244683525 Chapter 09

Prévia do material em texto

CHAPTER 9
9.1. In Fig. 9.4, let B = 0.2 cos 120⇡t 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 � = ⇡(0.15)2B = 1.41 ⇥
10�2 cos 120⇡t Wb. Now, emf = Vba(t) = �d�/dt = (120⇡)(1.41 ⇥ 10�2) sin 120⇡t.
Then Vab(t) = �Vba(t) = �5.33 sin 120⇡t V.
b) I(t) = Vba(t)/R = 5.33 sin(120⇡t)/250 = 21.3 sin(120⇡t) 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!t 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
�m =
Z
s
B · dS =
Z vt
0
Z d
0
B0 sin!t (az · az) dx dy = B0 v t d sin!t
Then the emf is
emf =
I
E · dL = �d�m
dt
= �B0 d v [sin!t + !t cos!t] V
9.3. Given H = 300az cos(3⇥ 108t� y) A/m in free space, find the emf developed in the general
a� 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:
� =
Z 1
0
Z 1
0
300µ0 cos(3⇥ 108t� y) dx dy = 300µ0 sin(3⇥ 108t� y)|10
= 300µ0
⇥
sin(3⇥ 108t� 1)� sin(3⇥ 108t)⇤ Wb
Then
emf = �d�
dt
= �300(3⇥ 108)(4⇡ ⇥ 10�7) ⇥cos(3⇥ 108t� 1)� cos(3⇥ 108t)⇤
= �1.13⇥ 105 ⇥cos(3⇥ 108t� 1)� cos(3⇥ 108t)⇤ V
b) corners at (0,0,0), (2⇡,0,0), (2⇡,2⇡,0), (0,2⇡,0): In this case, the flux is
� = 2⇡ ⇥ 300µ0 sin(3⇥ 108t� y)|2⇡0 = 0
The emf is therefore 0.
164
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 !, with the first-named corner moving in the
az direction at t = 0. Assume a uniform magnetic flux density B = B0az. 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
�m =
Z
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!taz � sin!tay
Therefore,
�m =
Z b/2
�b/2
Z a/2
�a/2
B0az · (cos!taz � sin!tay) dx dy = abB0 cos!t
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 =
I
E · dL = �d�m
dt
= ab!B0 sin!t V
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 = 0.8x2az T. Find the voltmeter reading at:
a) t = 0.4 s: The flux through the loop will be
� =
Z 0.2
0
Z x
0
0.8(x0)2 dx0 dy =
0.16
3
x3 =
0.16
3
(5t+ 2t3)3 Wb
Then
emf = �d�
dt
=
0.16
3
(3)(5t+2t3)2(5+6t2) = �(0.16)[5(.4)+2(.4)3]2[5+6(.4)2] = �4.32 V
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
165
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(!t� �y)az, where ! and
� are constants.
We begin by finding the net magnetic flux through the loop:
�m =
Z
s
B · dS =
Z b/2
�b/2
Z a/2
�a/2
B0 cos(!t� �y)az · az dx dy
=
B0a
�
[sin(!t+ �b/2)� sin(!t� �b/2)]
Now the emf is
emf =
I
E · dL = �d�m
dt
= �B0a!
�
[cos(!t+ �b/2)� cos(!t� �b/2)]
Using the trig identity, cos(a± b) = cos a cos b⌥ sin a sin b, we may write the above result
as
emf = +2B0 a
!
�
sin(!t) sin(�b/2) V
9.7. The rails in Fig. 9.7 each have a resistance of 2.2 ⌦/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 ⌦ resistor is present across the left end with the right end open-circuited: The flux
in the left-hand closed loop is
�l = B ⇥ area = (0.8)(0.2)(2 + 9t)
Then, emfl = �d�l/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) =
emfl
Rl(t)
=
�1.44
9.1 + 39.6t
A
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 ⌦ 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
�r = (0.8)(0.2)[(16� 2)� 9t]
and emfr = �d�r/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
61.9� 39.6t A
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
IT (t) = �1.44

1
61.9� 39.6t +
1
9.1 + 39.6t
�
A
166
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 inde-
pendently 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 thereforeI
E · dL = 0
This is just a statement of Kircho↵’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 a� direction is thus counter-clockwise around the
loop. The B field (out of the page as shown) can be written as B(t) = B0taz. With the
normal to the loop specified as az, the direction of dL is, by the right hand convention,
a�. Since the wire is perfectly-conducting, the only voltage appears across the resistor,
and is given as VR. Faraday’s law becomesI
E · dL = VR = �d�m
dt
= � d
dt
Z
s
B0taz · az da = �⇡a2B0
This indicates that the resistor voltage, VR = ⇡a2B0, 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 = ⇡a2B0/R.
