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Engineering Electromagnetics 9th solution
Engineering Electromagnetics (충남대학교)
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CHAPTER 9 – 9th Edition
9.1. In Fig. 9.4, let 𝐵 = 0.2 cos 120𝜋𝑡 T, and assume that the conductor joining the two ends of the resistor
is perfect. It may be assumed that the magnetic field produced by 𝐼(𝑡) is negligible. Find:
a) 𝑉𝑎𝑏 (𝑡): Since 𝐵 is constant over the loop area, the flux is Φ = 𝜋(0.15)2 𝐵 = 1.41 × 10−2 cos 120𝜋𝑡
Wb. Now, 𝑒𝑚𝑓 = 𝑉𝑏𝑎 (𝑡) = −𝑑Φ∕𝑑𝑡 = (120𝜋)(1.41 × 10−2 ) sin 120𝜋𝑡. Then 𝑉𝑎𝑏 (𝑡) = −𝑉𝑏𝑎 (𝑡) =
−5.33 sin 120𝜋𝑡 V.
b) 𝐼(𝑡) = 𝑉𝑏𝑎 (𝑡)∕𝑅 = 5.33 sin(120𝜋𝑡)∕250 = 21.3 sin(120𝜋𝑡) mA
9.2. In the example described by Fig. 9.1, replace the constant magnetic flux density by the time-varying
quantity 𝐁 = 𝐵0 sin 𝜔𝑡 𝐚𝑧 . Assume that 𝐯 is constant and that the displacement 𝑦 of the bar is zero at
𝑡 = 0. Find the emf at any time, 𝑡.
The magnetic flux through the loop area is
Φ𝑚 =
∫𝑠
𝑣𝑡
𝐁 ⋅ 𝑑𝐒 =
∫0 ∫0
𝑑
𝐵0 sin 𝜔𝑡 (𝐚𝑧 ⋅ 𝐚𝑧 ) 𝑑𝑥 𝑑𝑦 = 𝐵0 𝑣 𝑡 𝑑 sin 𝜔𝑡
Then the emf is
𝑒𝑚𝑓 =
∮
𝐄 ⋅ 𝑑𝐋 = −
𝑑Φ𝑚
= −𝐵0 𝑑 𝑣 [sin 𝜔𝑡 + 𝜔𝑡 cos 𝜔𝑡] V
𝑑𝑡
9.3. Given 𝐇 = 300 𝐚𝑧 cos(3 × 108 𝑡 − 𝑦) A/m in free space, find the emf developed in the general 𝐚𝜙
direction about the closed path having corners at
a) (0,0,0), (1,0,0), (1,1,0), and (0,1,0): The magnetic flux will be:
1
1
300𝜇0 cos(3 × 108 𝑡 − 𝑦) 𝑑𝑥 𝑑𝑦 = 300𝜇0 sin(3 × 108 𝑡 − 𝑦)|10
∫0 ∫0
[
]
= 300𝜇0 sin(3 × 108 𝑡 − 1) − sin(3 × 108 𝑡) Wb
Φ=
Then
[
]
𝑑Φ
= −300(3 × 108 )(4𝜋 × 10−7 ) cos(3 × 108 𝑡 − 1) − cos(3 × 108 𝑡)
𝑑𝑡
[
]
= −1.13 × 105 cos(3 × 108 𝑡 − 1) − cos(3 × 108 𝑡) V
emf = −
b) corners at (0,0,0), (2𝜋,0,0), (2𝜋,2𝜋,0), (0,2𝜋,0): In this case, the flux is
Φ = 2𝜋 × 300𝜇0 sin(3 × 108 𝑡 − 𝑦)|2𝜋
=0
0
The emf is therefore 0.
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9.4. A rectangular loop of wire containing a high-resistance voltmeter has corners initially at (𝑎∕2, 𝑏∕2, 0),
(−𝑎∕2, 𝑏∕2, 0), (−𝑎∕2, −𝑏∕2, 0), and (𝑎∕2, −𝑏∕2, 0). The loop begins to rotate about the 𝑥 axis at
constant angular velocity 𝜔, with the first-named corner moving in the 𝐚𝑧 direction at 𝑡 = 0. Assume a
uniform magnetic flux density 𝐁 = 𝐵0 𝐚𝑧 . Determine the induced emf in the rotating loop and specify
the direction of the current.
The magnetic flux though the loop is found (as usual) through
Φ𝑚 =
∫𝑠
𝐁 ⋅ 𝑑𝐒, where 𝐒 = 𝐧 𝑑𝑎
Because the loop is rotating, the direction of the normal, 𝐧, changing, and is in this case given by
𝐧 = cos 𝜔𝑡 𝐚𝑧 − sin 𝜔𝑡 𝐚𝑦
Therefore,
𝑏∕2
Φ𝑚 =
𝑎∕2
(
)
𝐵0 𝐚𝑧 ⋅ cos 𝜔𝑡 𝐚𝑧 − sin 𝜔𝑡 𝐚𝑦 𝑑𝑥 𝑑𝑦 = 𝑎𝑏𝐵0 cos 𝜔𝑡
∫−𝑏∕2 ∫−𝑎∕2
The integral is taken over the entire loop area (regardless of its immediate orientation). The
important result is that the component of 𝐁 that is normal to the loop area is varying sinusoidally,
and so it is fine to think of the 𝐁 field itself rotating about the 𝑥 axis in the opposite direction
while the loop is stationary. Now the emf is
𝑒𝑚𝑓 =
∮
𝐄 ⋅ 𝑑𝐋 = −
𝑑Φ𝑚
= 𝑎𝑏 𝜔𝐵0 sin 𝜔𝑡 V
𝑑𝑡
The direction of the current is the same as the direction of 𝐄 in the emf expression. It is easiest
to picture this by considering the 𝐁 field rotating and the loop fixed. By convention, 𝑑𝐋 will be
counter-clockwise when looking down on the loop from the upper half-space (in the opposite
direction of the normal vector to the plane). The current will be counter-clockwise whenever the
emf is positive, and will be clockwise whenever the emf is negative.
