BOUNDARY VALUE PROBLEMS IN ELECTROSTATICS
AND MAGNETOSTATICS
1 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
1. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
2. The potential inside the shell is constant and can be set to zero.
3. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
4. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
2 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
1. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
2. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
3. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
4. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
3 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
1. The potential can be obtained using the method of separation of variables in spherical
coordinates.
2. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
3. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
4. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
4 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
1. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
2. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
3. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
4. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
5 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
1. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
2. The magnetic field is obtained by integrating the current density using Ampère’s law.
3. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
4. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
6 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
1. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
2. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
3. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
4. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
7 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
1. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
2. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
3. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
4. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
8 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
1. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
2. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
3. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
4. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
9 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
5. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
6. The potential inside the shell is constant and can be set to zero.
7. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
8. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
10 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
9. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
10. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
11. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
12. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
11 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
13. The potential can be obtained using the method of separation of variables in spherical
coordinates.
14. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
15. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
16. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
12 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
17. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
18. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
19. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
20. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
13 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
21. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
22. The magnetic field is obtained by integrating the current density using Ampère’s law.
23. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
24. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
14 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
25. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
26. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
27. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
28. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
15 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
29. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
30. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
31. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
32. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
16 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
33. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
34. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
35. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
36. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
17 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
37. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
38. The potential inside the shell is constant and can be set to zero.
39. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
40. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
18 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
41. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
42. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
43. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
44. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
19 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
45. The potential can be obtained using the method of separation of variables in spherical
coordinates.
46. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
47. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
48. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
20 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
49. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
50. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
51. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
52. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
21 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
53. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
54. The magnetic field is obtained by integrating the current density using Ampère’s law.
55. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
56. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
22 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
57. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
58. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
59. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
60. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
23 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
61. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
62. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
63. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
64. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
24 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
65. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
66. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
67. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
68. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
25 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
69. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
70. The potential inside the shell is constant and can be set to zero.
71. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
72. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
26 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
73. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
74. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
75. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
76. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
27 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
77. The potential can be obtained using the method of separation of variables in spherical
coordinates.
78. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
79. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
80. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
28 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
81. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
82. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
83. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
84. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
29 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
85. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
86. The magnetic field is obtained by integrating the current density using Ampère’s law.
87. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
88. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
30 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
89. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
90. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
91. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
92. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
31 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
93. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
94. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
95. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
96. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
32 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
97. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
98. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
99. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
100. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
33 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
101. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
102. The potential inside the shell is constant and can be set to zero.
103. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
104. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
34 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
105. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
106. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
107. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
108. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
35 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
109. The potential can be obtained using the method of separation of variables in spherical
coordinates.
110. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
111. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
112. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
36 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
113. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
114. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
115. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
116. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
37 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
117. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
118. The magnetic field is obtained by integrating the current density using Ampère’s law.
119. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
120. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
38 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
121. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
122. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
123. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
124. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
39 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
125. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
126. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
127. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
128. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
40 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
129. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
130. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
131. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
132. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
41 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
133. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
134. The potential inside the shell is constant and can be set to zero.
135. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
136. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
42 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
137. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
138. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
139. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
140. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
43 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
141. The potential can be obtained using the method of separation of variables in spherical
coordinates.
142. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
143. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
144. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
44 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
145. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
146. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
147. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
148. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
45 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
149. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
150. The magnetic field is obtained by integrating the current density using Ampère’s law.
151. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
152. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
46 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
153. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
154. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
155. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
156. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
47 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
157. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
158. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
159. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
160. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
48 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
161. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
162. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
163. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
164. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
49 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
165. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
166. The potential inside the shell is constant and can be set to zero.
167. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
168. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
50 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
169. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
170. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
171. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
172. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
51 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
173. The potential can be obtained using the method of separation of variables in spherical
coordinates.
174. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
175. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
176. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
52 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
177. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
178. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
179. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
180. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
53 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
181. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
182. The magnetic field is obtained by integrating the current density using Ampère’s law.
183. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
184. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
54 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
185. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
186. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
187. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
188. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
55 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
189. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
190. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
191. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
192. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
56 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
193. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
194. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
195. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
196. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
57 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
197. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
198. The potential inside the shell is constant and can be set to zero.
199. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
200. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
58 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
201. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
202. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
203. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
204. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
59 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
205. The potential can be obtained using the method of separation of variables in spherical
coordinates.
206. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
207. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
208. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
60 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
209. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
210. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
211. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
212. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
61 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
213. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
214. The magnetic field is obtained by integrating the current density using Ampère’s law.
215. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
216. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
62 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
217. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
218. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
219. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
220. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
63 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
221. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
222. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
223. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
224. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
64 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
225. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
226. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
227. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
228. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
65 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
229. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
230. The potential inside the shell is constant and can be set to zero.
231. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
232. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
66 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
233. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
234. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
235. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
236. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
67 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
237. The potential can be obtained using the method of separation of variables in spherical
coordinates.
238. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
239. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
240. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
68 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
241. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
242. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
243. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
244. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
69 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
245. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
246. The magnetic field is obtained by integrating the current density using Ampère’s law.
247. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
248. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
70 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
249. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
250. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
251. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
252. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
71 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
253. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
254. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
255. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
256. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
72 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
257. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
258. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
259. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
260. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
73 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
261. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
262. The potential inside the shell is constant and can be set to zero.
263. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
264. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
74 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
265. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
266. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
267. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
268. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
75 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
269. The potential can be obtained using the method of separation of variables in spherical
coordinates.
270. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
271. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
272. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
76 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
273. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
274. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
275. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
276. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
77 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
277. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
278. The magnetic field is obtained by integrating the current density using Ampère’s law.
279. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
280. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
78 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
281. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
282. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
283. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
284. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
79 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
285. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
286. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
287. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
288. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
80 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
289. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
290. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
291. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
292. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
81 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
293. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
294. The potential inside the shell is constant and can be set to zero.
295. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
296. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
82 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
297. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
298. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
299. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
300. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
83 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
301. The potential can be obtained using the method of separation of variables in spherical
coordinates.
302. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
303. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
304. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
84 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
305. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
306. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
307. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
308. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
85 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
309. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
310. The magnetic field is obtained by integrating the current density using Ampère’s law.
311. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
312. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
86 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
313. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
314. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
315. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
316. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
87 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
317. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
318. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
319. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
320. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
88 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
321. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
322. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
323. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
324. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
89 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
325. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
326. The potential inside the shell is constant and can be set to zero.
327. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
328. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
90 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
329. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
330. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
331. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
332. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
91 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
333. The potential can be obtained using the method of separation of variables in spherical
coordinates.
334. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
335. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
336. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
92 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
337. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
338. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
339. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
340. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
93 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
341. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
342. The magnetic field is obtained by integrating the current density using Ampère’s law.
343. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
344. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
94 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
345. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
346. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
347. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
348. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
95 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
349. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
350. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
351. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
352. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
96 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
353. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
354. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
355. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
356. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
97 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
357. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
358. The potential inside the shell is constant and can be set to zero.
359. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
360. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
98 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
361. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
362. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
363. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
364. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
99 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
365. The potential can be obtained using the method of separation of variables in spherical
coordinates.
366. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
367. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
368. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
100 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
369. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
370. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
371. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
372. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
101 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
373. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
374. The magnetic field is obtained by integrating the current density using Ampère’s law.
375. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
376. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
102 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
377. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
378. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
379. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
380. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
103 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
381. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
382. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
383. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
384. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
104 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
385. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
386. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
387. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
388. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
105 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
389. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
390. The potential inside the shell is constant and can be set to zero.
391. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
392. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
106 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
393. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
394. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
395. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
396. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
107 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
397. The potential can be obtained using the method of separation of variables in spherical
coordinates.
398. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
399. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
400. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
108 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
401. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
402. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
403. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
404. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
109 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
405. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
406. The magnetic field is obtained by integrating the current density using Ampère’s law.
407. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
408. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
110 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
409. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
410. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
411. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
412. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
111 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
413. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
414. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
415. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
416. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
112 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
417. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
418. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
419. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
420. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
113 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
421. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
422. The potential inside the shell is constant and can be set to zero.
423. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
424. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
114 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
425. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
426. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
427. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
428. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
115 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
429. The potential can be obtained using the method of separation of variables in spherical
coordinates.
430. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
431. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
432. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
116 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
433. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
434. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
435. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
436. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
117 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
437. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
438. The magnetic field is obtained by integrating the current density using Ampère’s law.
439. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
440. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
118 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
441. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
442. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
443. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
444. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
119 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
445. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
446. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
447. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
448. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
120 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
449. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
450. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
451. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
452. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
121 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
453. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
454. The potential inside the shell is constant and can be set to zero.
455. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
456. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
122 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
457. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
458. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
459. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
460. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
123 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
461. The potential can be obtained using the method of separation of variables in spherical
coordinates.
462. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
463. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
464. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
124 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
465. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
466. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
467. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
468. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
125 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
469. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
470. The magnetic field is obtained by integrating the current density using Ampère’s law.
471. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
472. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
126 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
473. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
474. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
475. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
476. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
127 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
477. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
478. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
479. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
480. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
128 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
481. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
482. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
483. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
484. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
129 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
485. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
486. The potential inside the shell is constant and can be set to zero.
487. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
488. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
130 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
489. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
490. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
491. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
492. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
131 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
493. The potential can be obtained using the method of separation of variables in spherical
coordinates.
494. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
495. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
496. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
132 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
497. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
498. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
499. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
500. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
133 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
501. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
502. The magnetic field is obtained by integrating the current density using Ampère’s law.
503. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
504. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
134 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
505. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
506. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
507. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
508. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
135 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
509. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
510. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
511. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
512. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
136 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
513. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
514. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
515. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
516. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
137 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
517. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
518. The potential inside the shell is constant and can be set to zero.
519. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
520. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
138 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
521. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
522. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
523. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
524. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
139 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
525. The potential can be obtained using the method of separation of variables in spherical
coordinates.
526. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
527. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
528. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
140 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
529. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
530. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
531. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
532. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
141 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
533. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
534. The magnetic field is obtained by integrating the current density using Ampère’s law.
535. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
536. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
142 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
537. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
538. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
539. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
540. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
143 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
541. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
542. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
543. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
544. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
144 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
545. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
546. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
547. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
548. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
145 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
549. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
550. The potential inside the shell is constant and can be set to zero.
551. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
552. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
146 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
553. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
554. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
555. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
556. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
147 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
557. The potential can be obtained using the method of separation of variables in spherical
coordinates.
558. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
559. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
560. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
148 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
561. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
562. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
563. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
564. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
149 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
565. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
566. The magnetic field is obtained by integrating the current density using Ampère’s law.
567. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
568. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
150 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
569. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
570. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
571. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
572. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
151 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
573. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
574. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
575. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
576. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
152 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
577. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
578. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
579. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
580. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
153 PROBLEM 1
A spherical shell of radius 𝑎 carries a uniform surface charge density 𝜎. Find the electric
potential inside and outside the shell.
Solution:
581. Inside the shell, the electric field is zero due to the uniform charge distribution (Gauss’s
law).
582. The potential inside the shell is constant and can be set to zero.
583. Outside the shell, the potential is that of a point charge 𝑄 = 4𝜋𝑎2𝜎 located at the origin.
584. The potential outside the shell is given by 𝑉(𝑟)=1
4𝜋𝜖0
𝑄
𝑟 for 𝑟 > 𝑎.
154 PROBLEM 2
A long, straight wire of radius 𝑎 carries a uniform volume charge density 𝜌. Find the electric
potential inside and outside the wire.
Solution:
585. Inside the wire, the electric field is radial and given by 𝐸(𝑟)=𝜌
2𝜖0𝑟 for 𝑟 < 𝑎.
586. The potential inside the wire is obtained by integrating the electric field: 𝑉(𝑟)=𝜌
4𝜖0𝑟2+
𝐶1 for 𝑟 < 𝑎.
587. Outside the wire, the potential is that of a line charge with linear charge density 𝜆 =
𝜋𝑎2𝜌.
588. The potential outside the wire is given by 𝑉(𝑟)=𝜆
2𝜋𝜖0ln(𝑟
𝑎) + 𝐶2 for 𝑟 > 𝑎.
155 PROBLEM 3
A spherical shell of radius 𝑎 carries a surface charge density 𝜎 = 𝑘cos𝜃, where 𝑘 is a constant.
Find the electric potential inside and outside the shell.
Solution:
589. The potential can be obtained using the method of separation of variables in spherical
coordinates.
590. Inside the shell, the potential is given by 𝑉1(𝑟,𝜃)=∑𝐴𝑙
∞
𝑙=0 𝑟𝑙𝑃𝑙(cos𝜃).
591. Outside the shell, the potential is given by 𝑉2(𝑟,𝜃)=∑𝐵𝑙
∞
𝑙=0 𝑟−(𝑙+1)𝑃𝑙(cos𝜃).
592. The coefficients 𝐴𝑙 and 𝐵𝑙 are determined by applying the boundary conditions at 𝑟 = 𝑎.
156 PROBLEM 4
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
593. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
594. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)cos(𝑛𝜙).
595. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)cos(𝑛𝜙).
596. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
157 PROBLEM 5
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
597. Inside the wire, the magnetic field is given by 𝐻(𝑟)=𝑘𝑟2
4𝜇0 for 𝑟 < 𝑎.
598. The magnetic field is obtained by integrating the current density using Ampère’s law.
599. Outside the wire, the magnetic field is that of a line current with linear current density
𝐼 = 𝜋𝑎4𝑘/2.
600. The magnetic field outside the wire is given by 𝐻(𝑟)=𝐼
2𝜋𝑟 for 𝑟 > 𝑎.
158 PROBLEM 6
A long, straight wire of radius 𝑎 carries a surface current density 𝐾𝑧= 𝑘sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
601. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
602. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝐼𝑛(𝑘𝑟)sin((2𝑛 + 1)𝜙).
603. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝐾𝑛(𝑘𝑟)sin((2𝑛 +
1)𝜙).
604. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
159 PROBLEM 7
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2sin𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
605. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
606. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1sin((2𝑛 + 1)𝜙).
607. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)sin((2𝑛 +
1)𝜙).
608. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
160 PROBLEM 8
A long, straight wire of radius 𝑎 carries a volume current density 𝐽𝑧= 𝑘𝑟2cos𝜙, where 𝑘 is a
constant. Find the magnetic field inside and outside the wire.
Solution:
609. The magnetic field can be obtained using the method of separation of variables in
cylindrical coordinates.
610. Inside the wire, the magnetic field is given by 𝐻1(𝑟,𝜙)=∑𝐴𝑛
∞
𝑛=0 𝑟2𝑛+1cos((2𝑛 + 1)𝜙).
611. Outside the wire, the magnetic field is given by 𝐻2(𝑟,𝜙)=∑𝐵𝑛
∞
𝑛=0 𝑟−(2𝑛+2)cos((2𝑛 +
1)𝜙).
612. The coefficients 𝐴𝑛 and 𝐵𝑛 are determined by applying the boundary conditions at 𝑟 =
𝑎.
