1. Chapter 3 – Magnetostatic Field
Ali Othman
Universiti Teknologi MARA Pulau Pinang
HP No:
BKBA 4.7( )/BP 7.15 ( )

2. Objectives
The objective is to evaluate & understand the properties of magnetostatic field due to several scenarios of current distributions:
line current
Surface current
Volume current

3. Introduction
Electrostatic: Static charge will produce electrostatic field, E & D electric flux density.
Electric field characterized by E (electric field intensity) and D(electric flux density).
Magnetostatic: Moving charge (constant velocity) produce magnetostatic field, H & B magnetic.
magnetic field characterized by H (magnetic field intensity) and B (magnetic flux density). (i.e for linear material space B = μH) μ= permeability of material.

4. Introduction 2 major laws govern magnetostatic field:
Biot-Savart’s law – general law of magnetostatic.
Ampere’s law – special case of Biot-Savart especially applied in problem involving symmetrical current distribution.
Like Coulomb’s law, Biot-Savart’s law is the general law of magnetostatic.
Ampere’s law is a special case of Biot-Savart’s law, applied in problems involving symmetrical current distribution.

5. Magnetostatic field dH
Magnetic field dH at P due to current element Idl
field comes out of plane due to the cross product
point of interest
differential section of conductor
contributes to field at

6. Direction of dH Determining the direction of dH using
the right-hand rule or
the righthanded-screw rule.
Thumb pointing the direction of current flow
Fingers point is Direction of Magnetic Field

7. Current Representation
It is customary to represent the direction of magnetic field intensity , H or current I by a small circle in a dot or cross sign.
It depends whether H or I is out of, or into the page.
Conventional representation of H (or I)
(a) out of the page and (b) into the page.

8. Current (Source) Distributions
If in electrostatic, we have different charge distribution.
But in magnetostatic, we have different current distribution, which are line current, surface current and volume current.
 Distribution
Line
current
Surface
Volume
J = volume current density (A/m2)
K = surface current density (A/m)

9. Charge and Current Distributions
 Electric Field (random charge distributions)
symmetric charge distributions
Magnetic Field (random current distributions)
Apply Gauss’s law
Line charge D
Apply Coulomb’s law + +
= Random H
Apply Faraday’s law
Apply Biot-Savart law

10. The Biot-Savart Law
Jean-Baptiste Biot & Felix Savart developed the quantitative description for the magnetic field.
field comes out of plane due to the cross product
point of interest
differential section of conductor
contributes to field at

11. The Biot-Savart Law
The magnetic field intensity dH produced at a point P, by the differential current element Idl,
Is proportional to the product Idl and the sine of the angle between the element and the line joining P the element.
and is inversely proportional to the square of the distance between P and the element.
How to use Biot–Savart’s law to find H or B from a current distribution.

12. The Biot-Savart Law
If we define K = surface current density (A/m2) and J = volume current density (A/m3)
 Distribution
Line
current
Surface current
Volume current
Magnetic field
strength

13. Biot-Savart’s Law - Steps
Define an origin.
Write vector Rs from origin to Source (source in the current).
Write vector Rp from origin to H Field Point.
Find vector from source to point R or (Rsp) Also find Rsp/| Rsp |
Define the direction of the current and find
Apply Biot-Savart
Integrate.

14. The Biot-Savart Law – Example 1
The magnetic field intensity dH produced at a point P, by the differential current element Idl,

15. The Biot-Savart Law – Example 1
The magnetic field intensity dH produced at a point P (any place along the y-axis), by the differential current element Idl,

16. The Biot-Savart Law – Example 1
Solve the cross ( X ) product

17. The Biot-Savart Law – Example 2
The magnetic field intensity dH produced at a point P, by the differential current element I=KdS, where K = surface current density (A/m2)

18. The Biot-Savart Law – Example 2
Solve the cross ( X ) product

19. Exercise 1
A circular wire carrying 20 A current is placed on xy-plane and centered at origin. If the radius of the circular wire is 5 meter, determine H at (0,0,0)

20. Exercise 2
A rectangular loop carrying 20 A of current is place on z = 0 plane as shown in figure below. Using Biot-Savart’s Law, determine magnetic field intensity, H at (4,2,0).

21. Exercise 2 - Solution
A rectangular loop carrying 20 A of current is place on z = 0 plane as shown in figure below. Using Biot-Savart’s Law, determine magnetic field intensity, H at (4,2,0).
Let H = H1 + H2 + H3 + H4
where Hn is the conrtibution by side n

22. The Biot-Savart Law – Example 3
The magnetic field intensity dH produced at a point P, by the differential current element I=Jdv, where J = volume current density (A/m3)

23. The Biot-Savart Law – Example 3
Solve the cross ( X ) product

24. The Biot-Savart Law – Example 3
Substitute and integrate

25. Major Concepts : Gradient () = Directional Derivative
Gradient will tell us the slope of the line or the variation.

26. Major Concepts : Gradient () = Directional Derivative
Coulomb’s law & Gauss’s law are used to determine E when charge distribution is known, using
In practical electrostatic problems, only electrostatic conditions (charge & potential) at some boundaries are known.
From these boundary conditions, we want to find E & V throughout region.
Tackled using
i) Poisson’s Equation
ii) Laplace’s Equation
Both above are derived from Gauss’s law ( for a linear material medium)

27. Major Concepts : E and V
Gradient will tell us the slope of the line or the variation.
Electric field represent the changes of potential.
Potential different exactly the same.

28. Major Concepts : Laplacian Rate of Change
Telling how fast the potential are changing.

29. Major Concepts Telling how fast the potential are changing.
Poisson’s Equation is used whenever filled material between the paltes is semiconducting or conducting materials. Having volume charge density.
Laplace ‘s Equation (Charge Free Region v = 0) is used whenever the filled material between the plate is pure dielectric materials.