9.5. The location of the sliding bar in Fig. 9.5 is given by 𝑥 = 5𝑡 + 2𝑡3 , and the separation of the two rails
is 20 cm. Let 𝐁 = 0.8𝑥2 𝐚𝑧 T. Find the voltmeter reading at:
a) 𝑡 = 0.4 s: The flux through the loop will be
Φ=
∫0
0.2
∫0
𝑥
0.8(𝑥′ )2 𝑑𝑥′ 𝑑𝑦 =
0.16 3 0.16
𝑥 =
(5𝑡 + 2𝑡3 )3 Wb
3
3
Then
emf = −
𝑑Φ 0.16
=
(3)(5𝑡 + 2𝑡3 )2 (5 + 6𝑡2 ) = −(0.16)[5(.4) + 2(.4)3 ]2 [5 + 6(.4)2 ] = −4.32 V
𝑑𝑡
3
b) 𝑥 = 0.6 m: Have 0.6 = 5𝑡 + 2𝑡3 , from which we find 𝑡 = 0.1193. Thus
emf = −(0.16)[5(.1193) + 2(.1193)3 ]2 [5 + 6(.1193)2 ] = −.293 V
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9.6. Let the wire loop of Problem 9.4 be stationary in its 𝑡 = 0 position and find the induced emf that results
from a magnetic flux density given by 𝐁(𝑦, 𝑡) = 𝐵0 cos(𝜔𝑡 − 𝛽𝑦) 𝐚𝑧 , where 𝜔 and 𝛽 are constants.
We begin by finding the net magnetic flux through the loop:
Φ𝑚 =
=
∫𝑠
𝑏∕2
𝐁 ⋅ 𝑑𝐒 =
𝑎∕2
∫−𝑏∕2 ∫−𝑎∕2
𝐵0 cos(𝜔𝑡 − 𝛽𝑦) 𝐚𝑧 ⋅ 𝐚𝑧 𝑑𝑥 𝑑𝑦
]
𝐵0 𝑎 [
sin(𝜔𝑡 + 𝛽𝑏∕2) − sin(𝜔𝑡 − 𝛽𝑏∕2)
𝛽
Now the emf is
𝑒𝑚𝑓 =
∮
𝐄 ⋅ 𝑑𝐋 = −
]
𝑑Φ𝑚
𝐵 𝑎𝜔 [
=− 0
cos(𝜔𝑡 + 𝛽𝑏∕2) − cos(𝜔𝑡 − 𝛽𝑏∕2)
𝑑𝑡
𝛽
Using the trig identity, cos(𝑎 ± 𝑏) = cos 𝑎 cos 𝑏 ∓ sin 𝑎 sin 𝑏, we may write the above result as
𝜔
𝑒𝑚𝑓 = +2𝐵0 𝑎 sin(𝜔𝑡) sin(𝛽𝑏∕2) V
𝛽
9.7. The rails in Fig. 9.7 each have a resistance of 2.2 Ω∕m. The bar moves to the right at a constant speed
of 9 m/s in a uniform magnetic field of 0.8 T. Find 𝐼(𝑡), 0 < 𝑡 < 1 s, if the bar is at 𝑥 = 2 m at 𝑡 = 0
and
a) a 0.3 Ω resistor is present across the left end with the right end open-circuited: The flux in the
left-hand closed loop is
Φ𝑙 = 𝐵 × area = (0.8)(0.2)(2 + 9𝑡)
Then, emf 𝑙 = −𝑑Φ𝑙 ∕𝑑𝑡 = −(0.16)(9) = −1.44 V. With the bar in motion, the loop resistance is
increasing with time, and is given by 𝑅𝑙 (𝑡) = 0.3 + 2[2.2(2 + 9𝑡)]. The current is now
𝐼𝑙 (𝑡) =
emf 𝑙
−1.44
=
A
𝑅𝑙 (𝑡) 9.1 + 39.6𝑡
Note that the sign of the current indicates that it is flowing in the direction opposite that shown
in the figure.
b) Repeat part 𝑎, but with a resistor of 0.3 Ω across each end: In this case, there will be a contribution
to the current from the right loop, which is now closed. The flux in the right loop, whose area
decreases with time, is
Φ𝑟 = (0.8)(0.2)[(16 − 2) − 9𝑡]
and emf 𝑟 = −𝑑Φ𝑟 ∕𝑑𝑡 = (0.16)(9) = 1.44 V. The resistance of the right loop is 𝑅𝑟 (𝑡) =
0.3 + 2[2.2(14 − 9𝑡)], and so the contribution to the current from the right loop will be
𝐼𝑟 (𝑡) =
−1.44
A
61.9 − 39.6𝑡
The minus sign has been inserted because again the current must flow in the opposite direction
as that indicated in the figure, with the flux decreasing with time. The total current is found by
adding the part 𝑎 result, or
[
]
1
1
𝐼𝑇 (𝑡) = −1.44
A
+
61.9 − 39.6𝑡 9.1 + 39.6𝑡
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9.8. A perfectly-conducting filament is formed into a circular ring of radius 𝑎. At one point a resistance
𝑅 is inserted into the circuit, and at another a battery of voltage 𝑉0 is inserted. Assume that the loop
current itself produces negligible magnetic field.
a) Apply Faraday’s law, Eq. (4), evaluating each side of the equation carefully and independently to
show the equality: With no 𝐁 field present, and no time variation, the right-hand side of Faraday’s
law is zero, and so therefore
∮
𝐄 ⋅ 𝑑𝐋 = 0
This is just a statement of Kirchoff’s voltage law around the loop, stating that the battery voltage
is equal and opposite to the resistor voltage.
b) Repeat part 𝑎, assuming the battery removed, the ring closed again, and a linearly-increasing 𝐁
field applied in a direction normal to the loop surface: The situation now becomes the same as
that shown in Fig. 9.4, except the loop radius is now 𝑎, and the resistor value is not specified.
