1. CHAPTER 3
MAGNETOSTATICS

2. MAGNETOSTATICS 3.1 BIOT-SAVART’S LAW 3.2 AMPERE’S CIRCUITAL LAW
3.3 MAGNETIC FLUX DENSITY
3.4 MAGNETIC FORCES
3.5 BOUNDARY CONDITIONS

3. INTRODUCTION
Magnetism and electricity were considered distinct phenomena until 1820 when Hans Christian Oersted introduced an experiment that showed a compass needle deflecting when in proximity to current carrying wire.

4. INTRODUCTION (Cont’d)
He used compass to show that current produces magnetic fields that loop around the conductor. The field grows weaker as it moves away from the source of current.
A  represents current coming out of paper.
A represents current heading into the paper. 

5. INTRODUCTION (Cont’d)
The principle of magnetism is widely used in many applications:
Magnetic memory
Motors and generators
Microphones and speakers
Magnetically levitated high-speed vehicle.

6. INTRODUCTION (Cont’d)
Magnetic fields can be easily visualized by sprinkling iron filings on a piece of paper suspended over a bar magnet.

7. INTRODUCTION (Cont’d)
The field lines are in terms of the magnetic field intensity, H in units of amps per meter.
This is analogous to the volts per meter units for electric field intensity, E.
Magnetic field will be introduced in a manner paralleling our treatment to electric fields.

8. 3.1 BIOT-SAVART’S LAW
Jean Baptiste Biot and Felix Savart arrived a mathematical relation between the field and current.

9. BIOT-SAVART’S LAW (Cont’d)
To get the total field resulting from a current, sum the contributions from each segment by integrating:

10. BIOT-SAVART’S LAW (Cont’d)
Due to continuous current distributions:
Line current
Surface current
Volume current

11. BIOT-SAVART’S LAW (Cont’d)
In terms of distributed current sources, the Biot-Savart’s Law becomes:
Line current
Surface current
Volume current

12. DERIVATION Let’s apply
to determine the magnetic field, H everywhere due to straight current carrying filamentary conductor of a finite length AB .

13. DERIVATION (Cont’d)

14. DERIVATION (Cont’d)
We assume that the conductor is along the z-axis with its upper and lower ends respectively subtending angles and at point P where H is to be determined.
The field will be independent of z and φ and only dependant on ρ.

15. DERIVATION (Cont’d)
The term dL is simply and the vector from the source to the test point P is:
Where the magnitude is:
And the unit vector:

16. DERIVATION (Cont’d)
Combining these terms to have:

17. DERIVATION (Cont’d)
Cross product of :
This yields to:

18. DERIVATION (Cont’d) Trigonometry from figure, So,
Differentiate to get:

19. DERIVATION (Cont’d)
Remember!

20. DERIVATION (Cont’d)
Simplify the equation to become:

21. DERIVATION 1
Therefore,
This expression generally applicable for any straight filamentary conductor of finite length.

22. DERIVATION 2
As a special case, when the conductor is semifinite with respect to P,
The angle become:
So that,

23. DERIVATION 3
Another special case, when the conductor is infinite with respect to P,
The angle become:
So that,

24. HOW TO FIND UNIT VECTOR aφ ?
From previous example, the vector H is in direction of aφ, where it needs to be determine by simple approach:
Where,
unit vector along the line current
unit vector perpendicular from the line current to the field point

25. EXAMPLE 1
The conducting triangular loop carries of 10A. Find H at (0,0,5) due to side 1 of the loop.

26. SOLUTION TO EXAMPLE 1
Side 1 lies on the x-y plane and treated as a straight conductor.
Join the point of interest (0,0,5) to the beginning and end of the line current.

27. SOLUTION TO EXAMPLE 1 (Cont’d)
This will show how is applied for any straight, thin, current carrying conductor.
From figure, we know that
and from trigonometry and

28. SOLUTION TO EXAMPLE 1 (Cont’d)
To determine by simple approach:
and so that,

29. EXAMPLE 2
A ring of current with radius a lying in the x-y plane with a current I in the direction. Find an expression for the field at arbitrary point a height h on z axis.

30. SOLUTION TO EXAMPLE 2 Can we use ?
Solve for each term in the Biot-Savart’s Law

31. SOLUTION TO EXAMPLE 2 (Cont’d)
We could find:

32. SOLUTION TO EXAMPLE 2 (Cont’d)
It leads to:
The differential current element will give a field with:
from
from

33. SOLUTION TO EXAMPLE 2 (Cont’d)
However, consider the symmetry of the problem:
The radial components cancel but the components adds, so:

34. SOLUTION TO EXAMPLE 2 (Cont’d)
This can be easily solved to get:
At h=0 where at the center of the loop, this equation reduces to:

35. BIOT-SAVART’S LAW (Cont’d)
For many problems involving surface current densities and volume current densities, solving for the magnetic field using Biot-Savart’s Law can be quite cumbersome and require numerical integration.
There will be sufficient symmetry to be able to solve for the fields using Ampere’s Circuital Law.