9.9. A square filamentary loop of wire is 25 cm on a side and has a resistance of 125 ⌦ 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⇥108t�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:
�(t) =
Z .25
0
Z .25
0
8⇥ 10�6 cos(1.5⇥ 108t� 0.5x) dx dy
= �(4⇥ 10�6) ⇥sin(1.5⇥ 108t� 0.13)� sin(1.5⇥ 108t)⇤ Wb
Now, emf = V (t) = �d�/dt = 6.0⇥ 102 ⇥cos(1.5⇥ 108t� 0.13)� cos(1.5⇥ 108t)⇤, The total
loop resistance is R = 125(0.25 + 0.25 + 0.25 + 0.25) = 125⌦. Then the ohmic power is
P (t) =
V 2(t)
R
= 2.9⇥ 103 ⇥cos(1.5⇥ 108t� 0.13)� cos(1.5⇥ 108t)⇤2 Watts
167
9.10 a) Show that the ratio of the amplitudes of the conduction current density and the displacement
current density is �/!✏ for the applied field E = Em cos!t. Assume µ = µ0. First, D =
✏E = ✏Em cos!t. Then the displacement current density is @D/@t = �!✏Em sin!t. Second,
Jc = �E = �Em cos!t. Using these results we find |Jc|/|Jd| = �/!✏.
b) What is the amplitude ratio if the applied field is E = Eme�t/⌧ , where ⌧ is real? As before,
find D = ✏E = ✏Eme�t/⌧ , and so Jd = @D/@t = �(✏/⌧)Eme�t/⌧ . Also, Jc = �Eme�t/⌧ .
Finally, |Jc|/|Jd| = �⌧/✏.
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 ✏ = 10�11 F/m, µ = 10�5
H/m, and � = 10�5 S/m. If the electric field intensity is E = (106/⇢) cos(105t)a⇢ V/m, find:
a) J: Use
J = �E =
✓
10
⇢
◆
cos(105t)a⇢ A/m2
b) the total conduction current, Ic, through the capacitor: Have
Ic =
Z Z
J · dS = 2⇡⇢lJ = 20⇡l cos(105t) = 8⇡ cos(105t) A
c) the total displacement current, Id, through the capacitor: First find
Jd =
@D
@t
=
@
@t
(✏E) = �(10
5)(10�11)(106)
⇢
sin(105t)a⇢ = �1
⇢
sin(105t) A/m
Now
Id = 2⇡⇢lJd = �2⇡l sin(105t) = �0.8⇡ sin(105t) A
d) the ratio of the amplitude of Id to that of Ic, the quality factor of the capacitor: This will
be |Id|
|Ic| =
0.8
8
= 0.1
168
9.12. Find the displacement current density associated with the magnetic field (assume zero con-
duction current):
H = A1 sin(4x) cos(!t� �z)ax +A2 cos(4x) sin(!t� �z)az
The displacement current density is given by
@D
@t
= r⇥H = (4A2 + �A1) sin(4x) cos(!t� �z)ay A/m2
9.13. Consider the region defined by |x|, |y|, and |z| < 1. Let ✏r = 5, µr = 4, and � = 0. If
Jd = 20 cos(1.5⇥ 108t� bx)ay µA/m2;
a) find D and E: Since Jd = @D/@t, we write
D =
Z
Jddt+ C =
20⇥ 10�6
1.5⇥ 108 sin(1.5⇥ 10
8 � bx)ay
= 1.33⇥ 10�13 sin(1.5⇥ 108t� bx)ay C/m2
where the integration constant is set to zero (assuming no dc fields are present). Then
