1. Physics for Scientists and Engineers Chapter 27: Sources of the Magnetic Field Copyright © 2004 by W. H. Freeman & Company Paul A. Tipler Gene Mosca Fifth Edition

2. So far, we have learnt the effect of B fields on charged particles, on current carrying conductors, Torque on a current loop (magnetic moment) Next: B field from a current

3. Lectures 13 & 14: The Magnetic Field Related to its Cause: ELECTROMAGNETISM Learning Objectives: To Learn and Practice Methods used to Calculate the Magnetic Field of Steady Currents Method 1: The Biot-Savart Law (Tipler 27-1&2)

4. Hans Christian Oersted Danish Physicist (1777-1851) In 1820 Oersted demonstrated that a magnetic field exists near a current-carrying wire - first connection between electric and magnetic phenomena.

5. q v P r Magnetic Field of a Moving Charge  Perpendicular to the plane containing r and v Taking into account direction: B

6. In SI units the permeability of free space is a unit vector

8. Magnetic Field of a Current Element I ll r P   B into screen 

9. Magnetic Field of a Current Element In vector form

10. The Magnetic Fields Established by Complete Circuits Procedure Write down dB in terms of a single variable Integrate between the limits applicable to the problem Be careful about the directions of the vector quantities Use symmetry to simplify the problem

11. Magnetic Field of a Current Element The Biot-Savart Law In vector form dy P r Directions of the fields!!!

12. Magnetic Field of a Current Element I ll r P   B into screen 

13. I ll r P  ll I ll I

14. The B Field Due to a Long Straight Wire (Tipler pg 865)  I ll r P  a l  BB

15. The B Field Due to a Long Straight Wire I ll r P   a l 

16. Exercise: Find the magnitude of the magnetic field at the centre of a square current loop of side l = 50 cm carrying a current I = 1.5 A. l l/2 I First “picture” the problem. The magnetic field is the sum of the contributions from the four sides of the loop

17. The B Field Due to an Infinite Straight Wire I  a

18. E Field of a line charge

19. Lecture Questionnaire Summary A: 100, B: 80, C: 60, D: 40, E: 20 F: 0

20. 1.Could you clearly hear the lecturer?89 2.Was the lecturer enthusiastic?97 3.How helpful were the lecturer’s responses to questions? 86 4.Did the Lecturer use board/OHP/PPT in a clear way?93 5.Were the handouts useful?93 6.How interesting was the module?78 7.Introduction to each lecture88 8.How useful were the text books?79 9.Level of presentation95 10.Speed of the lecture94 11.Level of the problem sheets87

21. 7. How interesting was the module?78 Coulomb Gauss Ampere Faraday 78 Guo97 Tesla

22. B at any Point on the Axis of a Single Current Loop (Tipler pg. 859) dl parallel to z, B field at P has no z component. But it has both x and y components.dB perpendicular to r

23. I P dl1dl1 a dB1dB1  x dl2dl2 dB2dB2 r

24. x y dB1dB1 dB2dB2 dB total P By symmetry, components of dB perpendicular to the axis sum to zero

25. Each dB has a component parallel to the x-axis of magnitude Where and

26. Integrating:

27. B at any Point on the Axis of a Single Current Loop (Tipler pg. 859)

28. A “small” loop of current is called a magnetic dipole

29. Exercise: Obtain an expression for the magnetic field (i) At the centre of the loop (ii) At x >> a

30. Consider x >> a Magnetic dipole moment: Compare with electric dipole

31. The E-field due to an Electric Dipole - Calculation To simplify the calculation, we will only compute the field along the axis r E -q+q a a/2

32. For r >> a This applies to any points along the line of the dipole, and only for points along the line of the dipole. Along this particular direction, the E field from the positive charge is in opposite direction to that from the negative charge.

33. Electric Dipole Magnetic Dipole

34. B-field lines encircle the current that acts as their source. B-field lines are continuous loops (lecture 11 - Gauss’s Flux Law for Magnetism)

35. Review The magnetic field set up by a current-carrying conductor can be found from the Biot-Savart law. This law asserts that the contribution dB to the field set up by a current element Idl at a point P, a distance r from the current element, is REMEMBER THIS

36. Calculating Magnetic Field Strengths Method II Ampere’s Law (Tipler 27-4)

37. Thus it appears that The B Field Due to a Long Straight Wire We know that

38. Right hand rule and the direction of the B field from a current

39. In general, the line integral of B around any closed mathematical path equals  0 times the current intercepted by the area spanning the path. Ampere’s Law Left hand side is a line integral round a closed loop I is the current enclosed by the loop