Consider the loop as in the 𝑥-𝑦 plane with the positive 𝑧 axis directed out of the page. The 𝐚𝜙
direction is thus counter-clockwise around the loop. The 𝐁 field (out of the page as shown) can
be written as 𝐁(𝑡) = 𝐵0 𝑡 𝐚𝑧 . With the normal to the loop specified as 𝐚𝑧 , the direction of 𝑑𝐋 is, by
the right hand convention, 𝐚𝜙 . Since the wire is perfectly-conducting, the only voltage appears
across the resistor, and is given as 𝑉𝑅 . Faraday’s law becomes
∮
𝐄 ⋅ 𝑑𝐋 = 𝑉𝑅 = −
𝑑Φ𝑚
𝑑
=−
𝐵 𝑡 𝐚 ⋅ 𝐚 𝑑𝑎 = −𝜋𝑎2 𝐵0
𝑑𝑡
𝑑𝑡 ∫𝑠 0 𝑧 𝑧
This indicates that the resistor voltage, 𝑉𝑅 = 𝜋𝑎2 𝐵0 , has polarity such that the positive terminal
is at point 𝑎 in the figure, while the negative terminal is at point 𝑏. Current flows in the clockwise
direction, and is given in magnitude by 𝐼 = 𝜋𝑎2 𝐵0 ∕𝑅.
9.9. A square filamentary loop of wire is 25 cm on a side and has a resistance of 125 Ω per meter length.
The loop lies in the 𝑧 = 0 plane with its corners at (0, 0, 0), (0.25, 0, 0), (0.25, 0.25, 0), and (0, 0.25, 0)
at 𝑡 = 0. The loop is moving with velocity 𝑣𝑦 = 50 m/s in the field 𝐵𝑧 = 8 cos(1.5 × 108 𝑡 − 0.5𝑥) 𝜇T.
Develop a function of time which expresses the ohmic power being delivered to the loop: First, since
the field does not vary with 𝑦, the loop motion in the 𝑦 direction does not produce any time-varying
flux, and so this motion is immaterial. We can evaluate the flux at the original loop position to obtain:
∫0
.25
∫0
.25
8 × 10−6 cos(1.5 × 108 𝑡 − 0.5𝑥) 𝑑𝑥 𝑑𝑦
[
]
= −(4 × 10−6 ) sin(1.5 × 108 𝑡 − 0.13) − sin(1.5 × 108 𝑡) Wb
Φ(𝑡) =
[
]
Now, 𝑒𝑚𝑓 = 𝑉 (𝑡) = −𝑑Φ∕𝑑𝑡 = 6.0 × 102 cos(1.5 × 108 𝑡 − 0.13) − cos(1.5 × 108 𝑡) , The total loop
resistance is 𝑅 = 125(0.25 + 0.25 + 0.25 + 0.25) = 125 Ω. Then the ohmic power is
𝑃 (𝑡) =
[
]2
𝑉 2 (𝑡)
= 2.9 × 103 cos(1.5 × 108 𝑡 − 0.13) − cos(1.5 × 108 𝑡) Watts
𝑅
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9.10
a) Show that the ratio of the amplitudes of the conduction current density and the displacement
current density is 𝜎∕𝜔𝜖 for the applied field 𝐸 = 𝐸𝑚 cos 𝜔𝑡. Assume 𝜇 = 𝜇0 . First, 𝐷 =
𝜖𝐸 = 𝜖𝐸𝑚 cos 𝜔𝑡. Then the displacement current density is 𝜕𝐷∕𝜕𝑡 = −𝜔𝜖𝐸𝑚 sin 𝜔𝑡. Second,
𝐽𝑐 = 𝜎𝐸 = 𝜎𝐸𝑚 cos 𝜔𝑡. Using these results we find |𝐽𝑐 |∕|𝐽𝑑 | = 𝜎∕𝜔𝜖.
b) What is the amplitude ratio if the applied field is 𝐸 = 𝐸𝑚 𝑒−𝑡∕𝜏 , where 𝜏 is real? As before, find
𝐷 = 𝜖𝐸 = 𝜖𝐸𝑚 𝑒−𝑡∕𝜏 , and so 𝐽𝑑 = 𝜕𝐷∕𝜕𝑡 = −(𝜖∕𝜏)𝐸𝑚 𝑒−𝑡∕𝜏 . Also, 𝐽𝑐 = 𝜎𝐸𝑚 𝑒−𝑡∕𝜏 . Finally,
|𝐽𝑐 |∕|𝐽𝑑 | = 𝜎𝜏∕𝜖.
9.11. Let the internal dimension of a coaxial capacitor be 𝑎 = 1.2 cm, 𝑏 = 4 cm, and 𝑙 = 40 cm. The
homogeneous material inside the capacitor has the parameters 𝜖 = 10−11 F/m, 𝜇 = 10−5 H/m, and
𝜎 = 10−5 S/m. If the electric field intensity is 𝐄 = (106 ∕𝜌) cos(105 𝑡)𝐚𝜌 V/m, find:
a) 𝐉: Use
𝐉 = 𝜎𝐄 =
(
10
𝜌
)
cos(105 𝑡)𝐚𝜌 A∕m2
b) the total conduction current, 𝐼𝑐 , through the capacitor: Have
𝐼𝑐 =
∫ ∫
𝐉 ⋅ 𝑑𝐒 = 2𝜋𝜌𝑙𝐽 = 20𝜋𝑙 cos(105 𝑡) = 8𝜋 cos(105 𝑡) A
c) the total displacement current, 𝐼𝑑 , through the capacitor: First find
𝐉𝑑 =
(105 )(10−11 )(106 )
𝜕
1
𝜕𝐃
= (𝜖𝐄) = −
sin(105 𝑡)𝐚𝜌 = − sin(105 𝑡) A∕m
𝜕𝑡
𝜕𝑡
𝜌
𝜌
Now
𝐼𝑑 = 2𝜋𝜌𝑙𝐽𝑑 = −2𝜋𝑙 sin(105 𝑡) = −0.8𝜋 sin(105 𝑡) A
d) the ratio of the amplitude of 𝐼𝑑 to that of 𝐼𝑐 , the quality factor of the capacitor: This will be
|𝐼𝑑 | 0.8
=
= 0.1
|𝐼𝑐 |
8
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9.12. The magnetic flux density 𝐁 = 𝐵0 cos(𝜔𝑡) cos(𝑘0 𝑧) 𝐚𝑦 Wb∕m2 exists in free space. 𝐵0 and 𝑘0 are
constants, and 𝐉 = 0. Find:
a) the displacement current density: In free space, and with zero conduction current, Ampere’s Law
in point form becomes
𝜕𝐄
𝜕𝐃
∇×𝐇=
= 𝜖0
𝜕𝑡
𝜕𝑡
The given field field is 𝑦-directed and varies spatially only with 𝑧. So the above becomes
∇×𝐇=∇×
1 𝜕𝐵𝑦
𝜕𝐃 𝑘0 𝐵0
𝐁
=−
cos(𝜔𝑡) sin(𝑘0 𝑧) 𝐚𝑥 A∕m2
𝐚𝑥 =
=
𝜇0
𝜇0 𝜕𝑧
𝜕𝑡
𝜇0
b) the electric field intensity: We write
𝐄=
𝑘 𝐵
𝑘 𝐵
1
𝜕𝐃
cos(𝜔𝑡)𝑑𝑡 = 0 0 sin(𝑘0 𝑧) sin(𝜔𝑡) 𝐚𝑥
𝑑𝑡 + 𝐶 = 0 0 sin(𝑘0 𝑧) 𝐚𝑥
∫
𝜖0 ∫ 𝜕𝑡
𝜇0 𝜖0
𝜔𝜇0 𝜖0
where the integration constant, 𝐶, is set to zero because no steady electric field component is
expected with a purely time-varying 𝐇.
