MAGNETOSTATIC FIELD (MAGNETIC FORCE, MAGNETIC MATERIAL AND INDUCTANCE )

1. CHAPTER 8 MAGNETOSTATIC FIELD (MAGNETIC FORCE, MAGNETIC MATERIAL AND INDUCTANCE) 8.1 FORCE ON A MOVING POINT CHARGE 8.2 FORCE ON A FILAMENTARY CURRENT 8.3 FORCE BETWEEN TWO FILAMENTARY CURRENT 8.4 MAGNETIC MATERIAL 8.5 MAGNETIC BOUNDARY CONDITIONS 8.6 SELF INDUCTANCE AND MUTUAL INDUCTANCE 8.7 MAGNETIC ENERGY DENSITY

2. Force in electric field: Force in magnetic field: Total force: or Also known as Lorentz force equation. 8.1 FORCE ON A MOVING POINT CHARGE

3. Charge Condition Combination Stationary - Field Moving and Field Force on charge in the influence of fields:

4. The force on a differential current element , due to the uniform magnetic field, : 8.2 FORCE ON A FILAMENTARY CURRENT It is shown that the net force for any close current loop in the uniform magnetic field is zero.

5. Ex. 8.1: A semi-circle conductor carrying current I, is located in plane xy as shown in Fig. 8.1. The conductor is under the influence of uniform magnetic field, . Find: • Force on a straight part of the conductor. • Force on a curve part of the conductor. y (a) The straight part length = 2r. Current flows in the x direction. r x I Solution:

6. y r Hence, it is observed that and it is shown that the net force on a close loop is zero. x I (b) For curve part, will be in the –ve z direction and the magnitude is proportional to sin 

7. z Loop l2 Loop l1 R12 I2 I1 P1(x1,y1,z1) P2(x2,y2,z2) y x 8.3 FORCE BETWEEN TWO FILAMENTARY CURRENT

8. z (N) (A/m) Loop l2 Loop l1 R12 I2 I1 P1(x1,y1,z1) P2(x2,y2,z2) y x where is the force due to I2dl2 and due to the magnetic field of loop l1 We have : The magnetic field at point P2 due to the filamentary current I1dl1 :

9. Integrate: For surface current : For volume current :

10. at position conductor 2 Hence: Ex. 8.2: Find force per meter between two parallel infinite conductor carrying current, I Ampere in opposite direction and separated at a distance d meter. z Solution: I1 I2 y d I1 = I2 = I x

11. y Field created in the square loop due to filamentary current : (1,2,0) (3,2,0) 15 A 2 mA x z (3,0,0) (1,0,0) Ex. 8.3: A square conductor current loop is located in z = 0 plane with the edge given by coordinate (1,0,0), (1,2,0), (3,0,0) and (3,2,0) carrying a current of 2 mA in anti clockwise direction. A filamentary current carrying conductor of infinite length along the y axis carrying a current of 15 A in the –y direction. Find the force on the square loop. Solution:

12. y (1,2,0) (3,2,0) 15 A 2 mA x z (3,0,0) (1,0,0) Hence:

13. 8.4 MAGNETIC MATERIAL The prominent characteristic of magnetic material is magnetic polarization – the alignment of its magnetic dipoles when a magnetic field is applied. Through the alignment, the magnetic fields of the dipoles will combine with the applied magnetic field. The resultant magnetic field will be increased. 8.4.1 MAGNETIC POLARIZATION (MAGNETIZATION) Magnetic dipoles were the results of three sources of magnetic moments that produced magnetic dipole moments : (i) the orbiting electron about the nucleus (ii) the electron spin and (iii) the nucleus spin. The effect of magnetic dipole moment will produce bound current or magnetization current.

14. Am2 where is magnetic dipole moment in discrete and I is the bound current. A/m where is a magnetization and n is the volume dipole density when v -> 0. Magnetic dipole moment in microscopic view is given by : In macroscopic view, magnetic dipole moment per unit volume can be written as:

15. Am-1 ds If the dipole moments become totally aligned : Magnetic dipole moments in a magnetic material Macroscopic v tend to align themselves Microscopic base

16. Alignment of within a magnetic material under uniform conditions to form a non zero on the slab surfaces, and a within the material. 8.4.2 BOUND MAGNETIZATION CURRENT DENSITIES z y x

17. y 8.4.3 TO FIND

18. Am-1 through the loop l’ on the surface bound by the loop l’ Bound magnetization current : We have: Hence:

19. is the bound magnetization current density within the magnetic material. (Am-2) dIm = Mtan dl' loop l’ On the slab surface Using Stoke’s Theorem: Hence: And to find Jsm : From the diagram : is the surface bound magnetization current density (Am-1)

20. Define: 8.4.4 EFFECT OF MAGNETIZATION ON MAGNETIC FIELDS Due to magnetization in a material, we have seen the formation of bound magnetization and surface bound magnetization currents density. Maxwell’s equation: due to free charges and bound magnetization currents

21. Hence: Magnetization in isotropic material: Hence:

22. Ex. 8.4: A slab of magnetic material is found in the region given by 0 ≤ z ≤ 2 m with r = 2.5. If in the slab, determine: Solution:

23. (d) Because of z = 0 is under the slab region of 0 ≤ z ≤ 2 , therefore

24. Ex. 8.5: A closely wound long solenoid has a concentric magnetic rod inserted as shown in the diagram.In the center region, find: in both air and magnetic rod, (b) the ratio of the in the rod to the in the air, (c) on the surface of the rod and within the rod. Assume the permeability of the rod equals 5o . P3 0 P2 b magnetic rod a r P2’ P3’ z 0 P1 P4 Solution: (a) Using Ampere’s circuital law to the closed path P1 - P2 - P3- P4 . If using path P1 - P2’ - P3’ -P4 - P1, Hz in the rod will be the same as in the air since Ampere’s circuital law does not include any Im in its Ien term.

