CHAPTER 9
9.1. In
Fig. 9.4, let B let B = 0. 0 .2cos120π t T, and assume that the conductor joining the two ends of the resistor resistor is perfect. It may be assumed assumed that the magnetic magnetic field produced produced by I (t) is negligible. Find: a) V ab Sincee B is constant over the loop area, the flux is Φ = π(0. (0.15)2 B = 1.41 ab (t): Sinc 10 2 cos120π t Wb. Wb. No Now, w, emf = V ba dΦ/dt = (120π )(1. )(1.41 10 2 )sin120π t. ba (t) = Then V Then V ab ab (t) = V ba ba (t) = 5.33sin120 π t V. −
−
−
−
×
×
−
b) I (t) = V = V ba /R = 5.33 sin(120 sin(120π t)/250 = 21. 21.3 sin(120 sin(120π t) mA ba (t)/R = 9.2. In
the example described by Fig. 9.1, replace the constant magnetic flux density by the timevarying quantity B = B = B0 sin ωt az . Assume that v is constant and that the displacement y displacement y of the bar is zero at t at t = = 0. Find the emf at any time, t. t . The magnetic flux through the loop area is Φm =
vt
B · dS =
s
d
0
B0 sin ωt ( az · az ) dxdy = dxdy = B B 0 v t d sin ωt
0
Then the emf is
emf =
E · dL =
− dΦdtm = −B d v [sin ωt + ωt cos ωt] V 0
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), (0,0,0), (1,0,0), (1,0,0), (1,1,0), (1,1,0), and (0,1,0): (0,1,0): The magnetic magnetic flux will be:
9.3. Given H =
×
1
Φ
=
0
−
1
0
= 300µ 300µ0
8
300µ 300 µ0 cos(3
8
× 10 t − y) dxdy = dxdy = 300µ 300µ sin(3 × 10 t − y )| sin(3 × 10 t − 1) − sin(3 × 10 t) Wb
0
8
8
1 0
Then
− ddtΦ = −300(3 × 10 )(4π × 10 ) cos(3 × 10 t − 1) − cos(3 × 10 t) = −1.13 × 10 cos(3 × 10 t − 1) − cos(3 × 10 t) V 8
emf =
5
7
−
8
8
8
8
b) corners at (0,0,0), (2 π,0,0), (2π ,2π,0), (0,2π ,0): In this case, the flux is Φ
= 2π
8
2π 0
× 300 300µ µ sin(3 × 10 t − y)| 0
The emf is therefore 0.
164
=0
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 =
B · dS, where S = n da
s
Because the loop is rotating, the direction of the normal, n, changing, and is in this case given by n = cos ω t az sin ωt ay
−
Therefore,
b/2
Φm =
b/2
−
a/2
B0 az · (cos ωt az a/2
−
− sin ωt ay ) dxdy = abB cos ωt 0
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 =
− dΦdtm = ab ωB sin ωt V 0
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, d L 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. 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
9.5. The
0.2
Φ
=
0
x
0.8(x )2 dx dy =
0
0.16 3 0.16 x = (5t + 2t3 )3 Wb 3 3
Then emf =
− ddtΦ = 0.16 (3)(5t +2t ) (5+6t ) = −(0.16)[5(.4)+2(.4) ] [5+6(.4) ] = −4.32 V 3 3 2
2
3 2
2
b) x = 0.6 m: Have 0.6 = 5t + 2t3 , from which we find t = 0.1193. Thus emf =
3 2
2
−(0.16)[5(.1193) + 2(.1193) ] [5 + 6(.1193) ] = −.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(ω t β y) az , where ω and β are constants. We begin by finding the net magnetic flux through the loop:
−
Φm
=
b/2
B · dS =
s
=
b/2
−
a/2
B0 cos(ωt a/2
−
B0 a [sin(ωt + β b/2) β
− β y) az · az dxdy
− sin(ωt − β b/2)]
Now the emf is
− dΦdtm = − B β aω [cos(ωt + β b/2) − cos(ωt − β b/2)] Using the trig identity, cos(a ± b) = cos a cos b ∓ sin a sin b, we may write the above result emf =
0
E · dL =
as
ω β
emf = +2B0 a sin(ωt) sin(β b/2) V
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) = −dΦl /dt = −(0.16)(9) = − 1.44 V. With the bar in motion, the loop
Then, emf l resistance is increasing with time, and is given by Rl (t) = 0.3+2[2.2(2+9t)]. The current is now emf l 1.44 I l (t) = = 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