c) 𝑘0 : We find this by requiring consistency in applying both of the Maxwell curl equations. The
Faraday Law in point form reads
∇×𝐄=−
𝜕𝐁
⇒ 𝐁 = − ∇ × 𝐄 𝑑𝑡
∫
𝜕𝑡
where again the integration constant is set to zero. Using the result for 𝐄 in part 𝑏, and noting
that it is 𝑥 directed and spatially varies only with 𝑧, the curl simplifies to a single term, and the
above equation becomes:
𝐁=−
=
∫
∇ × 𝐄 𝑑𝑡 = −
𝑘20 𝐵0
𝜔2 𝜇0 𝜖0
−𝑘20 𝐵0
𝜕𝐸𝑥
cos(𝑘0 𝑧) sin(𝜔𝑡)𝐚𝑦 𝑑𝑡
𝐚𝑦 𝑑𝑡 =
∫ 𝜔𝜇0 𝜖0
∫ 𝜕𝑧
cos(𝑘0 𝑧) cos(𝜔𝑡) 𝐚𝑦
Requiring this result to be the same as the given field leads to the identification:
√
𝑘0 = 𝜔 𝜇0 𝜖0 m−1 .
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9.13. In free space it is known that 𝐄 = 𝐸0 ∕𝑟 sin 𝜃 cos(𝜔𝑡 − 𝑘0 𝑟) 𝐚𝜃 = 𝐸𝜃 𝐚𝜃 . Show that a component of
𝐇 in the 𝐚𝜙 direction arises, where 𝐻𝜙 = 𝐸𝜃 (𝜖0 ∕𝜇0 )1∕2 . Do this by applying Eqs. (20) and (21) and
requiring consistency:
Start with Eq. (20) in free space:
∇ × 𝐄 = −𝜇0
𝜕𝐇
𝜕𝑡
where
∇ × 𝐸𝜃 (𝑟, 𝜃) 𝐚𝜃 =
𝜕𝐻𝜙
1
1 𝜕(𝑟𝐸𝜃 )
𝐚𝜙 = − 𝐸0 𝑘0 sin 𝜃 sin(𝜔𝑡 − 𝑘0 𝑟)𝐚𝜙 = −𝜇0
𝐚
𝑟 𝜕𝑟
𝑟
𝜕𝑡 𝜙
After integrating the above result over time, we find
𝐇 = 𝐻𝜙 𝐚𝜙 =
𝐸0 𝑘0
sin 𝜃 cos(𝜔𝑡 − 𝑘0 𝑟)𝐚𝜙
𝜔𝜇0 𝑟
Now, use Eq. (21) in free space:
∇ × 𝐇 = 𝜖0
𝜕𝐄
𝜕𝑡
where
1 𝜕(𝐻𝜙 sin 𝜃)
1 𝜕(𝑟𝐻𝜙 )
𝐚𝑟 −
𝐚𝜃
𝑟 sin 𝜃
𝜕𝜃
𝑟 𝜕𝑟
𝐸0 𝑘2
2𝐸 𝑘
= 2 0 0 2 cos 𝜃 sin(𝜔𝑡 − 𝑘0 𝑟)𝐚𝑟 + 2 0 2 sin 𝜃 cos(𝜔𝑡 − 𝑘0 𝑟)𝐚𝜃
𝜔 𝜇 0 𝜖0 𝑟
𝜔 𝜇 0 𝜖0 𝑟
√
The second term in this field will be equal to the given field when 𝑘0 = 𝜔 𝜇0 𝜖0 . With this
substitution, the 𝜙 component of 𝐇 becomes:
√
𝜖0 𝐸0
sin 𝜃 cos(𝜔𝑡 − 𝑘0 𝑟) Q.E.D
𝐻𝜙 =
𝜇0 𝑟
∇ × 𝐻𝜙 (𝑟, 𝜃)𝐚𝜙 =
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9.14. A voltage source, 𝑉0 sin 𝜔𝑡, is connected between two concentric conducting spheres, 𝑟 = 𝑎 and 𝑟 = 𝑏,
𝑏 > 𝑎, where the region between them is a material for which 𝜖 = 𝜖𝑟 𝜖0 , 𝜇 = 𝜇0 , and 𝜎 = 0. Find the
total displacement current through the dielectric and compare it with the source current as determined
from the capacitance (Sec. 6.3) and circuit analysis methods: First, solving Laplace’s equation, we
find the voltage between spheres (see Eq. 39, Chapter 6):
𝑉 (𝑡) =
Then
𝐄 = −∇𝑉 =
(1∕𝑟) − (1∕𝑏)
𝑉 sin 𝜔𝑡
(1∕𝑎) − (1∕𝑏) 0
𝑉0 sin 𝜔𝑡
𝐚𝑟
2
𝑟 (1∕𝑎 − 1∕𝑏)
Now
𝐉𝐝 =
⇒ 𝐃=
𝜖𝑟 𝜖0 𝑉0 sin 𝜔𝑡
𝑟2 (1∕𝑎 − 1∕𝑏)
𝐚𝑟
𝜕𝐃 𝜖𝑟 𝜖0 𝜔𝑉0 cos 𝜔𝑡
𝐚𝑟
= 2
𝜕𝑡
𝑟 (1∕𝑎 − 1∕𝑏)
The displacement current is then
𝐼𝑑 = 4𝜋𝑟2 𝐽𝑑 =
4𝜋𝜖𝑟 𝜖0 𝜔𝑉0 cos 𝜔𝑡
𝑑𝑉
=𝐶
(1∕𝑎 − 1∕𝑏)
𝑑𝑡
where, from Eq. 6, Chapter 6,
𝐶=
4𝜋𝜖𝑟 𝜖0
(1∕𝑎 − 1∕𝑏)
9.15. Use each of Maxwell’s equations in point form to obtain as much information as possible about
a) 𝐇 if 𝐄 = 0: In this case we have
∇ × 𝐇 = 𝐉 (the source of 𝐇 is conduction current only)
𝜕𝐁
= 0 ⇒ 𝐁 is constant
𝜕𝑡
𝜌𝑣 = 0 (no charge exists)
∇ ⋅ 𝐁 = 0 (always true anyway)
As 𝐉 is non-zero, but 𝐄 = 0, the current-carrying medium would be a perfect conductor.