36. 3.2 AMPERE’S CIRCUITAL LAW
In magnetostatic problems with sufficient symmetry, we can employ Ampere’s Circuital Law more easily that the law of Biot-Savart.
The law says that the integration of H around any closed path is equal to the net current enclosed by that path. i.e.

37. AMPERE’S CIRCUITAL LAW (Cont’d)
The line integral of H around the path is termed the circulation of H.
To solve for H in given symmetrical current distribution, it is important to make a careful selection of an Amperian Path (analogous to gaussian surface) that is everywhere either tangential or normal to H.
The direction of the circulation is chosen such
that the right hand rule is satisfied.

38. DERIVATION 4
Find the magnetic field intensity everywhere resulting from an infinite length line of current situated on the z-axis using Ampere’s Circuital Law.

39. DERIVATION 4 (Cont’d)
Select the best Amperian path, where here are two possible Amperian paths around an infinite length line of current.
Choose path b which has a constant value of Hφ around the circle specified by the radius ρ

40. DERIVATION 4 (Cont’d)
Using Ampere’s circuital law:
We could find:
So,

41. DERIVATION 4 (Cont’d) Solving for Hφ:
Where we find that the field resulting from an infinite length line of current is the expected result:
Same as applying Biot-Savart’s Law!

42. DERIVATION 5
Use Ampere’s Circuital Law to find the magnetic field intensity resulting from an infinite extent sheet of current with current sheet in the x-y plane.

43. DERIVATION 5 (Cont’d)
Rectangular amperian path of height Δh and width Δw. According to right hand rule, perform the circulation in order of a  b  c  d  a

44. DERIVATION 5 (Cont’d) We have:
From symmetry argument, there’s only Hy component exists. So, Hz will be zero and thus the expression reduces to:

45. DERIVATION 5 (Cont’d)
So, we have:

46. DERIVATION 5 (Cont’d) The current enclosed by the path,
This will give:
Or generally,

47. EXAMPLE 3
An infinite sheet of current with exists on the x-z plane at y = 0. Find H at P (3,2,5).

48. SOLUTION TO EXAMPLE 3 Use previous expression, that is:
is a normal vector from the sheet to the test point P (3,4,5), where:
and
So,

49. EXAMPLE 4
Consider the infinite length cylindrical conductor carrying a radially dependent current Find H everywhere.

50. SOLUTION TO EXAMPLE 4 What components of H will be present?
Finding the field at some point P, the field has both and components.

51. SOLUTION TO EXAMPLE 4 (Cont’d)
The field from the second line current results in a cancellation of the components

52. SOLUTION TO EXAMPLE 4 (Cont’d)
To calculate H everywhere, two amperian paths are required:
Path #1 is for
Path #2 is for

53. SOLUTION TO EXAMPLE 4 (Cont’d)
Evaluating the left side of Ampere’s law:
This is true for both amperian path.
The current enclosed for the path #1:

54. SOLUTION TO EXAMPLE 4 (Cont’d)
Solving to get Hφ:
for Or
The current enclosed for the path #2:
Solving to get Hφ:
for

55. EXAMPLE 5
Find H everywhere for coaxial cable as shown.

56. SOLUTION TO EXAMPLE 5
Even current distributions are assumed in the inner and outer conductor.
Consider four amperian paths.

57. SOLUTION TO EXAMPLE 5 (Cont’d)
It will be four amperian paths:
Therefore, the magnetic field intensity, H will be determined for each amperian paths.

58. SOLUTION TO EXAMPLE 5 (Cont’d)
As previous example, only Hφ component is present, and we have the left side of ampere’s circuital law:
For the path #1:

59. SOLUTION TO EXAMPLE 5 (Cont’d)
We need to find current density, J for inner conductor because the problem assumes an event current distribution (ρ<a is a solid volume where current distributed uniformly).
Where,

60. SOLUTION TO EXAMPLE 5 (Cont’d)
We therefore have:

61. SOLUTION TO EXAMPLE 5 (Cont’d)
Equating both sides to get:
for
For the path #2:
The current enclosed is just I, Therefore:
for

62. SOLUTION TO EXAMPLE 5 (Cont’d)
For the path #3:
For total current enclosed by path 3, we need to find the current density, J in the outer conductor because the problem assumes an event current distribution (a<ρ<b is a solid volume where current distributed uniformly) given by:

63. SOLUTION TO EXAMPLE 5 (Cont’d)
We therefore have (for AP#3):
But, the total current enclosed is:

64. SOLUTION TO EXAMPLE 5 (Cont’d)
So we can solve for path #3:
for
For the path #4, the total current is zero. So,
This shows the shielding ability by coaxial cable!!
for

65. SOLUTION TO EXAMPLE 5 (Cont’d)
Summarize the results to have:

66. AMPERE’S CIRCUITAL LAW (Cont’d)
Expression for curl by applying Ampere’s Circuital Law might be too lengthy to derive, but it can be described as:
The expression is also called the point form of Ampere’s Circuital Law, since it occurs at some particular point.

