MAXWELL’S EQUATIONS AND ELECTROMAGNETIC
WAVE PROPAGATION PROBLEMS
1 PROBLEMS AND SOLUTIONS
1.1 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
1. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
2. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
3. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄(𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.2 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
1. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
2. Choose a circular path of radius 𝑟 centered on the wire
3. Due to symmetry, 𝐵
is tangential and uniform along the path
4. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
5. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.3 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
1. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
2. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
3. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
4. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.4 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
1. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
2. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
3. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
4. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
5. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.5 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
1. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
2. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
3. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
4. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
5. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
6. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
7. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
8. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.6 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
1. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
2. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
3. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
4. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
5. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.7 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
1. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
2. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
3. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
4. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
5. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
6. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
7. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.8 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
1. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
2. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
3. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
4. The cutoff frequency occurs when 𝛽 = 0
5. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
6. Therefore, 𝑘𝑐=𝜋
𝑎
7. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.9 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
8. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
9. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
10. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.10 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
11. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
12. Choose a circular path of radius 𝑟 centered on the wire
13. Due to symmetry, 𝐵
is tangential and uniform along the path
14. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
15. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.11 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
16. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
17. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
18. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
19. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.12 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
20. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
21. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
22. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
23. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
24. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.13 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
25. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
26. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
27. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
28. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
29. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
30. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
31. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
32. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.14 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
33. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
34. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
35. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
36. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
37. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.15 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
38. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
39. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
40. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
41. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
42. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
43. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
44. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.16 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
45. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
46. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
47. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
48. The cutoff frequency occurs when 𝛽 = 0
49. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
50. Therefore, 𝑘𝑐=𝜋
𝑎
51. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.17 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
52. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
53. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
54. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.18 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
55. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
56. Choose a circular path of radius 𝑟 centered on the wire
57. Due to symmetry, 𝐵
is tangential and uniform along the path
58. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
59. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.19 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
60. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
61. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
62. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
63. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.20 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
64. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
65. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
66. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
67. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
68. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.21 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
69. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
70. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
71. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
72. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
73. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
74. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
75. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
76. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.22 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
77. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
78. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
79. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
80. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
81. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.23 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
82. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
83. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
84. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
85. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
86. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
87. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
88. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.24 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
89. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
90. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
91. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
92. The cutoff frequency occurs when 𝛽 = 0
93. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
94. Therefore, 𝑘𝑐=𝜋
𝑎
95. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.25 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
96. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
97. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
98. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.26 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
99. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
100. Choose a circular path of radius 𝑟 centered on the wire
101. Due to symmetry, 𝐵
is tangential and uniform along the path
102. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
103. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.27 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
104. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
105. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
106. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
107. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.28 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
108. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
109. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
110. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
111. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
112. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.29 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
113. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
114. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
115. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
116. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
117. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
118. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
119. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
120. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.30 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
121. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
122. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
123. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
124. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
125. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.31 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
126. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
127. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
128. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
129. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
130. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
131. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
132. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.32 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
133. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
134. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
135. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
136. The cutoff frequency occurs when 𝛽 = 0
137. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
138. Therefore, 𝑘𝑐=𝜋
𝑎
139. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.33 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
140. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
141. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
142. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.34 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
143. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
144. Choose a circular path of radius 𝑟 centered on the wire
145. Due to symmetry, 𝐵
is tangential and uniform along the path
146. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
147. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.35 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
148. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
149. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
150. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
151. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.36 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
152. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
153. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
154. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
155. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
156. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.37 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
157. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
158. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
159. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
160. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
161. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
162. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
163. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
164. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.38 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
165. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
166. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
167. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
168. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
169. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.39 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
170. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
171. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
172. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
173. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
174. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
175. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
176. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.40 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
177. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
178. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
179. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
180. The cutoff frequency occurs when 𝛽 = 0
181. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
182. Therefore, 𝑘𝑐=𝜋
𝑎
183. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.41 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
184. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
185. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
186. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.42 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
187. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
188. Choose a circular path of radius 𝑟 centered on the wire
189. Due to symmetry, 𝐵
is tangential and uniform along the path
190. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
191. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.43 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
192. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
193. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
194. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
195. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.44 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
196. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
197. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
198. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
199. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
200. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.45 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
201. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
202. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
203. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
204. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
205. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
206. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
207. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
208. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.46 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
209. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
210. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
211. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
212. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
213. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.47 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
214. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
215. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
216. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
217. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
218. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
219. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
220. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.48 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
221. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
222. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
223. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
224. The cutoff frequency occurs when 𝛽 = 0
225. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
226. Therefore, 𝑘𝑐=𝜋
𝑎
227. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.49 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
228. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
229. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
230. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.50 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
231. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
232. Choose a circular path of radius 𝑟 centered on the wire
233. Due to symmetry, 𝐵
is tangential and uniform along the path
234. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
235. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.51 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
236. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
237. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
238. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
239. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.52 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
240. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
241. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
242. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
243. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
244. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.53 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
245. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
246. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
247. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
248. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
249. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
250. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
251. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
252. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.54 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
253. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
254. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
255. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
256. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
257. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.55 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
258. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
259. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
260. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
261. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
262. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
263. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
264. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.56 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
265. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
266. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
267. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
268. The cutoff frequency occurs when 𝛽 = 0
269. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
270. Therefore, 𝑘𝑐=𝜋
𝑎
271. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.57 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
272. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
273. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
274. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.58 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
275. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
276. Choose a circular path of radius 𝑟 centered on the wire
277. Due to symmetry, 𝐵
is tangential and uniform along the path
278. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
279. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.59 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
280. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
281. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
282. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
283. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.60 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
284. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
285. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
286. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
287. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
288. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.61 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
289. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
290. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
291. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
292. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
293. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
294. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
295. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
296. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.62 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
297. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
298. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
299. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
300. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
301. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.63 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
302. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
303. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
304. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
305. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
306. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
307. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
308. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.64 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
309. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
310. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
311. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
312. The cutoff frequency occurs when 𝛽 = 0
313. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
314. Therefore, 𝑘𝑐=𝜋
𝑎
315. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.65 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
316. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
317. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
318. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.66 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
319. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
320. Choose a circular path of radius 𝑟 centered on the wire
321. Due to symmetry, 𝐵
is tangential and uniform along the path
322. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
323. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.67 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
324. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
325. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
326. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
327. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.68 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
328. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
329. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
330. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
331. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
332. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.69 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
333. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
334. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
335. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
336. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
337. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
338. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
339. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
340. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.70 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
341. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
342. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
343. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
344. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
345. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.71 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
346. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
347. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
348. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
349. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
350. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
351. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
352. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.72 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
353. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
354. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
355. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
356. The cutoff frequency occurs when 𝛽 = 0
357. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
358. Therefore, 𝑘𝑐=𝜋
𝑎
359. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.73 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
360. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
361. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
362. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.74 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
363. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
364. Choose a circular path of radius 𝑟 centered on the wire
365. Due to symmetry, 𝐵
is tangential and uniform along the path
366. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
367. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.75 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
368. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
369. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
370. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
371. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.76 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
372. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
373. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
374. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
375. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
376. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.77 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
377. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
378. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
379. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
380. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
381. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
382. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
383. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
384. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.78 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
385. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
386. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
387. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
388. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
389. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.79 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
390. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
391. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
392. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
393. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
394. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
395. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
396. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.80 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
397. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
398. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
399. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
400. The cutoff frequency occurs when 𝛽 = 0
401. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
402. Therefore, 𝑘𝑐=𝜋
𝑎
403. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.81 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
404. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
405. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
406. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.82 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
407. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
408. Choose a circular path of radius 𝑟 centered on the wire
409. Due to symmetry, 𝐵
is tangential and uniform along the path
410. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
411. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.83 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
412. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
413. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
414. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
415. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.84 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
416. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
417. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
418. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
419. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
420. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.85 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
421. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
422. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
423. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
424. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
425. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
426. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
427. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
428. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.86 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
429. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
430. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
431. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
432. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
433. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.87 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
434. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
435. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
436. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
437. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
438. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
439. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
440. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.88 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
441. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
442. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
443. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
444. The cutoff frequency occurs when 𝛽 = 0
445. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
446. Therefore, 𝑘𝑐=𝜋
𝑎
447. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.89 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
448. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
449. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
450. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.90 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
451. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
452. Choose a circular path of radius 𝑟 centered on the wire
453. Due to symmetry, 𝐵
is tangential and uniform along the path
454. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
455. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.91 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
456. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
457. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
458. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
459. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.92 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
460. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
461. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
462. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
463. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
464. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.93 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
465. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
466. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
467. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
468. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
469. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
470. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
471. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
472. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.94 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
473. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
474. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
475. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
476. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
477. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.95 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
478. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
479. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
480. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
481. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
482. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
483. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
484. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.96 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
485. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
486. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
487. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
488. The cutoff frequency occurs when 𝛽 = 0
489. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
490. Therefore, 𝑘𝑐=𝜋
𝑎
491. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.97 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
492. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
493. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
494. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.98 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
495. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
496. Choose a circular path of radius 𝑟 centered on the wire
497. Due to symmetry, 𝐵
is tangential and uniform along the path
498. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
499. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.99 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
500. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
501. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
502. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
503. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.100 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
504. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
505. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
506. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
507. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
508. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.101 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
509. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
510. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
511. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
512. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
513. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
514. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
515. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
516. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.102 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
517. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
518. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
519. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
520. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
521. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.103 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
522. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
523. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
524. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
525. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
526. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
527. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
528. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.104 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
529. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
530. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
531. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
532. The cutoff frequency occurs when 𝛽 = 0
533. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
534. Therefore, 𝑘𝑐=𝜋
𝑎
535. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.105 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
536. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
537. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
538. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.106 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
539. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
540. Choose a circular path of radius 𝑟 centered on the wire
541. Due to symmetry, 𝐵
is tangential and uniform along the path
542. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
543. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.107 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
544. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
545. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
546. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
547. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.108 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
548. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
549. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
550. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
551. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
552. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.109 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
553. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
554. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
555. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
556. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
557. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
558. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
559. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
560. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.110 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
561. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
562. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
563. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
564. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
565. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.111 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
566. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
567. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
568. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
569. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
570. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
571. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
572. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.112 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
573. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
574. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
575. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
576. The cutoff frequency occurs when 𝛽 = 0
577. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
578. Therefore, 𝑘𝑐=𝜋
𝑎
579. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.113 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
580. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
581. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
582. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.114 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
583. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
584. Choose a circular path of radius 𝑟 centered on the wire
585. Due to symmetry, 𝐵
is tangential and uniform along the path
586. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
587. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.115 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
588. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
589. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
590. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
591. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.116 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
592. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
593. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
594. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
595. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
596. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.117 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
597. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
598. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
599. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
600. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
601. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
602. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
603. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
604. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.118 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
605. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
606. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
607. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
608. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
609. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.119 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
610. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
611. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
612. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
613. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
614. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
615. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
616. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.120 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
617. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
618. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
619. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
620. The cutoff frequency occurs when 𝛽 = 0
621. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
622. Therefore, 𝑘𝑐=𝜋
𝑎
623. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.121 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
624. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
625. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
626. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.122 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
627. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
628. Choose a circular path of radius 𝑟 centered on the wire
629. Due to symmetry, 𝐵
is tangential and uniform along the path
630. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
631. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.123 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
632. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
633. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
634. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
635. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.124 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
636. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
637. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
638. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
639. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
640. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.125 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
641. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
642. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
643. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
644. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
645. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
646. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
647. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
648. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.126 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
649. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
650. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
651. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
652. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
653. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.127 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
654. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
655. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
656. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
657. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
658. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
659. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
660. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.128 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
661. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
662. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
663. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
664. The cutoff frequency occurs when 𝛽 = 0
665. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
666. Therefore, 𝑘𝑐=𝜋
𝑎
667. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.129 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
668. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
669. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
670. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.130 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
671. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
672. Choose a circular path of radius 𝑟 centered on the wire
673. Due to symmetry, 𝐵
is tangential and uniform along the path
674. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
675. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.131 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
676. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
677. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
678. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
679. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.132 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
680. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
681. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
682. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
683. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
684. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.133 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
685. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
686. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
687. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
688. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
689. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
690. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
691. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
692. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.134 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
693. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
694. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
695. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
696. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
697. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.135 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
698. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
699. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
700. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
701. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
702. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
703. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
704. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.136 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
705. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
706. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
707. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
708. The cutoff frequency occurs when 𝛽 = 0
709. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
710. Therefore, 𝑘𝑐=𝜋
𝑎
711. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.137 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
712. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
713. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
714. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.138 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
715. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
716. Choose a circular path of radius 𝑟 centered on the wire
717. Due to symmetry, 𝐵
is tangential and uniform along the path
718. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
719. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.139 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
720. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
721. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
722. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
723. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.140 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
724. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
725. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
726. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
727. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
728. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.141 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
729. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
730. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
731. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
732. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
733. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
734. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
735. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
736. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.142 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
737. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
738. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
739. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
740. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
741. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.143 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
742. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
743. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
744. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
745. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
746. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
747. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
748. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.144 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
749. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
750. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
751. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
752. The cutoff frequency occurs when 𝛽 = 0
753. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
754. Therefore, 𝑘𝑐=𝜋
𝑎
755. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.145 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
756. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
757. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
758. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.146 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
759. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
760. Choose a circular path of radius 𝑟 centered on the wire
761. Due to symmetry, 𝐵
is tangential and uniform along the path
762. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
763. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.147 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
764. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
765. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
766. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
767. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.148 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
768. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
769. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
770. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
771. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
772. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.149 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
773. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
774. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
775. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
776. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
777. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
778. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
779. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
780. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.150 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
781. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
782. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
783. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
784. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
785. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.151 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
786. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
787. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
788. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
789. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
790. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
791. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
792. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.152 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
793. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
794. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
795. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
796. The cutoff frequency occurs when 𝛽 = 0
797. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
798. Therefore, 𝑘𝑐=𝜋
𝑎
799. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.153 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
800. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
801. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
802. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.154 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
803. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
804. Choose a circular path of radius 𝑟 centered on the wire
805. Due to symmetry, 𝐵
is tangential and uniform along the path
806. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
807. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.155 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
808. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
809. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
810. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
811. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.156 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
812. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
813. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
814. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
815. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
816. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.157 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
817. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
818. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
819. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
820. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
821. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
822. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
823. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
824. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.158 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
825. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
826. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
827. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
828. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
829. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.159 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
830. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
831. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
832. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
833. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
834. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
835. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
836. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.160 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
837. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
838. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
839. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
840. The cutoff frequency occurs when 𝛽 = 0
841. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
842. Therefore, 𝑘𝑐=𝜋
𝑎
843. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.161 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
844. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
845. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
846. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.162 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
847. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
848. Choose a circular path of radius 𝑟 centered on the wire
849. Due to symmetry, 𝐵
is tangential and uniform along the path
850. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
851. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.163 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
852. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
853. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
854. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
855. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.164 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
856. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
857. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
858. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
859. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
860. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.165 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
861. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
862. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
863. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
864. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
865. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
866. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
867. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
868. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.166 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
869. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
870. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
871. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
872. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
873. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.167 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
874. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
875. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
876. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
877. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
878. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
879. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
880. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.168 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
881. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
882. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
883. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
884. The cutoff frequency occurs when 𝛽 = 0
885. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
886. Therefore, 𝑘𝑐=𝜋
𝑎
887. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.169 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
888. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
889. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
890. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.170 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
891. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
892. Choose a circular path of radius 𝑟 centered on the wire
893. Due to symmetry, 𝐵
is tangential and uniform along the path
894. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
895. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.171 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
896. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
897. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
898. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
899. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.172 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
900. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
901. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
902. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
903. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
904. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.173 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
905. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
906. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
907. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
908. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
909. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
910. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
911. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
912. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.174 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
913. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
914. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
915. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
916. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
917. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.175 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
918. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
919. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
920. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
921. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
922. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
923. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
924. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.176 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
925. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
926. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
927. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
928. The cutoff frequency occurs when 𝛽 = 0
929. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
930. Therefore, 𝑘𝑐=𝜋
𝑎
931. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.177 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
932. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
933. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
934. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.178 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
935. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
936. Choose a circular path of radius 𝑟 centered on the wire
937. Due to symmetry, 𝐵
is tangential and uniform along the path
938. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
939. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.179 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
940. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
941. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
942. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
943. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.180 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
944. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
945. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
946. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
947. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
948. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.181 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
949. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
950. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
951. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
952. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
953. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
954. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
955. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
956. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.182 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
957. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
958. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
959. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
960. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
961. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.183 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
962. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
963. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
964. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
965. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
966. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
967. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
968. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.184 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
969. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
970. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
971. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+ (𝑛𝜋
𝑏)2+ 𝛽2
972. The cutoff frequency occurs when 𝛽 = 0
973. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
974. Therefore, 𝑘𝑐=𝜋
𝑎
975. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.185 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
976. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
977. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
978. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.186 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
979. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
980. Choose a circular path of radius 𝑟 centered on the wire
981. Due to symmetry, 𝐵
is tangential and uniform along the path
982. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
983. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.187 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
984. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
985. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
986. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
987. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.188 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
988. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
989. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
990. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
991. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
992. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.189 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
993. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
994. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
995. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
996. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
997. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
998. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
999. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
1000. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.190 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
1001. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
1002. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
1003. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
1004. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
1005. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.191 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
1006. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
1007. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
1008. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
1009. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
1010. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
1011. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
1012. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.192 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
1013. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
1014. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
1015. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+
(𝑛𝜋
𝑏)2+ 𝛽2
1016. The cutoff frequency occurs when 𝛽 = 0
1017. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
1018. Therefore, 𝑘𝑐=𝜋
𝑎
1019. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.193 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
1020. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
1021. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
1022. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.194 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
1023. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
1024. Choose a circular path of radius 𝑟 centered on the wire
1025. Due to symmetry, 𝐵
is tangential and uniform along the path
1026. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
1027. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.195 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
1028. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
1029. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
1030. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1031. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.196 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
1032. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
1033. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
1034. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
1035. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
1036. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.197 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
1037. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
1038. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
1039. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
1040. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
1041. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
1042. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
1043. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
1044. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.198 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
1045. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
1046. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
1047. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
1048. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
1049. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.199 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
1050. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
1051. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
1052. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
1053. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
1054. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
1055. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
1056. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.200 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
1057. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
1058. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
1059. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+
(𝑛𝜋
𝑏)2+ 𝛽2
1060. The cutoff frequency occurs when 𝛽 = 0
1061. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
1062. Therefore, 𝑘𝑐=𝜋
𝑎
1063. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.201 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
1064. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
1065. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
1066. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.202 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
1067. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
1068. Choose a circular path of radius 𝑟 centered on the wire
1069. Due to symmetry, 𝐵
is tangential and uniform along the path
1070. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
1071. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.203 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
1072. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
1073. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
1074. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1075. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.204 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
1076. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
1077. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
1078. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
1079. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
1080. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.205 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
1081. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
1082. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
1083. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
1084. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
1085. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
1086. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
1087. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
1088. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.206 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
1089. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
1090. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
1091. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
1092. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
1093. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.207 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
1094. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
1095. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
1096. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
1097. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
1098. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
1099. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
1100. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.208 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
1101. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
1102. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
1103. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+
(𝑛𝜋
𝑏)2+ 𝛽2
1104. The cutoff frequency occurs when 𝛽 = 0