b) 𝐄 if 𝐇 = 0: Here we have
∇ × 𝐄 = 0 (𝐄 is constant in time and conservative)
𝜕𝐃
𝜕𝐃
=0 ⇒ 𝐉=
=0
𝜕𝑡
𝜕𝑡
∇ ⋅ 𝐃 = 𝜌𝑣 (𝐄 arises f rom f ree charge only)
∇×𝐇=𝐉+
∇ ⋅ 𝐁 = 0 (irrelevant as 𝐁 = 0 anyway)
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9.16. Derive the continuity equation from Maxwell’s equations: First, take the divergence of both sides of
Ampere’s circuital law:
𝜕𝜌
𝜕
∇⋅∇×𝐇=∇⋅𝐉+ ∇⋅𝐃=∇⋅𝐉+ 𝑣 =0
⏟⏞⏞⏟⏞⏞⏟
𝜕𝑡
𝜕𝑡
0
where we have used ∇ ⋅ 𝐃 = 𝜌𝑣 , another Maxwell equation.
9.17. The electric field intensity in the region 0 < 𝑥 < 5, 0 < 𝑦 < 𝜋∕12, 0 < 𝑧 < 0.06 m in free space is
given by 𝐄 = 𝐶 sin(12𝑦) sin(𝑎𝑧) cos(2 × 1010 𝑡) 𝐚𝑥 V/m. Beginning with the ∇ × 𝐄 relationship, use
Maxwell’s equations to find a numerical value for 𝑎, if it is known that 𝑎 is greater than zero: In this
case we find
𝜕𝐸𝑧
𝜕𝐸𝑥
𝐚𝑦 −
𝐚
𝜕𝑧
𝜕𝑦 𝑧
[
]
𝜕𝐁
= 𝐶 𝑎 sin(12𝑦) cos(𝑎𝑧)𝐚𝑦 − 12 cos(12𝑦) sin(𝑎𝑧)𝐚𝑧 cos(2 × 1010 𝑡) = −
𝜕𝑡
∇×𝐄=
Then
1
∇ × 𝐄 𝑑𝑡 + 𝐶1
𝜇0 ∫
[
]
𝐶
=−
𝑎 sin(12𝑦) cos(𝑎𝑧)𝐚𝑦 − 12 cos(12𝑦) sin(𝑎𝑧)𝐚𝑧 sin(2 × 1010 𝑡) A∕m
10
𝜇0 (2 × 10
𝐇=−
where the integration constant, 𝐶1 = 0, since there are no initial conditions. Using this result, we now
find
]
[
𝜕𝐻𝑧 𝜕𝐻𝑦
𝐶(144 + 𝑎2 )
𝜕𝐃
−
𝐚𝑥 = −
sin(12𝑦) sin(𝑎𝑧) sin(2 × 1010 𝑡) 𝐚𝑥 =
∇×𝐇=
𝜕𝑦
𝜕𝑧
𝜕𝑡
𝜇0 (2 × 1010 )
Now
𝐄=
𝐶(144 + 𝑎2 )
1
𝐃
=
∇ × 𝐇 𝑑𝑡 + 𝐶2 =
sin(12𝑦) sin(𝑎𝑧) cos(2 × 1010 𝑡) 𝐚𝑥
𝜖0 ∫ 𝜖0
𝜇0 𝜖0 (2 × 1010 )2
where 𝐶2 = 0. This field must be the same as the original field as stated, and so we require that
𝐶(144 + 𝑎2 )
=1
𝜇0 𝜖0 (2 × 1010 )2
Using 𝜇0 𝜖0 = (3 × 108 )−2 , we find
[
(2 × 1010 )2
𝑎=
− 144
(3 × 108 )2
]1∕2
= 66 m−1
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9.18. The parallel plate transmission line shown in Fig. 9.7 has dimensions 𝑏 = 4 cm and 𝑑 = 8 mm, while
the medium between plates is characterized by 𝜇𝑟 = 1, 𝜖𝑟 = 20, and 𝜎 = 0. Neglect fields outside the
dielectric. Given the field 𝐇 = 5 cos(109 𝑡 − 𝛽𝑧)𝐚𝑦 A/m, use Maxwell’s equations to help find:
a) 𝛽, if 𝛽 > 0: Take
∇×𝐇=−
So
𝐄=
𝜕𝐻𝑦
𝜕𝑧
𝐚𝑥 = −5𝛽 sin(109 𝑡 − 𝛽𝑧)𝐚𝑥 = 20𝜖0
𝜕𝐄
𝜕𝑡
−5𝛽
𝛽
cos(109 𝑡 − 𝛽𝑧)𝐚𝑥
sin(109 𝑡 − 𝛽𝑧)𝐚𝑥 𝑑𝑡 =
∫ 20𝜖0
(4 × 109 )𝜖0
Then
∇×𝐄=
𝜕𝐸𝑥
𝛽2
𝜕𝐇
𝐚𝑦 =
sin(109 𝑡 − 𝛽𝑧)𝐚𝑦 = −𝜇0
𝜕𝑧
𝜕𝑡
(4 × 109 )𝜖0
So that
𝐇=
𝛽2
−𝛽 2
9
sin(10
𝑡
−
𝛽𝑧)𝐚
𝑑𝑡
=
cos(109 𝑡 − 𝛽𝑧)
𝑥
18
∫ (4 × 109 )𝜇0 𝜖0
(4 × 10 )𝜇0 𝜖0
= 5 cos(109 𝑡 − 𝛽𝑧)𝐚𝑦
where the last equality is required to maintain consistency. Therefore
𝛽2
= 5 ⇒ 𝛽 = 14.9 m−1
(4 × 1018 )𝜇0 𝜖0
b) the displacement current density at 𝑧 = 0: Since 𝜎 = 0, we have
∇ × 𝐇 = 𝐉𝑑 = −5𝛽 sin(109 𝑡 − 𝛽𝑧) = −74.5 sin(109 𝑡 − 14.9𝑧)𝐚𝑥
= −74.5 sin(109 𝑡)𝐚𝑥 A∕m at z = 0
c) the total displacement current crossing the surface 𝑥 = 0.5𝑑, 0 < 𝑦 < 𝑏, and 0 < 𝑧 < 0.1 m in
the 𝐚𝑥 direction. We evaluate the flux integral of 𝐉𝑑 over the given cross section:
𝐼𝑑 = −74.5𝑏
∫0
0.1
[
]
sin(109 𝑡 − 14.9𝑧) 𝐚𝑥 ⋅ 𝐚𝑥 𝑑𝑧 = 0.20 cos(109 𝑡 − 1.49) − cos(109 𝑡) A
9.19. In the first section of this chapter, Faraday’s law was used to show that the field 𝐄 = − 12 𝑘𝐵0 𝜌𝑒𝑘𝑡 𝐚𝜙
results from the changing magnetic field 𝐁 = 𝐵0 𝑒𝑘𝑡 𝐚𝑧 .
a) Show that these fields do not satisfy Maxwell’s other curl equation: Note that 𝐁 as stated is
constant with position, and so will have zero curl. The electric field, however, varies with time,
and so ∇×𝐇 = 𝜕𝐃
would have a zero left-hand side and a non-zero right-hand side. The equation
𝜕𝑡
is thus not valid with these fields.
b) If we let 𝐵0 = 1 T and 𝑘 = 106 𝑠−1 , we are establishing a fairly large magnetic flux density in 1
𝜇s. Use the ∇ × 𝐇 equation to show that the rate at which 𝐵𝑧 should (but does not) change with
𝜌 is only about 5 × 10−6 T/m in free space at 𝑡 = 0: Assuming that 𝐁 varies with 𝜌, we write
𝜕𝐻
1 𝑑𝐵0 𝑘𝑡
𝜕𝐄
1
𝑒 = 𝜖0
= − 𝜖0 𝑘2 𝐵0 𝜌𝑒𝑘𝑡
∇ × 𝐇 = − 𝑧 𝐚𝜙 = −
𝜕𝜌
𝜇0 𝑑𝜌
𝜕𝑡
2
Thus
𝑑𝐵0
1012 (1)𝜌
1
= 𝜇0 𝜖0 𝑘2 𝜌𝐵0 =
= 5.6 × 10−6 𝜌
𝑑𝜌
2
2(3 × 108 )2
which is near the stated value if 𝜌 is on the order of 1m.
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9.20. Given Maxwell’s equations in point form, assume that all fields vary as 𝑒𝑠𝑡 and write the equations
without explicitly involving time: Write all fields in the general form 𝐀(𝐫, 𝑡) = 𝐀0 (𝐫)𝑒𝑠𝑡 , where 𝐫 is a
position vector in any coordinate system. Maxwell’s equations become:
∇ × 𝐄0 (𝐫) 𝑒𝑠𝑡 = −
∇ × 𝐇0 (𝐫) 𝑒𝑠𝑡 = 𝐉0 (𝐫)𝑒𝑠𝑡 +
)
𝜕 (
𝐁0 (𝐫) 𝑒𝑠𝑡 = −𝑠𝐁0 (𝐫) 𝑒𝑠𝑡
𝜕𝑡
)
𝜕 (
𝐃0 (𝐫) 𝑒𝑠𝑡 = 𝐉0 (𝐫)𝑒𝑠𝑡 + 𝑠𝐃0 (𝐫) 𝑒𝑠𝑡
𝜕𝑡
∇ ⋅ 𝐃0 (𝐫) 𝑒𝑠𝑡 = 𝜌0 (𝐫) 𝑒𝑠𝑡
∇ ⋅ 𝐁0 (𝐫) 𝑒𝑠𝑡 = 0
In all cases, the 𝑒𝑠𝑡 terms divide out, leaving:
∇ × 𝐄0 (𝐫) = −𝑠𝐁0 (𝐫)
∇ × 𝐇0 (𝐫) = 𝐉0 (𝐫) + 𝑠𝐃0 (𝐫)
∇ ⋅ 𝐃0 (𝐫) = 𝜌0 (𝐫)
∇ ⋅ 𝐁0 (𝐫) = 0
9.21.
a) Show that under static field conditions, Eq. (55) reduces to Ampere’s circuital law. First use the
definition of the vector Laplacian:
∇2 𝐀 = −∇ × ∇ × 𝐀 + ∇(∇ ⋅ 𝐀) = −𝜇𝐉
which is Eq. (55) with the time derivative set to zero. We also note that ∇ ⋅ 𝐀 = 0 in steady state
(from Eq. (54)). Now, since 𝐁 = ∇ × 𝐀, (55) becomes
−∇ × 𝐁 = −𝜇𝐉 ⇒ ∇ × 𝐇 = 𝐉
b) Show that Eq. (51) becomes Faraday’s law when taking the curl: Doing this gives
∇ × 𝐄 = −∇ × ∇𝑉 −
𝜕
∇×𝐀
𝜕𝑡
The curl of the gradient is identially zero, and ∇ × 𝐀 = 𝐁. We are left with
∇ × 𝐄 = −𝜕𝐁∕𝜕𝑡
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9.22. In a sourceless medium, in which 𝐉 = 0 and 𝜌𝑣 = 0, assume a rectangular coordinate system in which
𝐄 and 𝐇 are functions only of 𝑧 and 𝑡. The medium has permittivity 𝜖 and permeability 𝜇.
a) If 𝐄 = 𝐸𝑥 𝐚𝑥 and 𝐇 = 𝐻𝑦 𝐚𝑦 , begin with Maxwell’s equations and determine the second order
partial differential equation that 𝐸𝑥 must satisfy.