− −
I r (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 I T (t) =
−1.44
1 61.9 166
−
1 + 39.6t 9.1 + 39.6t
A
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 V 0 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 linearlyincreasing 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) = B 0 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 V R . Faraday’s law becomes
E · dL = V R =
−
dΦm = dt
−
d dt
B0 t az · az da =
s
2
−πa B
0
This indicates that the resistor voltage, V R = π a2 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 = π a2 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 = 8cos(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:
×
−
.25
Φ(t)
=
0
=
.25
8
0
6
−
−(4 × 10
)
6
−
× 10
8
× 10 t − 0.5x) dxdy sin(1.5 × 10 t − 0.13x) − sin(1.5 × 10 t)
cos(1.5 8
8
Wb
Now, emf = V (t) = dΦ/dt = 6.0 102 cos(1.5 108 t 0.13x) 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
−
V 2 (t) P (t) = = 2.9 R
×
3
× 10
×
cos(1.5
8
−
−
×
8
× 10 t − 0.13x) − cos(1.5 × 10 t)
167
Watts
9.10
a) Show that the ratio of the amplitudes of the conduction current density and the displacement current density is σ/ω for the applied field E = E m cos ωt. Assume µ = µ0 . First, D = E = E m cos ω t. Then the displacement current density is ∂ D/∂ t = ωE m sin ω t. Second, J c = σ E = σ E m cos ωt. Using these results we find |J c |/|J d | = σ /ω.
−
b) What is the amplitude ratio if the applied field is E = E m e t/τ , where τ is real? As before, find D = E = E m e t/τ , and so J d = ∂ D/∂ t = (/τ )E m e t/τ . Also, J c = σ E m e t/τ . Finally, |J c |/|J d | = στ /. −
−
−
−
−
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 σ = 10 5 S/m. If the electric field intensity is E = (106 /ρ) cos(105 t)aρ V/m, find: a) J: Use 10 J = σ E = cos(105 t)aρ A/m2 −
−
−
ρ
b) the total conduction current, I c , through the capacitor: Have I c =
J · dS =
2πρlJ = 20 π l cos(105 t) = 8π cos(105 t) A
c) the total displacement current, I d , through the capacitor: First find ∂ D ∂ Jd = = (E) = ∂ t ∂ t
(105 )(10
11
−
−
ρ
)(106 ) sin(105 t)aρ =
− ρ1 sin(10 t) A/m 5
Now I d = 2πρlJ d =
5
5
−2πl sin(10 t) = −0.8π sin(10 t) A
d) the ratio of the amplitude of I d to that of I c , the quality factor of the capacitor: This will be |I d | 0.8 = = 0.1 |I c | 8
168
9.12. Find
the displacement current density associated with the magnetic field (assume zero conduction current): H = A 1 sin(4x)cos(ω t
− β z) ax + A cos(4x) sin(ωt − β z) az 2
The displacement current density is given by ∂ D = ∂ t
2
∇ × H = (4A + β A ) sin(4x)cos(ωt − β z) ay A/m 2
1
9.13. Consider
the region defined by |x|, |y|, and |z| < 1. Let r = 5, µr = 4, and σ = 0. If Jd = 20 cos(1.5 108 t bx)ay µA/m2 ;
×
a) find
D
−
and E: Since
Jd = ∂ D/∂ t,
D =
we write
20 10 6 sin(1.5 108 bx)ay 8 1.5 10 13 10 sin(1.5 108 t bx)ay C/m2
× × ×
Jd dt + C =
= 1.33
−
×
−
×
−
−
where the integration constant is set to zero (assuming no dc fields are present). Then 1.33 10 13 E = = sin(1.5 (5 8.85 10 12 ) D
= 3.0
×
3
−
× 10
×
−
−
×
8
× 10 t − bx)ay
8
sin(1.5
× 10 t − bx)ay V/m
b) use the point form of Faraday’s law and an integration with respect to time to find H: In this case,
∇ × E = ∂ ∂ Exy az = −b(3.0 × 10
3
−
Solve for B =
Now
