1. UNIVERSITI MALAYSIA PERLIS
EKT 241/4: ELECTROMAGNETIC THEORY
UNIVERSITI MALAYSIA PERLIS
CHAPTER 4 – MAGNETOSTATICS
PREPARED BY: NORDIANA MOHAMAD SAAID

2. Chapter Outline Maxwell’s Equations Magnetic Forces and Torques
The total electromagnetic force, known as Lorentz force
Biot- Savart’s law
Gauss’s law for magnetism
Ampere’s law for magnetism
Magnetic Field and Flux
Vector magnetic potential
Properties of 3 different types of material
Boundary conditions between two different media
Self inductance and mutual inductance
Magnetic energy

3. Maxwell’s equations Maxwell’s equations: Where;
E = electric field intensity
D = electric flux density
ρv = electric charge density per unit volume
H = magnetic field intensity
B = magnetic flux density

4. Maxwell’s equations For static case, ∂/∂t = 0.
Maxwell’s equations is reduced to:
Electrostatics Magnetostatics

5. Magnetic Force
B = Magnetic Flux Density B B q q I q B B

6. Magnetic Torque on a Current- Carrying Loop
Applied force vector F and distance vector d are used to generate a torque T
T = d× F (N·m)
Rotation direction is governed by right-hand rule.

7. Magnetic Forces and Torques
The electric force Fe per unit charge acting on a test charge placed at a point in space with electric field E.
When a charged particle moving with a velocity u passing through that point in space, the magnetic force Fm is exerted on that charged particle.
where B = magnetic flux density (Cm/s or Tesla T)

8. Magnetic Forces and Torques
If a charged particle is in the presence of both an electric field E and magnetic field B, the total electromagnetic force acting on it is:

9. Magnetic Force on a Current- Carrying Conductor
For closed circuit of contour C carrying I , total magnetic force Fm is:
In a uniform magnetic field, Fm is zero for a closed circuit.

10. Magnetic Force on a Current- Carrying Conductor
On a line segment, Fm is proportional to the vector between the end points.

11. Example 1
The semicircular conductor shown carries a current I. The closed circuit is exposed to a uniform magnetic field Determine (a) the magnetic force F1 on the straight section of the wire and (b) the force F2 on the curved section.

12. Solution to Example 1 a)

14. The Biot–Savart’s Law where:
The Biot–Savart law is used to compute the magnetic field generated by a steady current, i.e. a continual flow of charges, for example through a wire
Biot–Savart’s law states that:
where:
dH = differential magnetic field dl = differential length

15. The Biot–Savart’s Law
To determine the total H:

16. The Biot–Savart’s Law
Biot–Savart’s law may be expressed in terms of distributed current sources.

17. Example 2
Determine the magnetic field at the apex O of the pie-shaped loop as shown. Ignore the contributions to the field due to the current in the small arcs near O.

18. ? = -dl For segment AC, dl is in φ direction,
O  A
C  O
A C ?
For segment AC, dl is in φ direction,
Using Biot- Savart’s law:

19. QUIZ!! z
ε1=2ε0
ε2=8ε0 Ө2 E2 xy E1 Ө1
Find E1 if E2 = 2x -3y +3z with s = 3.54 x 10-11(C/m2) And find Ө1 and Ө2

20. Gauss’s Law for Magnetism
Gauss’s law for magnetism states that:
Magnetic field lines always form continuous closed loops.

22. Ampere’s law for magnetism
Ampere’s law states that:
true for an infinite length of conductor H
C, +aø dl
true for an infinite length of conductor
I, +az r

23. Magnetic Field of an infinite length of conductor
From then re-arrange the equation in terms of Hφ:
Hence, the magnetic field vector, H:

24. Example 3
A toroidal coil with N turns carrying a current I , determine the magnetic field H in each of the following three regions: r < a, a < r < b,and r > b, all in the azimuthal plane of the toroid.

25. Solution to Example 3
H = 0 for r < a as no current is flowing through the surface of the contour
H = 0 for r > b, as equal number of current coils cross the surface in both directions.
For a < r < b, we apply Ampere’s law:
Hence, H = NI/(2πr) .

26. Magnetic Flux
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:

27. Example 4
An infinite length coaxial cable with inner conductor radius of 0.01m and outer conductor radius of 0.05m carrying a current of 2.5A exists along the z axis in the + az direction.
Find the flux passing through the region between two conductors with height of 2 m in free space.

28. Solution to Example 4 inner conductor radius = r1 0.01m
outer conductor radius = r2 0.05m
current of 2.5A (in the +az direction)
Flux radius = 2m
Iaz=2.5A z aø
Flux,z xy r1 r2

29. Solution to Example 4
where dS is in the aø direction.
So,
Therefore,

30. Vector Magnetic Potential
For any vector of vector magnetic potential A:
We are able to derive:
Vector Poisson’s equation is given as:
where

31. Magnetic Properties of Materials
Diamagnetic
Paramagnetic
Ferromagnetic
The behavior of magnetic dipole moments & magnetic susceptibility, of its atoms with an external magnetic field is used as a basis for classifying magnetic materials.