67. AMPERE’S CIRCUITAL LAW (Cont’d)
The Ampere’s Circuital Law can be rewritten in terms of a current density, as:
Use the point form of Ampere’s Circuital Law to replace J, yielding:
This is known as Stoke’s Theorem.

68. 3.3 MAGNETIC FLUX DENSITY
In electrostatics, it is convenient to think in terms of electric flux intensity and electric flux density. So too in magnetostatics, where magnetic flux density, B is related to magnetic field intensity by:
Where μ is the permeability with:

69. MAGNETIC FLUX DENSITY (Cont’d)
The amount of magnetic flux, φ in webers from magnetic field passing through a surface is found in a manner analogous to finding electric flux:

70. MAGNETIC FLUX DENSITY (Cont’d)
Fundamental features of magnetic fields:
The field lines form a closed loops. It’s different from electric field lines, where it starts on positive charge and terminates on negative charge

71. MAGNETIC FLUX DENSITY (Cont’d)
The magnet cannot be divided in two parts, but it results in two magnets. The magnetic pole cannot be isolated.

72. MAGNETIC FLUX DENSITY (Cont’d)
The net magnetic flux passing through a gaussian surface must be zero, to get Gauss’s Law for magnetic fields:
By applying divergence theorem, the point form of Gauss’s Law for static magnetic fields:

73. EXAMPLE 6 Try to sketch this!
Find the flux crossing the portion of the plane φ=π/4 defined by 0.01m < r < 0.05m and 0 < z < 2m in free space. A current filament of 2.5A is along the z axis in the az direction.
Try to sketch this!

74. SOLUTION TO EXAMPLE 6 The relation between B and H is:
To find flux crossing the portion, we need to use:
where dS is in the aφ direction.

75. SOLUTION TO EXAMPLE 6 (Cont’d)
Therefore,

76. 3.4 MAGNETIC FORCES
Upon application of a magnetic field, the wire is deflected in a direction normal to both the field and the direction of current.

77. MAGNETIC FORCES (Cont’d)
The force is actually acting on the individual charges moving in the conductor, given by:
By the definition of electric field intensity, the electric force Fe acting on a charge q within an electric field is:

78. MAGNETIC FORCES (Cont’d)
A total force on a charge is given by Lorentz force equation:
The force is related to acceleration by the equation from introductory physics,

79. MAGNETIC FORCES (Cont’d)
To find a force on a current element, consider a line conducting current in the presence of magnetic field with differential segment dQ of charge moving with velocity u:
But,

80. MAGNETIC FORCES (Cont’d)
So,
Since corresponds to the current I in the line,
We can find the force from a collection of current elements

81. MAGNETIC FORCES (Cont’d)
Consider a line of current in +az direction on the z axis. For current element a,
But, the field cannot exert magnetic force on the element producing it. From field of second element b, the cross product will be zero since IdL and aR in same direction.

82. EXAMPLE 7
If there is a field from a second line of current parallel to the first, what will be the total force?

83. SOLUTION TO EXAMPLE 7 Why?!?!
The force from the magnetic field of line 1 acting on a differential section of line 2 is:
Where,
By inspection from figure,
Why?!?!

84. SOLUTION TO EXAMPLE 7
Consider , then:

85. MAGNETIC FORCES (Cont’d)
Generally,
Ampere’s law of force between a pair of current-carrying circuits.
General case is applicable for two lines that are not parallel, or not straight.
It is easier to find magnetic field B1 by Biot-Savart’s law, then use to find F12 .

86. EXAMPLE 8 Try to sketch this!
The magnetic flux density in a region of free space is given by B = −3x ax + 5y ay − 2z az T. Find the total force on the rectangular loop shown which lies in the plane z = 0 and is bounded by x = 1, x = 3, y = 2, and y = 5, all dimensions in cm.
Try to sketch this!

