1. Magnetics

2. Magnetic Circuit Concepts:

4. Electric machines and electromechanical devices are made up of coupled electric and magnetic circuits. By a magnetic circuit, we mean a path for magnetic flux, just as an electric circuit provides a path for the flow of electric current. Sources of magnetic fluxes are electric currents and permanent magnets. In electric machines, current-carrying conductors interact with magnetic fields (themselves arising from electric currents in conductors or from permanent magnets), resulting in electromechanical energy conversion.
Consider a conductor of length l placed between the poles of a magnet. Let the conductor carry a current I and be at right angles to the magnetic flux lines as shown in Fig. 1-1.

5. It is found experimentally that the conductor experiences a force F, the direction of which is shown in Fig. 1-1 and the magnitude of which is given by:
F= BIl
Here B is the magnitude of the magnetic flux density B, whose direction is given by the flux lines. The SI unit of B is tesla (T). Notice that B could be defined as the force per unit current moment. This is a statement of Ampere’s law, the more general statement which is:
F= Il ×B

6. Where I is a vector of magnitude l in the direction of the current
Where I is a vector of magnitude l in the direction of the current. The force is at right angles to both the conductor and the magnetic field. Ampere’s law provides as it does for the development of force or torque, underlies the operation of electric motors.

7. The magnetic flux 𝜙 through a given surface is the flux of B through that surface:
𝜙= 𝑩∙𝒅𝑺 = 𝑩∙𝒏𝑑𝑆
· Magnetic Induction in a Wire

8. 𝜙=𝐵𝐴
𝐵= 𝜙 𝐴
(expressed in Wb/m2)

9. The mutual relationship between an electric current and a magnetic field is given by Ampere’s circuital law:

10. 𝑯∙𝑑𝒍 =𝐼
𝑯∙𝑑𝒍 =𝑁𝐼=ℱ
ℱ: (Magnetomotive force)
units of At (amp turns)
𝜙= ℱ ℛ
ℛ (Reluctance of the magnetic circuit)

11. Permeability and Saturation
In an isotropic material medium, H, which is determined by moving charges (currents) only and B which depends also on the properties of the medium are related by:
𝐵=𝜇H
For free space:
𝐵= 𝜇 o H
𝜇 o =4π× 10 −7

12. The core material of an electric machine is generally ferromagnetic, and the variation of B with H is nonlinear, as shown by the typical saturation curve of Fig. 1-5(a). It is clear that the slope of the curve depends upon the operating flux density, as classified in regions I, II, and III. This leads us to the concept of different types of permeability:

13. We rewrite the equation above as:
𝐵=𝜇H= 𝜇 𝑟 𝜇 o H
𝜇 𝑟 is the relative permeability
𝜇 𝑟 = 1 𝜇 𝑜 𝑑𝐵 𝑑𝐻
differential permeability

14. Laws governing Magnetic Circuits:
Electric Circuit
Mmagnetic Circuit
Ohm’s law, I = V/R
𝜙=ℱ/ℛ
Resistance, R = l/σA
Reluctance, ℛ = l/𝜇A
Current, I
Flux, 𝜙
Voltage, V
mmf, ℱ
Conductivity, σ
Permeability, 𝜇
Conductance, G
Permeance, P

15. INDUCTANCE CALCULATIONS
Inductance is defined as flux linkage per unit current;
𝐿≡ 𝜆 𝑖 = 𝑁𝜙 𝑖
The unit of inductance is the henry (H).

16. 1.1 Find the magnetic field intensity due to an infinitely long, straight conductor carrying a current I amperes, at a point r meters away from the conductor.
𝑯∙𝑑𝒍 =2𝜋𝑟 𝐻 𝜙 or 𝐻 𝜙 = 𝐼 2𝜋𝑟
A/m
From the geometry of the problem, the radial and longitudinal components of H are zero.

17. 1. 2 The conductor of Problem 1
1.2 The conductor of Problem 1.1 carries 100 A current and is located in air. Determine the flux density at a point 0.05 m away from the conductor.
Since B= 𝜇 𝑜 𝐻 𝜙 , from Problem 1.1 we have
𝐵 𝜙 = 𝜇 𝑜 𝐻 𝜙 = 4𝜋× 10 −7 ×100 2𝜋×0.05 =0.04 mT

18. 1.3 A rectangular loop is placed in the field of the conductor of Problem 1.1 as shown in Fig What is the total flux linking the loop?

19. Assuming a medium of permeability g, we have from Problem 1.1
The flux d𝜙 in the elementary area dA = 𝑙 dr is given by
𝑑𝜙= 𝐵 𝜙 𝑑𝐴= 𝜇𝐼𝑙 2𝜋 𝑑𝑟 𝑟
and
𝜙= 𝜇𝐼𝑙 2𝜋 𝑟 1 𝑟 2 𝑑𝑟 𝑟 = 𝜇𝐼𝑙 2𝜋 𝑙𝑛 𝑟 2 𝑟 1 Wb

20. Problem 1.4 skipped.
There is an error in the problem with B = 7.07 and then .707.

21. 1. 5 For the magnetic circuit shown in Fig. 1-15, N=10 turns, 𝑙g=0
1.5 For the magnetic circuit shown in Fig. 1-15, N=10 turns, 𝑙g=0.1 mm, 𝑙m = 100 mm, stacking factor = 0.9; the core material is M-19. Calculate I required to establish a 1-T flux density in the airgap. Neglect fringing and leakage.