B by
b(3.0 1.5
× 10 t − bx)az = − ∂ ∂ Bt 8
integrating over time:
× 10 × 10
3
−
8
) sin(1.5
8
B
11
−
× 10 t − bx)az = (2.0)b × 10
(2.0)b 10 11 sin(1.5 µ 4 4π 10 7 = (4.0 10 6 )b sin(1.5 108 t
H =
−
× × ×
=
−
−
×
c) use case
)cos(1.5
B and
∇ × H = Jd + J to find Jd :
×
sin(1.5
8
× 10 t − bx)az − bx)az A/m
Since σ = 0, there is no conduction current, so in this
∇ × H = − ∂ ∂ Hxz ay = 4.0 × 10
6 2
−
b cos(1.5
8
2
× 10 t − bx)ay A/m
d) What is the numerical value of b? We set the given expression for of part c to obtain: 20
6
−
× 10
8
× 10 t − bx)az T
= 4.0
6 2
−
× 10 169
b
⇒
b =
√
Jd equal
1
−
5.0 m
= J d to the result
9.14. A
voltage source, V 0 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 = r 0 , 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): (1/r) (1/b) V (t) = V 0 sin ω t (1/a) (1/b)
− −
Then E =
Now
V sin ωt −∇V = r (1/a − 1/b) ar ⇒ 0
2
Jd =
r 0 V 0 sin ω t
D =
r2 (1/a
− 1/b) ar
∂ D r 0 ω V 0 cos ω t ar = 2 ∂ t r (1/a 1/b)
−
The displacement current is then I d = 4π r2 J d =
4πr 0 ωV 0 cos ωt dV = C (1/a 1/b) dt
−
where, from Eq. 6, Chapter 6, C = 9.15.
4πr 0 (1/a 1/b)
−
Let µ = 3 10 5 H/m, = 1.2 10 10 F/m, and σ = 0 everywhere. If H = 2 cos(1010 t β x)az A/m, use Maxwell’s equations to obtain expressions for B, D, E, and β : First, B = µH = 6 10 5 cos(1010 t β x)az T. Next we use −
×
−
×
−
×
−
−
∇ × H = − ∂ ∂ Hx a y = 2 β sin(10 t − β x)ay = ∂ ∂ Dt 10
from which D =
2β sin(1010 t
− β x) dt + C = − 102β cos(10 t − β x)ay C/m 10
2
10
where the integration constant is set to zero, since no dc fields are presumed to exist. Next, E =
D
=
− (1.2 × 102β )(10 10
10
−
Now
)
cos(1010 t
10
− β x)ay = −1.67β cos(10 t − β x)ay V/m
∇ × E = ∂ ∂ Exy az = 1.67β sin(10 t − β x)az = − ∂ ∂ Bt 2
So
−
10
)β 2 cos(1010 t
We require this result to be consistent with the expression for
B originally
B =
1.67β 2 sin(1010 t
10
(1.67
10
−
× 10
−
− β x)az dt = (1.67 × 10
)β 2 = 6
5
−
× 10 170
⇒
β = ±600 rad/m
− β x)az found. So
9.16. Derive
the continuity equation from Maxwell’s equations: First, take the divergence of both sides of Ampere’s circuital law:
∇ · ∇ × H = ∇ · J + ∂ ∂ t ∇ · D = ∇ · J + ∂ρ∂ tv = 0
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 E space is given by E = C sin(12y) sin(az) cos(2 1010 t) ax V/m. Beginning with the 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
×
∇ ×
∇ × E = ∂ ∂ Ezx ay − ∂ ∂ Eyz az = C [a sin(12y)cos(az)ay
10
− 12 cos(12y) sin(az)az ] cos(2 × 10
t) =
− ∂ ∂ Bt
Then H =
=
−
1 µ0
∇ ×
E dt + C 1
− µ (2 C × 10
10
0
[a sin(12y)cos(az)ay
10
− 12 cos(12y) sin(az)az ] sin(2 × 10
t) A/m
where the integration constant, C 1 = 0, since there are no initial conditions. Using this result, we now find
∇ × H =
∂ Hz ∂ y
=
∂ Hy ∂ z
−
2
ax =
) − µC ((2144×+ a sin(12y) sin(az) sin(2 × 10 10 )
10
10
0
t) ax =
∂ D ∂ t
Now E =
D
0
1 0
∇×
C (144 + a2 ) sin(12y) sin(az) cos(2 H dt + C 2 = µ0 0 (2 1010 )2
×
10
× 10
t) ax
where C 2 = 0. This field must be the same as the original field as stated, and so we require that C (144 + a2 ) =1 µ0 0 (2 1010 )2
×
Using µ0 0 = (3
8
× 10 )
2
−
, we find
(2 1010 )2 a = (3 108 )2
× ×
− 144
171
1/2 1
−
= 66 m