E =
D
✏
=
1.33⇥ 10�13
(5⇥ 8.85⇥ 10�12) sin(1.5⇥ 10
8t� bx)ay
= 3.0⇥ 10�3 sin(1.5⇥ 108t� bx)ay V/m
b) use the point form of Faraday’s law and an integration with respect to time to find B and
H: In this case,
r⇥E = @Ey
@x
az = �b(3.0⇥ 10�3) cos(1.5⇥ 108t� bx)az = �@B
@t
Solve for B by integrating over time:
B =
b(3.0⇥ 10�3)
1.5⇥ 108 sin(1.5⇥ 10
8t� bx)az = (2.0)b⇥ 10�11 sin(1.5⇥ 108t� bx)az T
Now
H =
B
µ
=
(2.0)b⇥ 10�11
4⇥ 4⇡ ⇥ 10�7 sin(1.5⇥ 10
8t� bx)az
= (4.0⇥ 10�6)b sin(1.5⇥ 108t� bx)az A/m
c) use r ⇥H = Jd + J to find Jd: Since � = 0, there is no conduction current, so in this
case
r⇥H = �@Hz
@x
ay = 4.0⇥ 10�6b2 cos(1.5⇥ 108t� bx)ay A/m2 = Jd
d) What is the numerical value of b? We set the given expression for Jd equal to the result
of part c to obtain:
20⇥ 10�6 = 4.0⇥ 10�6b2 ) b = p5.0 m�1
169
9.14. A voltage source, V0 sin!t, is connected between two concentric conducting spheres, r = a
and r = b, b > a, where the region between them is a material for which ✏ = ✏r✏0, µ = µ0, and
� = 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) =
(1/r)� (1/b)
(1/a)� (1/b)V0 sin!t
Then
E = �rV = V0 sin!t
r2(1/a� 1/b)ar ) D =
✏r✏0V0 sin!t
r2(1/a� 1/b)ar
Now
Jd =
@D
@t
=
✏r✏0!V0 cos!t
r2(1/a� 1/b) ar
The displacement current is then
Id = 4⇡r2Jd =
4⇡✏r✏0!V0 cos!t
(1/a� 1/b) = C
dV
dt
where, from Eq. 6, Chapter 6,
C =
4⇡✏r✏0
(1/a� 1/b)
9.15. Let µ = 3⇥10�5 H/m, ✏ = 1.2⇥10�10 F/m, and � = 0 everywhere. IfH = 2 cos(1010t��x)az
A/m, use Maxwell’s equations to obtain expressions for B, D, E, and �: First, B = µH =
6⇥ 10�5 cos(1010t� �x)az T. Next we use
r⇥H = �@H
@x
ay = 2� sin(1010t� �x)ay = @D
@t
from which
D =
Z
2� sin(1010t� �x) dt+ C = � 2�
1010
cos(1010t� �x)ay C/m2
where the integration constant is set to zero, since no dc fields are presumed to exist. Next,
E =
D
✏
= � 2�
(1.2⇥ 10�10)(1010) cos(10
10t� �x)ay = �1.67� cos(1010t� �x)ay V/m
Now
r⇥E = @Ey
@x
az = 1.67�2 sin(1010t� �x)az = �@B
@t
So
B = �
Z
1.67�2 sin(1010t� �x)azdt = (1.67⇥ 10�10)�2 cos(1010t� �x)az
We require this result to be consistent with the expression for B originally found. So
(1.67⇥ 10�10)�2 = 6⇥ 10�5 ) � = ±600 rad/m
170
9.16. Derive the continuity equation from Maxwell’s equations: First, take the divergence of both
sides of Ampere’s circuital law:
r ·r⇥H| {z }
0
= r · J+ @
@t
r ·D = r · J+ @⇢v
@t
= 0
where we have used r ·D = ⇢v, another Maxwell equation.