1105. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
1106. Therefore, 𝑘𝑐=𝜋
𝑎
1107. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.209 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
1108. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
1109. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
1110. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.210 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
1111. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
1112. Choose a circular path of radius 𝑟 centered on the wire
1113. Due to symmetry, 𝐵
is tangential and uniform along the path
1114. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
1115. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.211 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
1116. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
1117. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
1118. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1119. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.212 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
1120. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
1121. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
1122. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
1123. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
1124. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.213 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
1125. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
1126. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
1127. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
1128. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
1129. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
1130. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
1131. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
1132. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.214 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
1133. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
1134. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
1135. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
1136. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
1137. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.215 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
1138. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
1139. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
1140. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
1141. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
1142. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
1143. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
1144. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.216 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
1145. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
1146. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
1147. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+
(𝑛𝜋
𝑏)2+ 𝛽2
1148. The cutoff frequency occurs when 𝛽 = 0
1149. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
1150. Therefore, 𝑘𝑐=𝜋
𝑎
1151. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.217 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
1152. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
1153. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
1154. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.218 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
1155. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
1156. Choose a circular path of radius 𝑟 centered on the wire
1157. Due to symmetry, 𝐵
is tangential and uniform along the path
1158. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
1159. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.219 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
1160. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
1161. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
1162. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1163. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.220 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
1164. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
1165. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
1166. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
1167. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
1168. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.221 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
1169. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
1170. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
1171. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
1172. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
1173. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
1174. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
1175. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
1176. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.222 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
1177. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
1178. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
1179. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
1180. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
1181. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.223 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
1182. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
1183. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
1184. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
1185. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
1186. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
1187. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
1188. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.224 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
1189. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
1190. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
1191. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+
(𝑛𝜋
𝑏)2+ 𝛽2
1192. The cutoff frequency occurs when 𝛽 = 0
1193. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
1194. Therefore, 𝑘𝑐=𝜋
𝑎
1195. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.225 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
1196. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
1197. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
1198. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.226 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
1199. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
1200. Choose a circular path of radius 𝑟 centered on the wire
1201. Due to symmetry, 𝐵
is tangential and uniform along the path
1202. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
1203. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.227 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
1204. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
1205. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
1206. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1207. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.228 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
1208. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
1209. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
1210. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
1211. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
1212. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.229 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
1213. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
1214. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
1215. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
1216. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
1217. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
1218. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
1219. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
1220. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.230 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
1221. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
1222. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
1223. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
1224. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
1225. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.231 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
1226. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
1227. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
1228. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
1229. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
1230. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
1231. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
1232. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.232 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
1233. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
1234. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
1235. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+
(𝑛𝜋
𝑏)2+ 𝛽2
1236. The cutoff frequency occurs when 𝛽 = 0
1237. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
1238. Therefore, 𝑘𝑐=𝜋
𝑎
1239. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.233 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
1240. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
1241. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
1242. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.234 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
1243. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
1244. Choose a circular path of radius 𝑟 centered on the wire
1245. Due to symmetry, 𝐵
is tangential and uniform along the path
1246. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
1247. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.235 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
1248. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
1249. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
1250. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1251. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.236 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
1252. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
1253. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
1254. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
1255. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
1256. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.237 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
1257. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
1258. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
1259. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
1260. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
1261. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
1262. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
1263. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
1264. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.238 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
1265. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
1266. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
1267. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
1268. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
1269. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.239 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
1270. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
1271. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
1272. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
1273. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
1274. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
1275. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
1276. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.240 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
1277. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
1278. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
1279. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+
(𝑛𝜋
𝑏)2+ 𝛽2
1280. The cutoff frequency occurs when 𝛽 = 0
1281. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
1282. Therefore, 𝑘𝑐=𝜋
𝑎
1283. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.241 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
1284. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
1285. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
1286. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.242 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
1287. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
1288. Choose a circular path of radius 𝑟 centered on the wire
1289. Due to symmetry, 𝐵
is tangential and uniform along the path
1290. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
1291. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.243 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
1292. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
1293. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
1294. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1295. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.244 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
1296. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
1297. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
1298. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
1299. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
1300. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.245 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
1301. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
1302. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
1303. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
1304. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
1305. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
1306. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
1307. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
1308. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.246 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
1309. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
1310. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
1311. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
1312. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
1313. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.247 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
1314. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
1315. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
1316. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
1317. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
1318. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
1319. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
1320. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.248 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
1321. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
1322. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
1323. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+
(𝑛𝜋
𝑏)2+ 𝛽2
1324. The cutoff frequency occurs when 𝛽 = 0
1325. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
1326. Therefore, 𝑘𝑐=𝜋
𝑎
1327. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
1.249 PROBLEM 1: GAUSS’S LAW FOR ELECTRICITY
Calculate the electric field at a distance 𝑟 from a uniformly charged sphere of radius 𝑅 and total
charge 𝑄.