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 σ = 0. Neglect fields outside the dielectric. Given the field H = 5 cos(109 t β z )ay A/m, use Maxwell’s equations to help find: a) β , if β > 0: Take ∂ Hy ∂ E 9 β z )ax = 20 0 H = ax = 5β sin(10 t ∂ z ∂ t So β 5β E = sin(109 t β z )ax dt = cos(109 t β z )ax 9 200 (4 10 )0 Then ∂ Ex β 2 ∂ H 9 E = ay = β ay = µ0 sin(10 t z ) ∂ z ∂ t (4 109 )0 So that
−
∇×
−
−
−
−
H =
−
×
2
−β
9
sin(10 t
9
−
×
∇×
−
−
β 2
cos(109 t
− β z)ax dt = (4 × 10
18
(4 10 )µ0 0 )µ0 0 = 5 cos(109 t β z )ay where the last equality is required to maintain consistency. Therefore
×
−
β 2
(4
18
× 10
)µ0 0
=5
⇒
− β z)
1
−
β = 14.9 m
b) the displacement current density at z = 0: Since σ = 0, we have 9
9
∇ × H = J d = −5β sin(10 t − β z) = −74.5 sin(10 t − 14.9z)ax = −74.5 sin(10 t)ax A/m at z = 0 9
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:
0.1
I d = 9.19. In
−
1 2
−74.5b
sin(109 t
0
− 14.9z) ax · ax dz = 0.20
cos(109 t
9
− 1.49) − cos(10 t)
the first section of this chapter, Faraday’s law was used to show that the field kB0 ρekt aφ results from the changing magnetic field B = B 0 ekt az .
A E
=
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 ∂ D H = ∂ t would have a zero left-hand side and a non-zero right-hand with time, and so 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 H equation to show that the rate at which B z should (but density in 1 µs. Use the 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 ∂ Hz ∂ E 1 dB0 kt 1 2 H = aφ = 0 k B0 ρekt e = 0 = ∂ρ ∂ t µ0 dρ 2 Thus dB0 1 1012 (1)ρ = µ0 0 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. −
∇ ×
∇×
−
×
−
−
−
×
172
×
−
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:
9.20. Given
∇ × E (r) est = − ∂ ∂ t 0
∇ × H (r) est = J 0
( )est +
0 r
∂ ∂ t
st B0 (r) e
st D0 (r) e
=
−sB (r) est 0
= J 0 (r)est + sD0 (r) est
∇ · D (r) est = ρ (r) est 0
0
∇ · B (r) est = 0 0
In all cases, the est terms divide out, leaving:
∇ × E (r) = −sB (r) ∇ × H (r) = J (r) + sD (r) ∇ · D (r) = ρ (r) ∇ · B (r) = 0 0
0
0
0
0
0
0
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, (55) becomes state (from Eq. (54)). Now, since B =
∇× −∇ × B = −µJ ⇒ ∇ × H = J
∇ · A = 0 in steady
b) Show that Eq. (51) becomes Faraday’s law when taking the curl: Doing this gives
∇ × E = −∇×∇V − ∂ ∂ t ∇ × A The curl of the gradient is identially zero, and
∇ × A = B. We are left with ∇ × E = −∂ B/∂ t
173
9.22. In
a sourceless medium, in which J = 0 and ρ v = 0, assume a rectangular coordinate system in which E and H are functions only of z and t. The medium has permittivity and permeability µ. a) If E = E x ax and H = H y ay , begin with Maxwell’s equations and determine the second order partial diff erential equation that E x must satisfy. First use ∂ B ∂ Ex ∂ Hy E = ay = µ ay ∂ t ∂ z ∂ t in which case, the curl has dictated the direction that H must lie in. Similarly, use the other Maxwell curl equation to find
∇×
−
⇒
−
∇ × H = ∂ ∂ Dt ⇒ − ∂ ∂ Hzy ax = ∂ ∂ Etx ax Now, diff erentiate the first equation with respect to z, and the second equation with respect to t: ∂ 2 E x ∂ 2 H y ∂ 2 H y ∂ 2 E x = µ and = ∂ z 2 ∂ t∂ z ∂ z ∂ t ∂ t2 Combining these two, we find ∂ 2 E x ∂ 2 E x = µ 2 ∂ z 2 ∂ t
−
b) Show that E x = E 0 cos(ωt Substituting, we find ∂ 2 E x = ∂ z 2
2
−
− β z) is a solution of that equation for a particular value of β :
−β E cos(ωt − β z) 0
and µ
∂ 2 E x = ∂ t2
2
−ω µE cos(ωt − β z) 0
These two will be equal provided the constant multipliers of cos( ωt
− β z) are equal.