40. B-Field Inside a Long Solid Cylindrical Conductor Carrying Uniformly Distributed Current (Example 27-9 – Tipler) First Example

41. I R r Enclosed current:

42. r

43. B field inside a conductor, OK E field inside a conductor, Not OK.

44. B-Field Inside a Long Solenoid Second Example

46. n = number of turns per unit length

47. Field of a Toroidal Solenoid (Tipler pg. 872-873) Third Example

49. Path 2 Path 1 - no current enclosed B = 0 Path 2 - no net current enclosed B = 0 Path 3 - net current enclosed N = total number of turns Path 1

50. Ampere’s law is useful for calculating the magnetic field only when there is a high degree of symmetry. Need B to be constant in magnitude along the enclosed path and tangent to any such path.

51. (a)zero (b)into the screen (c)out of the screen (d)toward the top or bottom of the screen (e)toward one of the wires. Short Exercise 1:Two wires lie in the plane of the screen and carry equal currents in opposite directions. At a point midway between the wires, the magnetic field is

52. (a)zero (b)into the screen (c)out of the screen (d)toward the top or bottom of the screen (e)toward one of the wires.

53. The Force Between Two Long Parallel Currents

54. The B Field Due to a Long Straight Wire

55. Force F on a length L of the upper conductor is: I I F B  B Force F on a length L of the lower conductor is: F F = F Attraction!

56. F = F I I F B  B F Attraction! What happens when the currents are in opposite directions? force per unit length This fundamental magnetic effect was first studied by Ampere (1822) (Ans. Repulsion)

57. Short Exercise 2:Two parallel wires carry currents I 1 and I 2 (= 2I 1 ) in the same direction. The forces F 1 and F 2 on the wires are related by: (a)F 1 = F 2 (b)F 1 = 2F 2 (c)2F 1 = F 2 (d)F 1 = 4F 2 (e)4F 1 = F 2

58. Short Exercise 2:Two parallel wires carry currents I 1 and I 2 (= 2I 1 ) in the same direction. The forces F 1 and F 2 on the wires are related by: (a)F 1 = F 2 (b)F 1 = 2F 2 (c)2F 1 = F 2 (d)F 1 = 4F 2 (e)4F 1 = F 2

59. Definition of the Ampere The ampere is that steady current which, flowing in two infinitely long straight parallel conductors of negligible cross-sectional area placed 1 m apart in a vacuum, causes each wire to exert a force of 2 x 10 -7 N on each metre of the other wire.

61. A current of one ampere carries a charge of one coulomb per second Definition of the Coulomb

62. Review and Summary Ampere’s Law For current distributions involving a high degree of symmetry, Ampere’s law can be used (instead of the Biot-Savart law) to calculate the magnetic field where (i) is the line integral round a closed loop, and (ii) I is the current enclosed by the loop

63. Maxwell’s Equations – the story so far! Laws of Magnetostatics Laws of Electrostatics For E and B fields that do not vary with time

64. Exercise: The diagram shows two currents associated with infinitely long wires, one current of 8 A into the screen, the other current is 8 A out of the screen. Find for each path indicated.

65.  0  8 A -  0  8 A (field is opposite to the line of integration) 0 C1 Can we obtain B from this? C2: Is B=0 everywhere along the path?

66. Exercise: Two straight rods 50 cm long and 1.5 mm apart carry a current of 15 A in opposite directions. One rod lies vertically above the other. What mass must be placed on the upper rod to balance the magnetic force of repulsion. i.e. the magnetic force between two current- carrying wires is relatively small, even for currents as large as 15 A separated by only 1.5 mm

67. An extended,but brief, summary

68. Giving

69. A long straight cylindrical conductor has an outer radius b and a hollow core of radius a. The current density varies with radius r for a  r  b as j = j 0 r/a where j 0 is a constant. Find an expression for the total current flowing in the conductor. Find expressions for the strength of the magnetic field in the three regions (i) r b.

71. For r < a, B = 0 as no current linked a  r  b

72. r > b

73. The Relationship Between  0 and  0 (by definition) (by experiment)

75. Next Section of Course: Dealing with time varying electric and magnetic fields

76. Ampere’s Law What about ? For electrostatic electric fields In fact But B.dl does not represent work done! From Gauss’s Flux law for magnetism Static E fields are conservative

77. James Clerk Maxwell Sir Isaac Newton Jean-Baptiste Biot André Marie Ampère

78. Ludwig Boltzmann