9.17. The electric field intensity in the region 0 < x < 5, 0 < y < ⇡/12, 0 < z < 0.06 m in free
space is given by E = C sin(12y) sin(az) cos(2 ⇥ 1010t)ax V/m. Beginning with the r ⇥ 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
r⇥E = @Ex
@z
ay � @Ez
@y
az
= C [a sin(12y) cos(az)ay � 12 cos(12y) sin(az)az] cos(2⇥ 1010t) = �@B
@t
Then
H = � 1
µ0
Z
r⇥E dt+ C1
= � C
µ0(2⇥ 1010 [a sin(12y) cos(az)ay � 12 cos(12y) sin(az)az] sin(2⇥ 10
10t) A/m
where the integration constant, C1 = 0, since there are no initial conditions. Using this result,
we now find
r⇥H =

@Hz
@y
� @Hy
@z
�
ax = �C(144 + a
2)
µ0(2⇥ 1010) sin(12y) sin(az) sin(2⇥ 10
10t)ax =
@D
@t
Now
E =
D
✏0
=
Z
1
✏0
r⇥H dt+ C2 = C(144 + a
2)
µ0✏0(2⇥ 1010)2 sin(12y) sin(az) cos(2⇥ 10
10t)ax
where C2 = 0. This field must be the same as the original field as stated, and so we require
that
C(144 + a2)
µ0✏0(2⇥ 1010)2 = 1
Using µ0✏0 = (3⇥ 108)�2, we find
a =

(2⇥ 1010)2
(3⇥ 108)2 � 144
�1/2
= 66 m�1
171
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, ✏r = 20, and � = 0. Neglect fields
outside the dielectric. Given the field H = 5 cos(109t� �z)ay A/m, use Maxwell’s equations
to help find:
a) �, if � > 0: Take
r⇥H = �@Hy
@z
ax = �5� sin(109t� �z)ax = 20✏0 @E
@t
So
E =
Z �5�
20✏0
sin(109t� �z)ax dt = �(4⇥ 109)✏0 cos(10
9t� �z)ax
Then
r⇥E = @Ex
@z
ay =
�2
(4⇥ 109)✏0 sin(10
9t� �z)ay = �µ0 @H
@t
So that
H =
Z ��2
(4⇥ 109)µ0✏0 sin(10
9t� �z)ax dt = �
2
(4⇥ 1018)µ0✏0 cos(10
9t� �z)
= 5 cos(109t� �z)ay
where the last equality is required to maintain consistency. Therefore
�2
(4⇥ 1018)µ0✏0 = 5 ) � = 14.9 m
�1
b) the displacement current density at z = 0: Since � = 0, we have
r⇥H = Jd = �5� sin(109t� �z) = �74.5 sin(109t� 14.9z)ax
= �74.5 sin(109t)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
Z 0.1
0
sin(109t� 14.9z)ax · ax dz = 0.20
⇥
cos(109t� 1.49)� cos(109t)⇤ A
9.19.In the first section of this chapter, Faraday’s law was used to show that the field E =
�12kB0⇢ekta� results from the changing magnetic field B = B0ektaz.
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 r⇥H = @D@t would have a zero left-hand side and a non-zero right-hand
side. The equation 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 r⇥H equation to show that the rate at which Bz should (but
does not) change with ⇢ is only about 5 ⇥ 10�6 T/m in free space at t = 0: Assuming
that B varies with ⇢, we write
r⇥H = �@Hz
@⇢
a� = � 1
µ0
dB0
d⇢
ekt = ✏0
@E
@t
= �1
2
✏0k
2B0⇢e
kt
Thus
dB0
d⇢
=
1
2
µ0✏0k
2⇢B0 =
1012(1)⇢
2(3⇥ 108)2 = 5.6⇥ 10
�6⇢
which is near the stated value if ⇢ is on the order of 1m.
172
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:
r⇥E0(r) est = � @
@t
�
B0(r) est
�
= �sB0(r) est
r⇥H0(r) est = J0(r)est + @
@t
�
D0(r) est
�
= J0(r)est + sD0(r) est
r ·D0(r) est = ⇢0(r) est
r ·B0(r) est = 0
In all cases, the est terms divide out, leaving:
r⇥E0(r) = �sB0(r)
r⇥H0(r) = J0(r) + sD0(r)
r ·D0(r) = ⇢0(r)
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:
r2A = �r⇥r⇥A+r(r ·A) = �µJ
which is Eq. (55) with the time derivative set to zero. We also note that r ·A = 0 in steady
state (from Eq. (54)). Now, since B = r⇥A, (55) becomes
�r⇥B = �µJ ) r⇥H = J
b) Show that Eq. (51) becomes Faraday’s law when taking the curl: Doing this gives
r⇥E = �r⇥rV � @
@t
r⇥A
The curl of the gradient is identially zero, and r⇥A = B. We are left with
r⇥E = �@B/@t
173
9.22. In a sourceless medium, in which J = 0 and ⇢v = 0, assume a rectangular coordinate system in
which E andH are functions only of z and t. The medium has permittivity ✏ and permeability
µ.
a) If E = Exax and H = Hyay, begin with Maxwell’s equations and determine the second
order partial di↵erential equation that Ex must satisfy.