First use
𝜕𝐻𝑦
𝜕𝐸𝑥
𝜕𝐁
⇒
𝐚𝑦 = −𝜇
𝐚
𝜕𝑡
𝜕𝑧
𝜕𝑡 𝑦
in which case, the curl has dictated the direction that 𝐇 must lie in. Similarly, use the other
Maxwell curl equation to find
∇×𝐄=−
∇×𝐇=
𝜕𝐻𝑦
𝜕𝐸
𝜕𝐃
⇒ −
𝐚𝑥 = 𝜖 𝑥 𝐚𝑥
𝜕𝑡
𝜕𝑧
𝜕𝑡
Now, differentiate the first equation with respect to 𝑧, and the second equation with respect to 𝑡:
𝜕 2 𝐸𝑥
𝜕𝑧2
= −𝜇
𝜕 2 𝐻𝑦
𝜕𝑡𝜕𝑧
and
𝜕 2 𝐻𝑦
𝜕𝑧𝜕𝑡
= −𝜖
𝜕 2 𝐸𝑥
𝜕𝑡2
Combining these two, we find
𝜕 2 𝐸𝑥
𝜕𝑧2
= 𝜇𝜖
𝜕 2 𝐸𝑥
𝜕𝑡2
b) Show that 𝐸𝑥 = 𝐸0 cos(𝜔𝑡 − 𝛽𝑧) is a solution of that equation for a particular value of 𝛽: Substituting, we find
𝜕 2 𝐸𝑥
𝜕𝑧2
= −𝛽 2 𝐸0 cos(𝜔𝑡 − 𝛽𝑧) and 𝜇𝜖
𝜕 2 𝐸𝑥
𝜕𝑡2
= −𝜔2 𝜇𝜖𝐸0 cos(𝜔𝑡 − 𝛽𝑧)
These two will be equal provided the constant multipliers of cos(𝜔𝑡 − 𝛽𝑧) are equal.
√
c) Find 𝛽 as a function of given parameters. Equating the two constants in part 𝑏, we find 𝛽 = 𝜔 𝜇𝜖.
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9.23. Referring to the parallel plate transmission line of Fig. 9.7, the electric field between plates is given
as 𝐄(𝑧, 𝑡) = −𝐸0 cos(𝜔𝑡 − 𝛽𝑧)𝐚𝑦 . Find the vector potential 𝐀(𝑦, 𝑧, 𝑡) if it is given that 𝐀(0, 𝑧, 𝑡) = 0.
Assume 𝑏 >> 𝑑, so that 𝑥 variation is negligible:
Start with
∇×𝐄=−
from which
𝐇=
𝜕𝐸𝑦
𝜕𝑧
𝐚𝑥 = +𝛽𝐸0 sin(𝜔𝑡 − 𝛽𝑧)𝐚𝑥 = −𝜇0
𝜕𝐇
𝜕𝑡
𝛽𝐸
−𝛽𝐸0
sin(𝜔𝑡 − 𝛽𝑧)𝐚𝑥 𝑑𝑡 + 𝐶 = + 0 cos(𝜔𝑡 − 𝛽𝑧)𝐚𝑥
𝜇0 ∫
𝜔𝜇0
where the integration constant is zero. Next we have
𝛽𝐸0
∇ × 𝐀 = 𝐁 = 𝜇0 𝐇 =
cos(𝜔𝑡 − 𝛽𝑧)𝐚𝑥 =
𝜔
(
𝜕𝐴𝑧 𝜕𝐴𝑦
−
𝜕𝑦
𝜕𝑧
)
𝐚𝑥
where the last term is the 𝑥 component of the curl of 𝐀, within which we expect 𝐴𝑦 = 0 because
the current will be entirely 𝑧 directed. The equation to solve is thus
𝜕𝐴𝑧
𝛽𝐸0
𝛽𝐸0
=
cos(𝜔𝑡 − 𝛽𝑧) ⇒ 𝐀 =
𝑦 cos(𝜔𝑡 − 𝛽𝑧)𝐚𝑧
𝜕𝑦
𝜔
𝜔
which is seen to satisfy the boundary condition at 𝑦 = 0.
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9.24. A vector potential is given as 𝐀 = 𝐴0 cos(𝜔𝑡 − 𝑘𝑧) 𝐚𝑦 . a) Assuming as many components as possible
are zero, find 𝐇, 𝐄, and 𝑉 ;
With 𝐀 𝑦-directed only, and varying spatially only with 𝑧, we find
𝐇=
𝑘𝐴
1 𝜕𝐴𝑦
1
∇×𝐀=−
𝐚𝑥 = − 0 sin(𝜔𝑡 − 𝑘𝑧) 𝐚𝑥 A∕m
𝜇
𝜇 𝜕𝑧
𝜇
Now, in a lossless medium we will have zero conductivity, so that the point form of Ampere’s
circuital law involves only the displacement current term:
∇×𝐇=
𝜕𝐃
𝜕𝐄
=𝜖
𝜕𝑡
𝜕𝑡
Using the magnetic field as found above, we find
∇×𝐇=
Now,
𝜕𝐻𝑥
𝑘 2 𝐴0
𝑘 2 𝐴0
𝜕𝐄
𝐚𝑦 =
cos(𝜔𝑡 − 𝑘𝑧) 𝐚𝑦 = 𝜖
⇒ 𝐄=
sin(𝜔𝑡 − 𝑘𝑧) 𝐚𝑦 V∕m
𝜕𝑧
𝜇
𝜕𝑡
𝜔𝜇𝜖
[
]
𝜕𝐀
𝜕𝐀
⇒ ∇𝑉 = −
+𝐄
𝜕𝑡
𝜕𝑡
]
[
𝜕𝑉
𝑘2
sin(𝜔𝑡 − 𝑘𝑧) 𝐚𝑦 =
∇𝑉 = 𝐴0 𝜔 1 − 2
𝐚
𝜕𝑦 𝑦
𝜔 𝜇𝜖
𝐄 = −∇𝑉 −
or
Integrating over 𝑦 we find
[
]
𝑘2
sin(𝜔𝑡 − 𝑘𝑧) + 𝐶
𝑉 = 𝐴0 𝜔 𝑦 1 − 2
𝜔 𝜇𝜖
√
where 𝐶, the integration constant, can be taken as zero. In part 𝑏, it will be shown that 𝑘 = 𝜔 𝜇𝜖,
which means that 𝑉 = 0.