c) Find β as a function of given parameters. Equating the two constants in part b, we find β = ω µ.
√
174
region 1, z < 0, 1 = 2 10 11 F/m, µ1 = 2 10 6 H/m, and σ1 = 4 10 3 S/m; in region 2, z > 0, 2 = 1 /2, µ2 = 2µ1 , and σ2 = σ 1 /4. It is known that E1 = (30ax + 20 ay + 10az ) cos(109 t) V/m at P 1 (0, 0, 0 ).
9.23. In
−
×
×
−
×
−
−
a) Find
EN 1 , Et1 , DN 1 ,
and
DN 1 = 1 EN 1 =
b) Find
JN 1
(2
c) Find
× 10
11
−
× 10
11
−
(2
Et1 =
(30ax + 20 ay ) cos(109 t) V/m
)(10)cos(109 t)az C/m2 = 200cos(109 t)az pC/m2
)(30ax + 20 ay )cos(109 t) = (600ax + 400ay ) cos(109 t) pC/m2
and Jt1 at P 1 : JN 1 = σ 1 EN 1
Jt1
These will be
10 cos(109 t)az V/m
EN 1 =
Dt1 = 1 Et1 =
Dt1 :
= (4
× 10
Et2 , Dt2 ,
Jt2
and
3
−
= σ 1 Et1 = (4
3
−
× 10
)(10cos(109 t))az = 40 cos(109 t)az mA/m2
)(30ax + 20ay ) cos(109 t) = (120ax + 80 ay ) cos(109 t) mA/m2
at P 1 : By continuity of tangential Et2 = E t1 =
E,
(30ax + 20 ay ) cos(109 t) V/m
Then Dt2
= 2 Et2 = (10
Jt2 = σ 2 Et2 =
11
−
(10
)(30ax + 20 ay ) cos(109 t) = (300ax + 200ay ) cos(109 t) pC/m2
3
−
)(30ax + 20ay ) cos(109 t) = (30ax + 20ay ) cos(109 t) mA/m2
d) (Harder) Use the continuity equation to help show that J N 1 J N 2 = ∂ DN 2 /∂ t ∂ DN 1 /∂ t and then determine EN 2 , DN 2 , and JN 2 : We assume the existence of a surface charge layer at the boundary having density ρs 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
−
− J N )∆a = − ∂ρ∂ ts ∆a
(J N 2 where ρs = DN 2
−
1
− DN . Therefore, 1
J N 1
− J N = ∂ ∂ t (DN − DN ) 2
2
1
In terms of the normal electric field components, this becomes σ1 E N 1
− σ E N 2
2
=
∂ (2 E N 2 ∂ t
− E N ) 1
1
Now let E N 2 = A cos(109 t) + B sin(109 t), while from before, E N 1 = 10 cos(109 t).