First use
r⇥E = �@B
@t
) @Ex
@z
ay = �µ@Hy
@t
ay
in which case, the curl has dictated the direction that H must lie in. Similarly, use the
other Maxwell curl equation to find
r⇥H = @D
@t
) �@Hy
@z
ax = ✏
@Ex
@t
ax
Now, di↵erentiate the first equation with respect to z, and the second equation with
respect to t:
@2Ex
@z2
= �µ@
2Hy
@t@z
and
@2Hy
@z@t
= �✏@
2Ex
@t2
Combining these two, we find
@2Ex
@z2
= µ✏
@2Ex
@t2
b) Show that Ex = E0 cos(!t��z) is a solution of that equation for a particular value of �:
Substituting, we find
@2Ex
@z2
= ��2E0 cos(!t� �z) and µ✏@
2Ex
@t2
= �!2µ✏E0 cos(!t� �z)
These two will be equal provided the constant multipliers of cos(!t� �z) are equal.
c) Find � as a function of given parameters. Equating the two constants in part b, we find
� = !pµ✏.
174
9.23. In region 1, z < 0, ✏1 = 2 ⇥ 10�11 F/m, µ1 = 2 ⇥ 10�6 H/m, and �1 = 4 ⇥ 10�3 S/m; in
region 2, z > 0, ✏2 = ✏1/2, µ2 = 2µ1, and �2 = �1/4. It is known that E1 = (30ax + 20ay +
10az) cos(109t) V/m at P1(0, 0, 0�).
a) Find EN1, Et1, DN1, and Dt1: These will be
EN1 = 10 cos(109t)az V/m Et1 = (30ax + 20ay) cos(109t) V/m
DN1 = ✏1EN1 = (2⇥ 10�11)(10) cos(109t)az C/m2 = 200 cos(109t)az pC/m2
Dt1 = ✏1Et1 = (2⇥ 10�11)(30ax + 20ay) cos(109t) = (600ax + 400ay) cos(109t) pC/m2
b) Find JN1 and Jt1 at P1:
JN1 = �1EN1 = (4⇥ 10�3)(10 cos(109t))az = 40 cos(109t)az mA/m2
Jt1 = �1Et1 = (4⇥ 10�3)(30ax + 20ay) cos(109t) = (120ax + 80ay) cos(109t) mA/m2
c) Find Et2, Dt2, and Jt2 at P1: By continuity of tangential E,
Et2 = Et1 = (30ax + 20ay) cos(109t) V/m
Then
Dt2 = ✏2Et2 = (10�11)(30ax + 20ay) cos(109t) = (300ax + 200ay) cos(109t) pC/m2
Jt2 = �2Et2 = (10�3)(30ax + 20ay) cos(109t) = (30ax + 20ay) cos(109t) mA/m2
d) (Harder) Use the continuity equation to help show that JN1�JN2 = @DN2/@t�@DN1/@t
and then determine EN2, DN2, and JN2: We assume the existence of a surface charge
layer at the boundary having density ⇢s C/m2. If we draw a cylindrical “pillbox” whose
top and bottom surfaces (each of area �a) are on either side of the interface, we may use
the continuity condition to write
(JN2 � JN1)�a = �@⇢s
@t
�a
where ⇢s = DN2 �DN1. Therefore,
JN1 � JN2 = @
@t
(DN2 �DN1)
In terms of the normal electric field components, this becomes
�1EN1 � �2EN2 = @
@t
(✏2EN2 � ✏1EN1)
Now let EN2 = A cos(109t) +B sin(109t), while from before, EN1 = 10 cos(109t).