25. Hence: P3 0 P2 b magnetic rod a P3’ P2’ z 0 P1 P4 (since m of air is zero) in air in the rod

26. 0 a a flux Js b Hz=NI/l Hz Bz 50Hz Bz=0Hz flux Jsm Mz Mz=4Hz Jsm Js Js=Hz=NI/l Jsm=4Hz Plots of H, B and M, Js and Jsm along the cross section of the solenoid and the magnetic rod

27. Group A – has a zero dipole moment diamagnetic material eg. Bismuth Group B – has a non zero dipole moment (a) Paramagnetic material - ; Eg. Aluminum - (b) Ferromagnetic material : has strong magnetic moment in the absence of an applied field. When is applied, there will be a slight alignment of the atomic dipole moment to produce Eg: metals such as nickel, cobalt and iron. 8.4.5 MAGNETIC MATERIAL CLASSIFICATION Magnetic material can be classified into two main groups:

28. To find the relationship between To find normal component of at the boundary 8.5 MAGNETIC BOUNDARY CONDITIONS Region 1: a b Boundary d c Region 2: and use Consider a small cylinder as

29. = Region 1: a b Boundary d c Region 2: x Consider a closed abcd as and use where is perpendicular to the directions of and To find tangential component of at the boundary z x y In vector form : is a normal unit vector from region 2 to region 1

30. We have: We have: Hence: Hence: We have: We have: Hence: Hence: If Js = 0 :

31. y Region 1: Boundary at y = 0 plane x Region 2: If the fields were defined by an angle normal to the interface (1) (2) Divide (2) to (1):

32. Normal component #1 Z = 0 #2 z y x Ex. 8.6: Region 1 defined by z > 0 has 1 = 4 H/m and 2 = 7 H/m in region 2 defined by z < 0. A/m on the surface at z = 0. Given , find Solution:

33. z The unit normal: #2 #1  r2 = 6 y O r1 = 4 Ex. 8.7: Region 1, where r1 = 4 is the side of the plane y + z < 1 . In region 2 , y + z > 1 has r2 = 6 . If find Solution: y + z = 1

34. Magnetic flux Coil _ I + VL 8.6 SELF INDUCTANCE AND MUTUAL INDUCTANCE Simple electric circuit that shows the effect of energy stored in a magnetic field of an inductor : From circuit theory the induced potential across a wire wound coil such as solenoid or a toroid : where L is the inductance of the coil, I is the time varying current flowing through the coil – inductor.

35. Magnetic flux Coil _ I + VL switch where (lambda) is the total flux linkage of the inductor In a capacitor, the energy is stored in the electric field : In an inductor, the energy is stored in the electric field, as suggested in the diagram : Define the inductance of an inductor :

36. I2 Weber turns Circuit 2 N2 turns Hence : H Circuit 1 N1 turns I1 Two circuits coupled by a common magnetic flux that leads to mutual inductance. Mutual inductance : is the linkage of circuit 2 produced by I1 in circuit 1 For linear magnetic medium M12 = M21

37. flux N turns Assume all the flux links all N turns and that does not vary over the cross section area of the solenoid. Ex. 8.8: Obtain the self inductance of the long solenoid shown in the diagram. Solution:

38. Mean path Cross sectional area S b c a I Feromagnetic core N turns Ex. 8.9: Obtain the self inductance of the toroid shown in the diagram. Solution:

39. b a The will exist only between a and b and will link all the current I Ex. 8.10: Obtain the expression for self inductance per meter of the coaxial cable when the current flow is restricted to the surface of the inner conductor and the inner surface of the outer conductor as shown in the diagram. Solution:

40. I2 Circuit 2 N2 turns c b a Circuit 1 N1 turns I1 Two circuits coupled by a common magnetic flux that leads to mutual inductance. Ex. 8.11: Find the expression for the mutual inductance between circuit 1 and circuit 2 as shown in the diagram. Solution: Let us assume the mean path : 2b >> (c-a)

41. b c S a I N turns 8.7 MAGNETIC ENERGY DENSITY We have : Consider a toroidal ring : The energy in the magnetic field : Multiplying the numerator and denominator by 2b :

42. Hence : In vector form : Hence the inductance :

43. Solution: b a Ex. 8.12: Derive the expression for stored magnetic energy density in a coaxial cable with the length l and the radius of the inner conductor a and the inner radius of the outer conductor is b. The permeability of the dielectric is  .