2. L P X dL r Biot-Savard Law

3. L P X dL r Biot-Savard Law

4. L P X dL r

5. L P X r

6. L P X r infinite length

7. r RHR gives

8. This is how it is done RHR for field direction

9. 9 Lecture 16 slide 3 Area enclosed A Current density flowing through loop.

10. 10 Ampere’s Law: Area enclosed A Current density flowing through loop. Using Stoke’s theorem Differential equation for Can we solve this equation?

11. Current out of page Current into page Infinite coil of wire carrying a current I Axis of solenoid P Evaluate B field here

12. 1 In the vicinity of the point P P 2 3 4 5 3 1 2 4 5 Axis of solenoid resultant Expect B to lie along axis of the solenoid Current out of page Implies that B field has no radial component. I.e. no component pointing towards or away from the solenoid axis.

13. Current out of page Current into page P Claim: The magnetic field outside of the solenoid is zero. Closed path

14. Current out of page Current into page P Must have since there is no net enclosed current. Closed path Conceivably there might be some non-zero field components outside of the solenoid, and as shown.

15. Current out of page Current into page P We would need to have to give. Closed path Conceivably there might be some non-zero field components outside of the solenoid, and as shown.

16. Current out of page Current into page P Now we distort the path by moving one side away from the solenoid. Closed path Still valid

17. Current out of page Current into page P To have,B would have to have the same magnitude no matter how far we moved away from the solenoid. This is possible only if B = 0 outside of the solenoid. Closed path Still valid

18. Current out of page Current into page Using the closed path shown we can now obtain an expression for the magnetic field inside the long solenoid. Closed path P L

19. Current out of page Current into page Closed path P L Ampere’s Law N : number of turns enclosed by length L

20. Current out of page Current into page P N : number of turns enclosed by length L B is independent of distance from the axis of the long solenoid as we are inside the solenoid! B is uniform inside the long solenoid.

21. 21

22. 22 To find for general problems, we need more sophisticated techniques than Ampere’s law and a few postulates.

23. Suppose we can find a function such that: Is called the Magnetic Vector Potential. It is a function of the coordinates. It has direction. It is possible to show that for any:

24. 24 Recall: And: Then: Add in mathematical manipulations

25. 25 But for any vector function : And if use here

26. 26 GIVES In component form: We have managed to generate independent equations governing the x, y and z components of

27. 27 Is called the Magnetic Vector Potential. It is a function of the coordinates. It has direction. Is not unique. Three scalar equations involving derivatives Natural progression of the student’s perspective of the magnetic vector potential TodayNext weekBefore exam After exam I’m doing the deferred exam

28. 28 Similar to Poisson’s equation: Solution of this form in (x, y, z) Integration over volume containing charge . Same form of equation, same form of solution Figure next page Need to obtain all three components General solution for scalar potential

29. 29 x y z Charge distribution Could write solution in short hand form as:

30. 30 In the short hand vector notation

31. 31 Find and for a long wire aligned along the z-axis. y z I wire Evaluate here By geometry: Simplifies solving for

32. 32 y z I wire Becomes Where Current in the wire

33. 33 y z I wire A z should have the same value for any z 1. (Once again by a symmetry argument) Chose to evaluate for z 1 = 0 for simplicity. Wire extends for -  to + 

34. 34 y z I wire Since is the distance from the point P to the wire

35. 35 y z I wire Now for We have and Then we can evaluate the curl of A to get B. Note that A has only a z component.

36. 36 y z I wire Now for

37. 37 y z I wire Now for Similarly for

38. 38 y z I wire Now for We have and Then we can evaluate the curl of A to get B. Note that A has only a z component.

39. 39 y z I wire Gives As would be obtained from Ampere’s Law circles wire

41. Magnetic dipole = product of current in loop with surface area of loop

42. Consider a circular ring of current I placed at the end of a solenoid as shown in the figure. The current in the solenoid produces a magnetic field in which the current loop is placed into. By postulates 1 and 2 of magnetic fields, the current ring will be subjected to a magnetic force. out of page into page

43. Circular ring Cancel in pairs around the ring Will add in same direction on ring giving a net force.

44. Circular ring Will add in same direction on ring giving a net force. Using postulate 1: We need for find B r Gives:

45. Gaussian cylinder We will relate B r to Total magnetic flux through Gaussian cylindrical surface must be zero. As many magnetic field lines that enter the surface, leave the surface. No magnetic charges or monopoles. 3-D view

46. Flux through side: 3-D view Flux through top: Flux through bottom:

47. 3-D view We can now use this in our force on current ring expression

48. Circular ring We have found B r

49. Circular ring Force pulls dipole into region of stronger magnetic field

50. 3-D view In general

51. We will consider a dipole in a uniform magnetic field. We can use any shape we want for the dipole. Here we will select a square loop of wire. I out of page I into page Side view Top view Wire loop

52. Top view Wire loop

53. Side view Top view Wire loop Torque attempts to align dipole moment with. Pivot point Pivot line

54. Side view Torque attempts to align dipole moment with. Pivot point Total torque F => Magnetic force on wire of length a

55. Side view Pivot point F => Magnetic force on wire of length a Through postulate 1 for magnetic fields Then

56. Side view Pivot point Wire loop