175
9.23d (continued)
These, along with the permittivities and conductivities, are substituted to obtain
(4⇥ 10�3)(10) cos(109t)� 10�3[A cos(109t) +B sin(109t)]
=
@
@t
⇥
10�11[A cos(109t) +B sin(109t)]� (2⇥ 10�11)(10) cos(109t)⇤
= �(10�2A sin(109t) + 10�2B cos(109t) + (2⇥ 10�1) sin(109t)
We now equate coe�cients of the sin and cos terms to obtain two equations:
4⇥ 10�2 � 10�3A = 10�2B
�10�3B = �10�2A+ 2⇥ 10�1
These are solved together to find A = 20.2 and B = 2.0. Thus
EN2 =
⇥
20.2 cos(109t) + 2.0 sin(109t)
⇤
az = 20.3 cos(109t+ 5.6�)az V/m
Then
DN2 = ✏2EN2 = 203 cos(109t+ 5.6�)az pC/m2
and
JN2 = �2EN2 = 20.3 cos(109t+ 5.6�)az mA/m2
176
9.24. A vector potential is given as A = A0 cos(!t � 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 =
1
µ
r⇥A = � 1
µ
@Ay
@z
ax = �kA0
µ
sin(!t� kz)ax A/m
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:
r⇥H = @D
@t
= ✏
@E
@t
Using the magnetic field as found above, we find
r⇥H = @Hx
@z
ay =
k2A0
µ
cos(!t� kz)ay = ✏@E
@t
) E = k
2A0
!µ✏
sin(!t� kz)ay V/m
Now,
E = �rV � @A
@t
) rV = �

@A
@t
+E
�
or
rV = A0!

1� k
2
!2µ✏
�
sin(!t� kz)ay = @V
@y
ay
Integrating over y we find
V = A0 ! y

1� k
2
!2µ✏
�
sin(!t� kz) + C
where C, the integration constant, can be taken as zero. In part b, it will be shown that
k = !pµ✏, which means that V = 0.
b) Specify k in terms of A0, !, and the constants of the lossless medium, ✏ and µ. Use the
other Maxwell curl equation:
r⇥E = �@B
@t
= �µ@H
@t
so that
@H
@t
= � 1
µ
r⇥E = 1
µ
@Ey
@z
ax = �k
3A0
!µ2✏
cos(!t� kz)ax
Integrate over t (and set the integration constant to zero) and require the result to be
consistant with part a:
H = � k
3A0
!2µ2✏
sin(!t� kz)ax = �kA0
µ
sin(!t� kz)ax| {z }
from part a
We identify
k = !
p
µ✏
177
9.25. In a region where µr = ✏r = 1 and � = 0, the retarded potentials are given by V = x(z � ct)
V and A = x[(z/c)� t]az Wb/m, where c = 1/pµ0✏0.
a) Show that r ·A = �µ✏(@V/@t):
First,
r ·A = @Az
@z
=
x
c
= x
p
µ0✏0
Second,
@V
@t
= �cx = � xp
µ0✏0
so we observe that r ·A = �µ0✏0(@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 = r⇥A = �@Ax
@x
ay =
⇣
t� z
c
⌘
ay T
Then
H =
B
µ0
=
1
µ0
⇣
t� z
c
⌘
ay A/m
Now,
E = �rV � @A
@t
= �(z � ct)ax � xaz + xaz = (ct� z)ax V/m
Then
D = ✏0E = ✏0(ct� z)ax C/m2
c) Show that these results satisfy Maxwell’s equations if J and ⇢v are zero:
i. r ·D = r · ✏0(ct� z)ax = 0
ii. r ·B = r · (t�z/c)ay = 0
iii.
r⇥H = �@Hy
@z
ax =
1
µ0c
ax =
r
✏0
µ0
ax
which we require to equal @D/@t:
@D
@t
= ✏0cax =
r
✏0
µ0
ax
iv.
r⇥E = @Ex
@z
ay = �ay
which we require to equal �@B/@t:
@B
@t
= ay
So all four Maxwell equations are satisfied.
178
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 ⇢v are both zero. Replace ✏ by µ, µ by ✏, 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:
r⇥E = �µ@H
@t
(1)
r⇥H = ✏@E
@t
(2)
r · ✏E = 0 (3)
r · µH = 0 (4)
In making the above substitutions, we find that (1) converts to (2), (2) converts to (1),
and (3) and (4) convert to each other.
179