32. Magnetic Properties of Materials
Magnetization in a material is associated with atomic current loops generated by two principal mechanisms:
Orbital motions of the electrons around the nucleus, i.e orbital magnetic moment, mo
Electron spin about its own axis, i.e spin magnetic moment, ms

33. Magnetic Permeability
Magnetization vector M is defined as
where = magnetic susceptibility (dimensionless)
Magnetic permeability is defined as:
and to define the magnetic properties in term of relative permeability is defined as:

34. Magnetic Materials - Diamagnetic
metals have a very weak and negative susceptibility ( ) to magnetic field
slightly repelled by a magnetic field and the material does not retain the magnetic properties when the external field is removed
Most elements in the periodic table, including copper, silver, and gold, are diamagnetic.

35. Magnetic Materials - Paramagnetic
Paramagnetic materials have a small and positive susceptibilities to magnetic fields.
slightly attracted by a magnetic field and the material does not retain the magnetic properties when the external field is removed.
Paramagnetic materials include magnesium, molybdenum, lithium, and tantalum.

36. Magnetic Materials – Diamagnetic, Paramagnetic
However, the absolute susceptibilities value of both materials is in the order Thus, can be ignored. Hence, we have Magnetic permeability:
Diamagnetic and paramagnetic materials include dielectric materials and most metals.

37. Magnetic Materials – Ferromagnetic Materials
Ferromagnetic materials is characterized by magnetized domain - a microscopic region within which the magnetic moments of all its atoms are aligned parallel to each other.
Hysteresis – “to lag behind”. It determines how easy/hard for a magnetic material to be magnetized and demagnetized.

38. Process of Magnetic Hysteresis
material is magnetized
and can serve as
permanent magnet! B
material is demagnetize

39. Magnetic Hysteresis of Ferromagnetic Materials
Comparison of hysteresis curves for (a) a hard and (b) a soft ferromagnetic material is shown.
Hard magnetic material- cannot be
easily magnetized & demagnetized by an external magnetic field.
Soft magnetic material – easily
magnetized & demagnetized.

40. Magnetic Hysteresis of Ferromagnetic Materials
Properties of magnetic materials as follows:

41. Magnetic boundary conditions
Boundary between medium 1 with μ1 and medium 2 with μ2

42. Magnetic boundary conditions
Boundary condition related to normal components of the electric field;
By analogy, application of Gauss’s law for magnetism, we get first boundary condition:
Since ,
For linear, isotropic media, the first boundary condition which is related to H;

43. z xy
By applying Ampere’s law

44. Magnetic boundary conditions
The result is generalized to a vector form:
Where
However, surface currents can exist only on the surfaces of perfect conductors and perfect superconductors (infinite conductivities).
Hence, at the interface between media with finite conductivities, Js=0. Thus:

45. Example
xy (plane)

46. Solution: H1t = H2t thus, H2t = 6ax + 2ay Hn1 = 3az, but, Hn2 = ??
6000μ0(3az) = 3000 μ0(Hn2)
Hn2 = 6az
thus, H2 =6ax + 2ay + 6az
μr1 = 6000 ; μr2 = 3000 ;

47. Inductance
An inductor is the magnetic analogue of an electrical capacitor.
Capacitor can store electric energy in the electric field present in the medium between its conducting surfaces.
Inductor can store magnetic energy in the volume comprising the inductors.

48. calculate magnetic field in solenoid Calculate Magnetic Energy
Mutual inductance:
produced by magnetic coupling between two different conducting structures.
Calculate Magnetic Energy
Self inductance:
is the ratio of the magnetic flux linkage, Λ to the current I flowing through the structure.
INDUCTANCE
store magnetic
energy

49. Inductance
Example of an inductor is a solenoid - a coil consisting of multiple turns of wire wound in a helical geometry around a cylindrical core.

50. Magnetic Field in a Solenoid
For one cross section of solenoid,
When l >a, θ1≈−90° and θ2≈90°,
Where, N=nl
=total number of turns over the length l

51. (The ratio of the magnetic flux to the current)
Self Inductance
The self-inductance of a circuit is used to describe the reaction of the circuit to a changing current in the circuit,
(The ratio of the magnetic flux to the current)

52. Self Inductance
Self-inductance of any conducting structure is the ratio of the magnetic flux linkage, Λ to the current I flowing through the structure.
Magnetic flux linkage, Λ is the total magnetic flux linking a given conducting structure.

53. Self Inductance Magnetic flux, linking a surface S is given by:
In a solenoid with uniform magnetic field, the flux linking a single loop is:

54. Self Inductance – magnetic flux in solenoid

55. Self Inductance Magnetic flux, linking a surface S is given by:
In a solenoid with uniform magnetic field, the flux linking a single loop is:

56. Self Inductance
For a solenoid:
For two conductor configuration:

57. Self Inductance for a solenoid
Thus,

58. Mutual Inductance
Mutual inductance – produced by magnetic coupling between two different conducting structures.

59. Mutual Inductance
Magnetic field B1 generated by current I1 results in a flux Φ12 through loop 2:
If loop 2 consists of N2 turns all coupled by B1 in exactly the same way, the total magnetic flux linkage through loop 2 due to B1 is:

60. Mutual Inductance
Hence, the mutual inductance:

61. Magnetic Energy
Consider an inductor with an inductance L connected to a current source.
The current I flowing through the inductor is increased from zero to a final value I.
The energy expended in building up the current in the inductor:
i.e the magnetic energy stored in the inductor

62. Magnetic Energy Magnetic energy density (for solenoid):
i.e magnetic energy per unit volume
Magnetic energy in magnetic field:

63. Example: Magnetic Energy in a Coaxial Cable
Derive an expression for the magnetic energy stored in a coaxial cable of length l and inner and outer a and b. The insulation material has magnetic permeability μ.