87. SOLUTION TO EXAMPLE 8
The figure is as shown.

88. SOLUTION TO EXAMPLE 8 (Cont’d)
First, note that in the plane z = 0, the z component of the given field is zero, so will not contribute to the force. We use:
Which in our case becomes with,
and

89. SOLUTION TO EXAMPLE 8 (Cont’d)

90. SOLUTION TO EXAMPLE 8 (Cont’d)
Simplifying these becomes:

91. 3.5 BOUNDARY CONDITIONS
We could see how the fields behave at the boundary between a pair of magnetic materials which derived using Ampere’s Circuital Law and Gauss’s Law for magnetostatic fields:

92. BOUNDARY CONDITIONS (Cont’d)
Consider,

93. BOUNDARY CONDITIONS (Cont’d)
A pair of magnetic media separated by a sheet current density K. Choose a rectangular Amperian path of width Δw and height Δh centered at the interface. The current enclosed by the path is:

94. BOUNDARY CONDITIONS (Cont’d)
The sheet current is heading into the page and use the right hand rule to determine the direction of integration around the loop. So,

95. BOUNDARY CONDITIONS (Cont’d)
For first and second integral,

96. BOUNDARY CONDITIONS (Cont’d)
For third and fourth integral,

97. BOUNDARY CONDITIONS (Cont’d)
Combining the result, we get the first boundary condition for magnetostatic field,
In more general case,
Where a21 is unit vector normal from medium to medium 1

98. BOUNDARY CONDITIONS (Cont’d)
Second boundary condition can be determined by applying Gauss’s Law over a small pillbox shaped Gaussian surface,

99. BOUNDARY CONDITIONS (Cont’d)
The Gauss’s Law,
Where,
The pillbox is short enough, so the flux out of the side is negligible.

100. BOUNDARY CONDITIONS (Cont’d)
We have
Since ΔS can be chosen unequal to zero, it follows that:

101. EXAMPLE 9 The magnetic field intensity is given as:
In a medium with µr1=6000 that exist for z < 0. Find H2 in a medium with µr2=3000 for z>0.

102. SOLUTION TO EXAMPLE 9

103. BOUNDARY CONDITIONS (Cont’d)
Recall that, for a conductor-dielectric interface:
Generally, it is not exist for magnetostatic fields. If one of the media is superconductor, where the magnetic field rapidly attenuates away from the surface, such that:

104. BOUNDARY CONDITIONS (Cont’d)
If medium 2 is superconductor, the equations for magnetostatic fields become:
The second expression is logical since the magnetic field lines must form closed loops and cannot suddenly terminate even on a superconductor.

105. CHAPTER 3
END

106. Maglev (Magnetically Levitated Trains)
PRACTICAL APPLICATION
Loudspeakers
Maglev (Magnetically Levitated
Trains)

107. LOUDSPEAKERS
Paper or plastic cone affixed to a voice coil (electromagnet) suspended in a magnetic field.
AC Signals to the voice coil  moves back and forth, resulting vibration of the cone and producing sound waves of the same frequency as the AC signal

108. MAGLEV

109. MAGLEV (Cont’d)
Interaction between electromagnets in the train and the current carrying coils in the guide rail provide levitation.
By sending waves along the guide rail coils, the train magnet pushed/pulled in the direction of travel. The train is guided by magnet on the side of guide rail.
Computer algorithms maintain the separation distance.

110. SUMMARY (1)
For a differential current element I1dL1 at point 1, the magnetic field intensity H at point 2 is given by the law of Biot-Savart,
Where is a vector from the source element at point 1 to the location where the field is desired at point 2. By summing all the current elements, it can rewritten as:

111. SUMMARY (2)
The Biot-Savart law can be written in terms of surface and volume current densities:
Surface current
Volume current
The magnetic field intensity resulting from an infinite length line of current is:

112. SUMMARY (3) and from a current sheet of extent it is:
Where aN is a unit vector normal from the current sheet to the test point.
An easy way to solve the magnetic field intensity in problems with sufficient current distribution symmetry is to use Ampere’s Circuital Law, which says that the circulation of H is equal to the net current enclosed by the circulation path

113. SUMMARY (4)
The point or differential form of Ampere’s circuital Law is:
A closed line integral is related to surface integral by Stoke’s Theorem:
Magnetic flux density, B in Wb/m2 or T, is related to the magnetic field intensity by

114. SUMMARY (5) Material permeability µ can be written as:
and the free space permeability is:
The amount of magnetic flux Φ in webers through a surface is:
Since magnetic flux forms closed loops, we have Gauss’s Law for static magnetic fields:

115. SUMMARY (6)
The total force vector F acting on a charge q moving through magnetic and electric fields with velocity u is given by Lorentz Force equation:
The force F12 from a magnetic field B1 on a current carrying line I2 is:

116. SUMMARY (7)
The magnetic fields at the boundary between different materials are given by:
Where a21 is unit vector normal from medium to medium 1, and:

117. VERY IMPORTANT!
From electrostatics and magnetostatics, we can now present all four of Maxwell’s equation for static fields:
Integral Form
Differential Form