Exploring Magnetostatics in Electrodynamics Lecture 11

phy 712 electrodynamics 9 9 50 am mwf olin 103 l.w
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Dive into the world of magnetostatics in this lecture as we discuss magnetic flux density, current densities, comparison with electrostatics, and alternative forms of magnetostatic equations. Discover the relationships between current densities, magnetic fields, and Ampere's law in the absence of magnetic monopoles.

  • Magnetostatics
  • Electrodynamics
  • Amperes Law
  • Current Densities
  • Magnetic Fields
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  1. PHY 712 Electrodynamics 9-9:50 AM MWF Olin 103 Plan for Lecture 11: Start reading Chapter 5 A. Magnetostatics B. Vector potential C. Example: current loop 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 1

  2. 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 2

  3. Magnetostatics Magnetic flux density or magnetic induction field B Steady state (time constant) current density J y ( ) i = 3 J r v r r ( ) iq i i qi vi ri x = J Note that " statics" implies that . 0 continuity the from follows This J t equation : + = 0 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 3

  4. Comparison of electrostatics and magnetostatics ( ) r charge to due field tic Electrosta density : r r 1 ' ( ) r ( ) r = ' 3 E ' d r 3 4 r r ' 0 ( ) r J Magnetosta current to due field tic density : r r ' ( ) r ( ) r = 3 B J 0 ' ' d r 3 4 r r ' 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 4

  5. Alternative forms magnetostatic equations ( ) J r Magnetostatic field due to current density : r r r r r r r r ' 1 ( ) ( ) J r ( ) J r = = 3 3 B r ' ' ' ' 0 0 d r d r 3 r r 4 4 ' ' 1 ' = Note that: 3 r r ' ' ( ) ( ) ( ) r V r J r r ( ) r ( ) ( ) r ( ) = + V r V r Also note that: s s s ( ) ' ( ) = 3 B r ' 0 d r = V r V r ( ) J r In this case ( so that ) r 4 ' ( ) 0 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 5

  6. Alternative forms magnetostatic equations -- continued ( ) ' 4 = r J r B ( ) J r ' = 3 B r 0 ' d r r r ( ) ( ) of B r 0 = magnetic No monopoles ( ) Ampere' law s 0 " Proof" Ampere' magnetosta for law s J r d system tic : ( ) r ' ( ) r = 3 B 0 ' r r 4 ' ( ) = 2 V V V Note that : 1 ( ) ( ) r = = 2 3 r r J Recall that : 4 ' and 0 r PHY 712 Spring 2017 -- Lecture 11 ' r 02/06/2017 6

  7. Differential forms of magnetostatic equations: ( ) ( ) 0 r B = = B r 0 J magnetic No monopoles ( ) r Ampere' law s Magnetostatic vector potential ( ) ( ) r B = 4 ( ) r = B r A ( ) r ' J ' 3 0 ' d r r r ' ( ) r J ( ) r ( ) r = + 3 A 0 ' d r s r r 4 ' 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 7

  8. Non uniqueness of the magnetostatic vector potential ( ) ( ) r ( ) r ( ) r = = B r A A Note that : = ' ( ) r ( ) r + A A if ' s ( ) r = = B z Example : for B 0 ( ) ( ) ( ) r ( y y ) A r y x B x y 1 0 2 B = A r or x 0 B = A x or 0 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 8

  9. Differential form of Amperes law in terms of vector potential: ( ) ( ( ) ( ) r A ( ) ( ) gauge) = = B r A r J r 0 ( ) ( ) r ) = 2 A r J r 0 ( ) r ( ) r = = 2 A A J If (Coulomb 0 0 ( ) r J ' ( ) r = 3 A 0 ' d r r r 4 ' 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 9

  10. Magnetostatics example: current loop z r y x I ( ) r ( ) ( )( ) = + J x y ' ' ' sin ' cos ' ' sin cos r a a ( ) r J ' ( ) r = 3 A 0 ' d r r r 4 ' 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 10