175
9.23d
(continued) These, along with the permittivities and conductivities, are substituted to obtain (4
3
−
× 10
=
∂ 10 ∂ t
=
−(10
)(10) cos(109 t) 11
−
2
−
3
−
− 10
[A cos(109 t) + B sin(109 t)]
[A cos(109 t) + B sin(109 t)]
A sin(109 t) + 10
2
−
11
−
− (2 × 10 B cos(10 t) + (2 × 10 9
)(10) cos(109 t)
1
−
) sin(109 t)
We now equate coefficients of the sin and cos terms to obtain two equations: 2
−
4
3
A = 10
2
A + 2
−
× 10 − 10 −10 B = −10 3
−
−
2
−
B 1
−
× 10
These are solved together to find A = 20.2 and B = 2.0. Thus EN 2 =
20.2 cos(109 t) + 2.0 sin(109 t)
az
= 20.3 cos(109 t + 5.6 )az V/m ◦
Then DN 2 = 2 EN 2 =
203 cos(109 t + 5.6 )az pC/m2 ◦
and JN 2 = σ 2 EN 2
= 20.3 cos(109 t + 5.6 )az mA/m2 ◦
176
9.24. A
vector potential is given as A = A 0 cos(ω t possible are zero, find H, E, and V ; With
A
− kz) ay .
a) Assuming as many components as
y-directed only, and varying spatially only with z, we find H =
1 µ
∇ × A = − µ1 ∂ ∂ Azy ax = − kAµ sin(ωt − kz) ax A/m 0
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 = ∂ ∂ Dt = ∂ ∂ Et Using the magnetic field as found above, we find
∇×
∂ Hx H = ∂ z
ay
k2 A0 = cos(ωt µ
− kz) ay
∂ E = ∂ t
⇒
Now, E =
−∇V −
or
∂ A ∂ t
∇V = A ω 1 − 0
⇒ ∇V = −
k2 sin(ωt ω 2 µ
k2 A0 E = sin(ωt ω µ
− kz) ay V/m
∂ A +E ∂ t
− kz) ay = ∂ ∂ Vy ay
Integrating over y we find
V = A0 ω y 1
−
k2 sin(ωt ω 2 µ
− kz) + C
where C , the integration constant, can be taken as zero. In part b, it will be shown that k = ω µ, which means that V = 0.
√
b) Specify k in terms of A0 , ω, and the constants of the lossless medium, and µ. Use the other Maxwell curl equation:
∇ × E = − ∂ ∂ Bt = −µ ∂ ∂ Ht so that
∂ H = ∂ t
3
− µ1 ∇ × E = µ1 ∂ ∂ Ezy ax = − kωµA cos(ωt − kz) ax 0
2
Integrate over t (and set the integration constant to zero) and require the result to be consistant with part a: 3
H =
− ωk µA sin(ωt − kz) ax = − kAµ sin(ωt − kz) ax 2
0 2
0
We identify
√
k = ω µ 177
from part a
9.25. In
a region where µ r = r = 1 and σ = 0, the retarded potentials are given by V = x(z V and A = x[(z/c) t]az Wb/m, where c = 1/ µ0 0 .
√
−
− ct)
a) Show that
∇ · A = −µ(∂ V /∂ t):
First,
∇ · A = ∂ ∂ Azz = xc = x√ µ
0 0
Second,
∂ V = ∂ t
−cx = − √ µx
0 0
so we observe that · 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 =
∇ × A = −
Then H =
Now, E =
B
µ0
=
− −
∂ Ax ay = t ∂ x
1 t µ0
z c
z c
ay
ay
T
A/m
−∇V − ∂ ∂ At = −(z − ct)ax − xaz + xaz = (ct − z)ax V/m
Then D = 0 E = 0 (ct
2
− z)ax C/m
c) Show that these results satisfy Maxwell’s equations if J and ρ v are zero: i. · D = · 0 (ct z)ax = 0
∇ ∇ − ii. ∇ · B = ∇ · (t − z/c)ay = 0
iii.
∇ × H = −
∂ Hy 1 ax = ax = ∂ z µ0 c
which we require to equal ∂ D/∂ t: ∂ D = 0 cax = ∂ t
iv.
0
µ0
ax
∇ × E = ∂ ∂ Ezx ay = −ay which we require to equal
−∂ B/∂ t: ∂ B = a y ∂ t
So all four Maxwell equations are satisfied. 178
0
µ0
ax
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 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:
∇ × E = −µ ∂ ∂ Ht ∇ × H = ∂ ∂ Et ∇ · E = 0 ∇ · µH = 0
(1) (2) (3) (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