30. Example 5.1 pg 173
Current-carrying components in high-voltage power equipment can be cooled to carry away the heat caused by ohmic losses. A means of pumping is based on the force transmitted to the cooling fluid by charges in an electric field. Electrohydrodynamic (EHD) pumping is modeled in Figure. The region between the electrodes contains a uniform charge o, which is generated at the left electrode and collected at the right electrode. Calculate the pressure of pump if o = 25mC/m3 and Vo = 22kV

31. Example 5.1 pg 173 Since v  0, apply Poisson’s equation.
The boundary conditions V(z=0)=Vo and V(z=d)=0 show that V depends only on z (there is no  or  dependence)

32. Example 5.1 pg 173 Integrating once gives Integrating again yeilds
Where A and B are integration constants to be determined by applying the boundary conditions.

33. Example 5.1 pg 173
Where z=0, V=Vo
Where z=d, V=0 or

34. Example 5.1 pg 173
The electric field is given by

35. Exercise
For the parallel plate shown in figure, a charge density between the plates is given by v = o sin (2z/d) C/m3. Determine the potential.

36. Exercise Since v  0, apply Poisson’s equation.
The boundary conditions V(z=0)=Vo and V(z=d)=0 show that V depends only on z (there is no  or  dependence)

37. Exercise Integrating once gives Integrating again yeilds
Where A and B are integration constants to be determined by applying the boundary conditions.

38. The Ampere’s Law
Similar to Gauss’s Law. Ampere stated that all magnetic field are generated by currents. The elementary magnetic dipoles are current loops.
Ampere’s law states that the line integral of the tangential components of H around a closed path is the same as the net current Ienc enclosed by the path.
The integral form of Ampere’s Circuit Law:
Ampere’s Circuit Law is used when we want to determine H when the current distribution is symmetrical.

39. The Ampere’s Law Apply Stoke’s theorem Since
Compare blue and green equations and we get third maxwell equation

40. The Ampere’s Law : What do these equations mean?
Compare blue and green equations and we get third maxwell equation

41. Steps to Ampere’s Law
Be sure your current is SYMMETRIC!! (H must be equal in magnitude everywhere on your Gaussian Contour) (if not, use Biot-Savart Law).
Choose a Gaussian contour (think of a belt) that has your field point (H or B) on it. The surface is inside the contour.
Do the integral on the left side (this is always the same for a given geometry).
Do the integral on the right side to find I.
Solve for H or B.

42. Ampere’s Law application
Some symmetrical current Distribution:
Infinite line current
Infinite sheet of current
Infinite long coaxial transmission line

43. Infinite line current
Be sure your current is SYMMETRIC!! (H must be equal in magnitude everywhere on your Gaussian Contour).
To determine H at P, allow a close path to pass through P (Amperian path). H is constant provided  is constant. (LHS)

44. Infinite line current How much current is enclosed. (RHS)
Ampere’s law states that the line integral of the tangential components of H around a closed path is the same as the net current Ienc enclosed by the path.

45. Infinite sheet of current
Consider an infinite current sheet in the z=0 plane with a uniform current density K = Kyay

46. Infinite sheet of current
Evaluating the Ampere’s law along the closed path gives:-

47. Infinite sheet of current
Applying Ampere’s law to the rectangular closed path (Amperian path): and compare with We obtain Regard the infinite sheet as comprising of filaments. The resultant dH has only an x-component. Also, H on one side of the sheet is the negative of that on the aother side:

48. Infinite sheet of current
Thus In general

49. Infinite long coaxial transmission line
Consider an infinitely long transmission consisting of two concentric cylinders having their axes along z-axis.
(Cross section of the transmission line; the +ve direction is out of page)

50. Infinite long coaxial transmission line
The inner conductor has a radius a carries current I , while the outer conductor has inner radius b and thickness t and carries return current –I
Since the current distribution is symmetrical,
apply Ampere’s Law for the Amperian path
for each of the four possible regions:
(L1) 0    a
(L2) a    b
(L3) b    b+t
(L4)   b+t

51. Infinite long coaxial transmission line
Apply Ampere’s law inside inner conductor (L1) 0    a
Since the current is uniformly distributed over the cross section.

52. Infinite long coaxial transmission line
Thus or

53. Infinite long coaxial transmission line
2. For region (L2) a    b or

54. Infinite long coaxial transmission line
3. For region (L3) b    b+t Where And J in this case is the current density (current per unit area) of the outer conductor and is along –az :

55. Infinite long coaxial transmission line
Thus Substituting into

56. Infinite long coaxial transmission line
4. For region (L4)   b+t or

57. Infinite long coaxial transmission line
Putting it all together gives H:
Since the current distribution is symmetrical,
apply Ampere’s Law for the Amperian path
for each of the four possible regions:
(L1) 0    a
(L2) a    b
(L3) b    b+t
(L4)   b+t

58. Infinite long coaxial transmission line
Plot of H against :

59. Example Find H at the center of a square current loop of side L
Choose a cartesian coordinate system such that the loop is located as shown in figure:
By symmetry, each half-side contributes the same amount to H at the center. For the half-side 0  x  L/2, y = -L/2, the Bit-Savart law gives for the field at the origin:

60. Example
Determine H for a solid cylindrical conductor of radius a, where the current I is uniformly distributed over the cross section.
Applying Ampere’s law to contour

61. Example
A current filament of 5.0 A in the ay direction is parallel to the y axis at x = 2m, z = -2m. Find H at the origin.

62. Example
A current filament of 5.0 A in the ay direction is parallel to the y axis at x = 2m, z = -2m. Find H at the origin. The expression for H due to a straight current filament applies where and use RHR

63. To be continued
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