Solution:
1328. We use Gauss’s law: ∮𝐸
⋅ 𝑑𝐴
=𝑄𝑒𝑛𝑐
𝜖0
1329. For 𝑟 > 𝑅:
a. The enclosed charge is the total charge 𝑄
b. Due to spherical symmetry, 𝐸
is radial and uniform over the Gaussian surface
c. ∮ 𝐸
⋅ 𝑑𝐴
= 𝐸(4𝜋𝑟2)=𝑄
𝜖0
d. Solving for 𝐸: 𝐸 = 𝑄
4𝜋𝜖0𝑟2
1330. For 𝑟 < 𝑅:
a. The enclosed charge is 𝑄𝑒𝑛𝑐 = 𝑄 (𝑟
𝑅)3
b. 𝐸(4𝜋𝑟2)=𝑄(𝑟
𝑅)3
𝜖0
c. Solving for 𝐸: 𝐸 = 𝑄𝑟
4𝜋𝜖0𝑅3
1.250 PROBLEM 2: AMPÈRE’S LAW
Find the magnetic field at a distance 𝑟 from a long, straight wire carrying a current 𝐼.
Solution:
1331. We use Ampère’s law: ∮𝐵
⋅ 𝑑𝑙
= 𝜇0𝐼𝑒𝑛𝑐
1332. Choose a circular path of radius 𝑟 centered on the wire
1333. Due to symmetry, 𝐵
is tangential and uniform along the path
1334. ∮ 𝐵
⋅ 𝑑𝑙
= 𝐵(2𝜋𝑟)= 𝜇0𝐼
1335. Solving for 𝐵: 𝐵 = 𝜇0𝐼
2𝜋𝑟
1.251 PROBLEM 3: FARADAY’S LAW
A circular loop of radius 𝑟 is in a uniform magnetic field 𝐵
= 𝐵0cos(𝜔𝑡)𝑧. Find the induced EMF
in the loop.
Solution:
1336. We use Faraday’s law: ℰ = −𝑑𝛷𝐵
𝑑𝑡
1337. The magnetic flux through the loop is 𝛷𝐵= 𝐵
⋅ 𝐴
= 𝐵0cos(𝜔𝑡)(𝜋𝑟2)
1338. Taking the time derivative: 𝑑𝛷𝐵
𝑑𝑡 = −𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1339. Therefore, the induced EMF is ℰ = 𝐵0𝜔sin(𝜔𝑡)(𝜋𝑟2)
1.252 PROBLEM 4: DISPLACEMENT CURRENT
A parallel-plate capacitor with circular plates of radius 𝑅 is being charged. The current in the
wire is 𝐼 = 𝐼0sin(𝜔𝑡). Find the displacement current between the plates.
Solution:
1340. The displacement current is given by 𝐼𝑑= 𝜖0𝑑𝛷𝐸
𝑑𝑡
1341. The electric flux 𝛷𝐸 is related to the charge 𝑄 on the plates: 𝛷𝐸=𝑄
𝜖0
1342. The current in the wire is 𝐼 = 𝑑𝑄
𝑑𝑡 = 𝐼0sin(𝜔𝑡)
1343. Therefore, 𝑑𝛷𝐸
𝑑𝑡 =1
𝜖0
𝑑𝑄
𝑑𝑡 =𝐼0
𝜖0sin(𝜔𝑡)
1344. The displacement current is 𝐼𝑑= 𝐼0sin(𝜔𝑡), equal to the current in the wire
1.253 PROBLEM 5: WAVE EQUATION
Derive the electromagnetic wave equation for the electric field in vacuum using Maxwell’s
equations.