22. For the air gap:
𝐻 𝑔 = 𝐵 𝑔 𝜇 𝑜 = 1.0 4𝜋× 10 −7 =7.95× 10 5 𝐴𝑚
ℱ 𝑔 = 𝐻 𝑔 𝑙 𝑔 = 7.95× −4 =79.5𝐴𝑡
For the core:
𝐵 𝑚 = 𝐵 𝑔 stacking factor = =1.11 𝑇

23. and from Appendix C, Fig. C-1, at 1.11 T, we have
𝐻 𝑚 =130 𝐴 𝑚
and
ℱ= =13. 𝐴𝑡
Then the total required mmf is
ℱ 𝑔 + ℱ 𝑚 = =92.5 𝐴𝑡
From which
𝐼= =9.25 𝐴

24. 1.6 From Appendix C, determine the relative amplitude permeability for (a) AISI 1020 and (b) M-19 at a flux density of 1 T. Remember that 𝜇 𝑜 =4𝜋× 10 −7 H/m.
𝜇 𝑎 = 1 𝜇 𝑜 ( )≈500
(a)
𝜇 𝑎 = 1 𝜇 𝑜 ( )≈8800
(b)

26. 1.7 Assuming an ideal core ( 𝜇 𝑖 →∞), calculate the flux density of the air gap of magnetic circuit shown in Fig. 1-16(a). The electrical analog may be reduced to the simpler Fig. 1-16(c). From the latter and Table 1-1,
ℛ 𝑔 ≡airgap reluctance= 5× 10 −3 𝜇 𝑜 (20×40× 10 −6 ) = 50 8 𝜇 𝑜
ℛ 𝑠 ≡sleeve reluctance= 2× 10 −3 𝜇 𝑜 (20×20× 10 −6 ) = 20 4 𝜇 𝑜
ℛ 𝑡 ≡total reluctance= ℛ 𝑔 ℛ 𝑠 = 70 8 𝜇 𝑜

27. 𝜙 𝑔 ≡airgap flux= ℱ ℛ 𝑔 = (50)(10) 70/8 𝜇 𝑜 = 400 𝜇 𝑜 7
Substituting 𝜇 𝑜 =4𝜋× 10 −7 , we get 𝐵 𝑔 =90 𝑚𝑇 or 900 gauss

28. 1.8 A composite magnetic circuit of varying cross section is shown below in Fig. 1-17A; the iron portion has the B-H characteristic of Fig. 1-17(b). Given N=100 turns; 𝑙 1 =4* 𝑙 2 =40 cm; A1=2*A2=10 cm2; 𝑙 𝑔 =2 mm; leakage flux, 𝜙 1 = 0.01 mWb.
Calculate I required to establish an air-gap flux density of 0.6 T.
Corresponding to Bg=0.6T,
𝐻 𝑔 = 0.6 𝜇 𝑜 =4.78× 10 5 A/m
ℱ 𝑔 = 4.78× × 10 −3 =956 At
𝐵 𝐻 = 𝐵 𝑔 =0.6T

30. From Fig. 1-17(b), at B=0. 6 T, H=100 A/m
From Fig. 1-17(b), at B=0.6 T, H=100 A/m. Thus for the two lengths, 𝑙 1 , we have:
ℱ 𝑙 1 = =80 At
The flux in the air gap, 𝜙g, is given by
𝜙 𝑔 = B g A 1 = × 10 −4 =0.6 mWb
The total flux produced by the coil, 𝜙 𝑐 , is the sum of the air gap flux and the leakage flux:
𝜙 𝑐 = 𝜙 𝑔 + 𝜙 𝑙 = =0.61 mWb
The density in the portion 𝑙 2 is therefore,
𝐵 2 = 𝜙 𝑐 A 2 = 0.61× 10 −3 5× 10 −4 =1.22 T

31. For this flux density, from Fig. 1-17(b), H = 410 A/m and
ℱ 𝑙 2 = =41 At
The total required mmf, ℱ is thus
ℱ= ℱ 𝑔 + ℱ 𝑙 1 + ℱ 𝑙 2 = =1077 At
For N=100 turns, the desired current is, finally,
𝐼= =10.77 A

34. Magnetomotive Force:
also represented by script F: F

37. Exercise 1-17

38. The B-H Curve and Magnetic Domains

41. Exercise 1-18

42. Reluctance

44. The Concept of Generated Voltage
Transformer Voltage:

48. B, U, and l perpendicular to each other:
(flux density, velocity, length of conductor)

51. Voltage-Flux Relationship

53. Exercise 1-21

54. Principles of Magnetic Circuits

55. Exercise 1-22

62. Exercise 1-23

63. Equivalent Circuit of a Coil

64. Force, Energy and Torque