [ (1−
ǫE0
0 √ ) √ ]
ǫα cos[kz −ωt]− ǫαsin[kz −ωt] e y . (5)
(b) Let the transmitted wave be E 2 (r,t) = E 0 2 ei(kmz−ωt) e x . Then
E 0 2
E 0 0
to leading order. The complex wavenumber
= 2n 1
n m +n 1
√ ǫ(1−i)α
=
1+ √ ǫα(1−i)/2
≃ √ ǫ(1−i)α = (1−i) √ ǫω/2πσ m (6)
k m = n m
ω
c = ω c (1+i)α−1 = (1+i) √ 2πσ m ω/c 2 ≡ (1+i)δ −1 . (7)
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Phys 3327 Fall ’11 Solutions 8
(c) The magnetic field of the transmitted wave B 2 = n m e z ×E 2 . Hence
in terms of only E 0 2 and k m.
B 2 (r,t) = k mc
ω E0 2e i(kmz−ωt) e y , (8)
(d) Applying Ohm’s Law, the current density in the conductor j = σ m E 2 . Hence,
j = σ m E 0 2e i(kmz−ωt) e x . (9)
After taking real parts we have
j(r,t) = σ m E0√ 0 ǫω/2πσm e −z/δ[ ]
cos(δ −1 z −ωt)+sin(δ −1 z −ωt) . (10)
(e) The time-averaged Poynting vector
〈S〉 = c
8π E 2 ×B ∗ 2
= c
8π e−2z/δ |E 0 2 |2 n ∗ m e z
= ǫ
8π 2 c 2
δσ m
(E 0 0 )2 e −2z/δ e z
(11)
after taking the real part only. Since there are free currents flowing in the conductor, then applying
Ohm’s Law j = σ m E, the energy conservation law is
∂
∂t E +∇·S +σ mE 2 = 0 . (12)
Ex. 8.2: Parallel Conducting Planes
Consider two conducting planes, located at x = 0 and x = d, so their common surface normal is e x . The
boundary condition for the two planes is
e x ×E = 0 , and e x ·B = 0 . (13)
Clearly, a TEM mode with k = ke z , E = Ee x and B = Be y satisfies the boundary conditions, so these
plane support TEM modes.
The proof that uniform cross-section hollow conductors do not support TEM modes depends on the conductor’s
surface being closed (i.e. having no boundary) and connected. However, in the case of the two planes,
the conducting surface is clearly not connected, so we do not expect TEM modes to be forbidden.
Ex. 8.3: E & M Fields for a Transmission Line
(a) Consider a coaxial cable with inner radius a and outer radius b, with the inner conductor at a potential
V 0 with respect to the outer one. As found in previous homework, in the cylindrical coordinates (r,θ,z)
the potential for a ≤ r ≤ b is
and hence the electric field
Φ(r) = −V 0
ln(r/a)
ln(b/a) , (14)
E(r) = −∇Φ(r) = V 0 e r
ln(b/a) r . (15)
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Phys 3327 Fall ’11 Solutions 8
(b) Assuming the the current I 0 flows along the inner conductor in the +z direction, then by Ampère’s
Law in integral form, clearly the magnetic field for a ≤ r ≤ b is
B(r) = 2I 0
c
e θ
r . (16)
(c) Let V(z,t) = V 0 e i(kz−ωt) and I(z,t) = I 0 e i(kz−ωt) , and Z 0 = V 0 /I 0 . Consider the dynamical fields
E(r,t) = V(z,t) e r
ln(b/a) r = V 0
ln(b/a)
B(r,t) = 2I(z,t) e θ
c r = 2I 0
c
e i(kz−ωt)
r
e i(kz−ωt)
r
e r (17)
e θ . (18)
which are obtained by näively replacing V 0 → V(z,t) and I 0 → I(z,t). Note these fields have not been
derived from Maxwell’s equations, so the idea is to check that they satisfy the usual wave equations,
and hence are the correct fields generated by V(z,t) and I(z,t).
In the cylindrical coordinates chosen, we may write the electric field E = E r (r,z)e r . Using the hint
provided, the Laplacian
∇ 2 E = ∇ ( ∇·E ) −∇× ( ∇×E )
) ∂
= 0−∇×(
∂z E r(r,z) e θ
(
= ik ∂ [
rEr (r,z) ]) e r
∂z r
noting that ∂/∂r(rE r ) = 0. Hence
∇ 2 E − 1 c 2 ∂ 2
= −k 2 E , (19)
∂t 2E = −k2 E + ω2
c 2 E = 0 , (20)
provided k = ω/c, and so the electric field (17) satisfies the wave equation.