b) Specify 𝑘 in terms of 𝐴0 , 𝜔, and the constants of the lossless medium, 𝜖 and 𝜇. Use the other
Maxwell curl equation:
𝜕𝐁
𝜕𝐇
∇×𝐄=−
= −𝜇
𝜕𝑡
𝜕𝑡
so that
𝑘3 𝐴
𝜕𝐇
1
1 𝜕𝐸𝑦
=− ∇×𝐄=
𝐚𝑥 = − 2 0 cos(𝜔𝑡 − 𝑘𝑧) 𝐚𝑥
𝜕𝑡
𝜇
𝜇 𝜕𝑧
𝜔𝜇 𝜖
Integrate over 𝑡 (and set the integration constant to zero) and require the result to be consistant
with part 𝑎:
𝑘𝐴
𝑘3 𝐴
𝐇 = − 2 20 sin(𝜔𝑡 − 𝑘𝑧) 𝐚𝑥 = − 0 sin(𝜔𝑡 − 𝑘𝑧) 𝐚𝑥
𝜇
𝜔 𝜇 𝜖
⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
f rom part a
We identify
√
𝑘 = 𝜔 𝜇𝜖
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9.25. In a region where 𝜇𝑟 = 𝜖𝑟 = 1 and 𝜎 = 0, the retarded potentials are given by 𝑉 = 𝑥(𝑧 − 𝑐𝑡) V and
√
𝐀 = 𝑥[(𝑧∕𝑐) − 𝑡]𝐚𝑧 Wb/m, where 𝑐 = 1∕ 𝜇0 𝜖0 .
a) Show that ∇ ⋅ 𝐀 = −𝜇𝜖(𝜕𝑉 ∕𝜕𝑡):
First,
∇⋅𝐀=
Second,
√
𝜕𝐴𝑧
𝑥
= = 𝑥 𝜇0 𝜖0
𝜕𝑧
𝑐
𝜕𝑉
𝑥
= −𝑐𝑥 = − √
𝜕𝑡
𝜇0 𝜖0
so we observe that ∇ ⋅ 𝐀 = −𝜇0 𝜖0 (𝜕𝑉 ∕𝜕𝑡) in free space, implying that the given statement would
hold true in general media.
b) Find 𝐁, 𝐇, 𝐄, and 𝐃:
Use
𝐁=∇×𝐀=−
Then
𝐇=
)
(
𝜕𝐴𝑥
𝑧
𝐚 T
𝐚𝑦 = 𝑡 −
𝜕𝑥
𝑐 𝑦
)
(
1
𝑧
𝐁
=
𝐚 A∕m
𝑡−
𝜇0
𝜇0
𝑐 𝑦
Now,
𝐄 = −∇𝑉 −
𝜕𝐀
= −(𝑧 − 𝑐𝑡)𝐚𝑥 − 𝑥𝐚𝑧 + 𝑥𝐚𝑧 = (𝑐𝑡 − 𝑧)𝐚𝑥 V∕m
𝜕𝑡
Then
𝐃 = 𝜖0 𝐄 = 𝜖0 (𝑐𝑡 − 𝑧)𝐚𝑥 C∕m2
c) Show that these results satisfy Maxwell’s equations if 𝐉 and 𝜌𝑣 are zero:
i. ∇ ⋅ 𝐃 = ∇ ⋅ 𝜖0 (𝑐𝑡 − 𝑧)𝐚𝑥 = 0
ii. ∇ ⋅ 𝐁 = ∇ ⋅ (𝑡 − 𝑧∕𝑐)𝐚𝑦 = 0
iii.
∇×𝐇=−
𝜕𝐻𝑦
𝜕𝑧
𝐚𝑥 =
1
𝐚 =
𝜇0 𝑐 𝑥
√
𝜖0
𝐚
𝜇0 𝑥
which we require to equal 𝜕𝐃∕𝜕𝑡:
𝜕𝐃
= 𝜖0 𝑐𝐚𝑥 =
𝜕𝑡
iv.
∇×𝐄=
√
𝜖0
𝐚
𝜇0 𝑥
𝜕𝐸𝑥
𝐚 = −𝐚𝑦
𝜕𝑧 𝑦
which we require to equal −𝜕𝐁∕𝜕𝑡:
𝜕𝐁
= 𝐚𝑦
𝜕𝑡
So all four Maxwell equations are satisfied.
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9.26. Write Maxwell’s equations in point form in terms of 𝐄 and 𝐇 as they apply to a sourceless medium,
where 𝐉 and 𝜌𝑣 are both zero. Replace 𝜖 by 𝜇, 𝜇 by 𝜖, 𝐄 by 𝐇, and 𝐇 by −𝐄, and show that the
equations are unchanged. This is a more general expression of the duality principle in circuit theory.
Maxwell’s equations in sourceless media can be written as:
∇ × 𝐄 = −𝜇
𝜕𝐇
𝜕𝑡
(1)
𝜕𝐄
𝜕𝑡
∇ ⋅ 𝜖𝐄 = 0
(3)
∇ ⋅ 𝜇𝐇 = 0
(4)
∇×𝐇=𝜖
(2)
In making the above substitutions, we find that (1) converts to (2), (2) converts to (1), and (3)
and (4) convert to each other.
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