Solution:
1345. Start with Faraday’s law: ∇ × 𝐸
= −∂𝐵
∂𝑡
1346. Take the curl of both sides: ∇ × (∇ × 𝐸
)= − ∂
∂𝑡 (∇ × 𝐵
)
1347. Use the vector identity ∇ × (∇ × 𝐸
)= ∇(∇ ⋅ 𝐸
)− ∇2𝐸
1348. In vacuum, ∇ ⋅ 𝐸
= 0, so ∇ × (∇ × 𝐸
)= −∇2𝐸
1349. Use Ampère’s law: ∇ × 𝐵
= 𝜇0𝜖0∂𝐸
∂𝑡
1350. Substituting, we get: −∇2𝐸
= −𝜇0𝜖0∂2𝐸
∂𝑡2
1351. Rearranging: ∇2𝐸
= 𝜇0𝜖0∂2𝐸
∂𝑡2
1352. This is the wave equation for 𝐸
with wave speed 𝑐 = 1
√𝜇0𝜖0
1.254 PROBLEM 6: POYNTING VECTOR
Calculate the Poynting vector for a plane electromagnetic wave with 𝐸
= 𝐸0cos(𝑘𝑧 −𝜔𝑡)𝑥.
Solution:
1353. The Poynting vector is given by 𝑆
=1
𝜇0𝐸
× 𝐵
1354. For a plane wave, 𝐵
=1
𝑐𝑧 × 𝐸
1355. 𝐵
=𝐸0
𝑐cos(𝑘𝑧 −𝜔𝑡)𝑦
1356. 𝑆
=1
𝜇0𝐸
× 𝐵
=𝐸0
2
𝑐𝜇0cos2(𝑘𝑧 −𝜔𝑡)𝑧
1357. The time-averaged Poynting vector is ⟨𝑆
⟩ = 𝐸0
2
2𝑐𝜇0𝑧
1.255 PROBLEM 7: REFLECTION AND TRANSMISSION
An electromagnetic wave in medium 1 (𝑛1) is incident on medium 2 (𝑛2) at normal incidence.
Find the reflection and transmission coefficients.
Solution:
1358. Define the reflection coefficient 𝑟 = 𝐸𝑟
𝐸𝑖 and transmission coefficient 𝑡 = 𝐸𝑡
𝐸𝑖
1359. At the boundary, the tangential components of 𝐸
and 𝐵
must be continuous
1360. For 𝐸
: 𝐸𝑖+ 𝐸𝑟= 𝐸𝑡
1361. For 𝐵
: 1
𝑣1(𝐸𝑖− 𝐸𝑟)=1
𝑣2𝐸𝑡, where 𝑣1 and 𝑣2 are wave speeds
1362. Divide the second equation by the first: 1−𝑟
1+𝑟 =𝑛1
𝑛2
1363. Solve for 𝑟: 𝑟 = 𝑛1−𝑛2
𝑛1+𝑛2
1364. From continuity of 𝐸
: 𝑡 = 1 + 𝑟 = 2𝑛1
𝑛1+𝑛2
1.256 PROBLEM 8: WAVEGUIDE MODES
Find the cutoff frequency for the TE10 mode in a rectangular waveguide of width 𝑎 and height
𝑏.
Solution:
1365. The wave equation in the waveguide is ∇2𝐸
+ 𝑘2𝐸
= 0
1366. For TE modes, 𝐸𝑧= 0 and 𝐻𝑧= 𝐴sin(𝑚𝜋𝑥
𝑎)sin(𝑛𝜋𝑦
𝑏)𝑒−𝑗𝛽𝑧
1367. The wavenumber 𝑘 is related to the propagation constant 𝛽 by 𝑘2= (𝑚𝜋
𝑎)2+
(𝑛𝜋
𝑏)2+ 𝛽2
1368. The cutoff frequency occurs when 𝛽 = 0
1369. For TE10 mode, 𝑚 = 1 and 𝑛 = 0
1370. Therefore, 𝑘𝑐=𝜋
𝑎
1371. The cutoff frequency is 𝑓𝑐=𝑘𝑐
2𝜋√𝜇𝜖 =𝑐
2𝑎