Similarly writing the magnetic field as B = B θ (r,z)e θ , then
noting that ∂/∂r(rB θ ) = 0. Hence
∇ 2 B = ∇ ( ∇·B ) −∇× ( ∇×B )
) ∂
= 0+∇×(
∂z B θ(r,z) e r
(
= ik ∂ )
∂z B θ(r,z)
∇ 2 B − 1 c 2 ∂ 2
e θ
= −k 2 B , (21)
∂t 2B = −k2 B + ω2
c 2 B = 0 , (22)
provided k = ω/c, and so the magnetic field (18) satisfies the wave equation.
(d) The time-averaged Poynting vector
〈S〉 = c
8π E ×B∗
= 1
4π
V 0 I 0
ln(b/a)
e z
r 2 . (23)
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Phys 3327 Fall ’11 Solutions 8
Hence the power transmitted through the interior cross section of the cable by the EM field is
∫
P = S ·dA
= 1 V 0 I 0
4π ln(b/a)
= V 0I 0
2
∫ 2π ∫ b
0
a
dr
r
. (24)
Since the electric and magnetic fields outside the cable are zero, then P is the total power transmitted
down the line by the EM wave. Note that P is the same as the rms power dissipated by the impedance
Z 0 , which is I 2 0 Z 0/2.
Ex. 8.4: Discrete Element Representation of a Transmission Line
(a) Let the nth inductor be L n . The current running through L n is I n , and the impedance Z L = −iωL,
assuming the ladder is driven by an external sinusoidal voltage or current source ∼ e −iωt . Note that
the − sign in the exponential means that Z L is the opposite sign to the convention you may have seen
elsewhere! From the defined direction of I n , the voltage drop over L n is V n −V n+1 , so by Ohm’s Law
(V = IZ) we have
V n+1 −V n = iωLI n . (25)
Similarly, let the nth capacitor be C n , which has impedance Z C = i/(ωC) (different again by a sign
to the usual convention). Clearly the voltage drop over C n is by definition V n+1 , and by Kirchoff’s
junction rule, the current flowing through C n over the voltage drop is I n −I n+1 . Hence by Ohms’ Law
I n+1 −I n = iωCV n+1 . (26)
(b) First, note that to discretize the ladder we define the location of the nth node of the ladder to be z n ,
and let z n+1 = z n +∆z. Letting V n = V(z n ) [I n = I(z n )] be the voltage at [current into] the n node,
then we have the Taylor expansions
V n+1 = V n + ∂V
∂z ∆z +O(∆z2 ) , (27)
I n+1 = I n + ∂I
∂z ∆z +O(∆z2 ) . (28)
Let the capacitance (inductance) per unit length be C l (L l ), so C = C l ∆z and L = L l ∆z. The
recursion relation (25) then becomes
V n+1 −V n
∆z
= iωI n L l = − ∂I
∂t (z n)L l . (29)
since I ∼ e −iωt . But in the limit of a continuous transmission line, ∆z → 0, clearly from (27)
(V n+1 −V n )/∆z → ∂V/∂z. Hence
Similarly, we find from (26) and (28) that
∂V
∂z = −∂I ∂t L l . (30)
∂I
∂z = −∂V ∂t C l . (31)
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Phys 3327 Fall ’11 Solutions 8
(c) The impedance of the ladder may be expressed as a repeating continued fraction, so adding an extra
ladder element to the infinite ladder does not change its limiting impedance, assuming that such limit
exists. Hence, consider the impedance of the circuit shown, in which an extra ladder element is added
to the ladder.
Since Z is the impendance of an infinite ladder, then it is unchanged by the addition of the element,
so it follows that
Z = Z L + Z CZ
Z C +Z , (32)
as Z L is in series with the parallel combination of Z C and Z. Hence Z satisfies
Z 2 −Z L Z −Z L Z C = 0 (33)
which has solution
Z = 1 2
( )
Z L ±
√ZL 2 +4Z LZ C . (34)
Now, Z L = −iωL l ∆z, and Z C = i(ωC l ∆z), so in the limit ∆z → 0 only the Z L Z C term survives.
Hence for the continuum limit ∆z → 0,
Z → √ L l /C l . (35)
(d) If a resistance R = R l ∆z is added in series with the inductor, the the impedance Z L = −iωL+R. The
recursion relation (25) then becomes
and (26) remains the same.
In the continuum limit, the differential equations are then modified to be
which may be rewritten as
V n+1 −V n = (iωL−R)I n , (36)
∂V
∂z = −∂I
(
Ll +iR l /ω ) , ∂I
∂t ∂z = −∂V ∂t C l , (37)
∂ 2 V
∂z 2 −( L l +iR l /ω ) ∂ 2 V
C l
∂t 2 = 0 ,
∂ 2 I
∂z 2 −( L l +iR l /ω ) ∂ 2 I
C l
∂t 2 = 0 (38)
The solution to these are V(z,t) = V 0 e i(kz−ωt) and I(z,t) = I 0 e i(kz−ωt) , with dispersion relation
√ (
k = C l ω2 L l +iωR) . (39)
Note that k is complex, so that I and V are exponentially suppressed - or damped - in the z direction, as
expected.
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