  11. Magnetostatics example: current loop -- continued ( ) ( ) ( )( ( ) 3 0 ' ' 4 ' r r I a ) = + J r x y ' sin ' cos ' ' sin ' cos ' r a J r ( ) = A r d r ( ) ( )( ) + x y sin ' cos ' ' sin ' cos ' r a I a ( ) = 2 A r ' ' cos ' ' 0 r dr d d ( ) ( ) 1/2 ( ) 4 + ' cos cos ' sin sin 'cos + 2 2 ' 2 ' r r rr cos ' Completing integration over ' and ' : r sin ' ( ) + 2 x y 2 Ia ( ) = A r ' 0 d ( ) ( ) 1/2 ( ) 4 a + sin cos 2 2 2 ' r a ra 0 Let sin ' ' ( ) = = = sin cos = cos sin + sin ( ) cos ' Remaining non-trivial terms = A r cos cos cos sin sin 2 cos Ia ( ) ( ) x y sin cos 0 d ( ) 1/2 ( ) 4 + sin co 2 2 2 s r a ra 0 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 11

  12. Magnetostatics example: current loop -- continued ( ) ( sin cos 4 2 cos Ia ) = A r x y 0 d ( ) 1/2 ( ) + sin cos 2 2 2 r a ra 0 Elliptic integrals: /2 du = ( ) K m ( ) 1/2 2 1 sin m u 0 /2 ( ) 1/2 = 2 ( ) 1 sin E m m u du 0 ( ) ( ) 2 2 ( ) 2 ( ) E k k K k x y sin cos ( ) = A r 4 0 Ia ( ) 1/2 2 4 k + + 2 2 2 sin r a ra 4 a sin 2 + ar 2 where: k + 2 2 sin r ra ( ) ( ) = B r A r 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 12

  13. /2 du = ( ) K m ( ) 1/2 2 1 sin m u 0 /2 ( ) 1/2 = 2 ( ) 1 sin E m m u du 0 m 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 13

  14. Magnetostatics example: current loop -- continued ( ( 2 4 sin where: 2 sin r a ra + + For 0: sin , 0 x r y = = = ( ) ) 2 2 ( ) 2 ( ) E k k K k x y sin cos ( ) = A r 4 0 Ia ) 1/2 2 4 k + + 2 2 sin r a ra ar 2 k 2 2 ( ) 2 2 ( ) 2 ( ) E k k K k 1 ( ) = = A r y y ( , ) A x z 4 0 Ia ( ) y 1/2 2 4 k + + + 2 2 2 2 x z a ax 4 + ax a 2 where: k + + 2 2 2 2 x z ax Ay(x,z) 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 14

  15. Magnetostatics example: current loop -- continued ( ( r a + ( ) ) 2 2 ( ) 2 ( ) E k k K k x y sin cos ( ) = A r 4 0 Ia ) 1/2 2 4 k + 2 2 2 sin ra ( ) 2 2 ( ) 2 ( ) E k k K k 4 Ia = 0 ( ) 1/2 2 4 k + + 2 2 2 sin r a ra 4 a sin 2 + ar 2 where: k + 2 2 sin r ra ( ) , ) ( r r A r 1 in 1 r ( ) = = ( , ) A r r B A sin r s r Evaluation for special cases 2 For 0: k ( ) 2 2 ( ) 2 ( ) E k k K k 2 k 2 16 k 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 15

  16. Evaluation for special case k2 0 2 sin + Ia r ( ) = A r 0 ( ) 3/2 4 + 2 2 2 sin r a ra ( ) ( , ) r a r A r 1 1 r ( ) = = ( , ) A r r B A sin r sin r ( ) ( ) ( 5/2 ) + + + 2 2 2 2 cos r 2 2 sin sin 2 sin a r ar r ar 2 Ia = 0 ( ) 4 + + 2 2 2 sin r a ra 4 a = sin 2 + z ar 2 where 0 k + 2 2 sin r ra = For 0: r ( ) ( , ) r r A r 1 1 r ( ) = = ( , ) A r r B A sin r sin r 2 1 Ia z = 0 ( ) 3/2 2 + 2 2 z a 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 16

  17. Evaluation for special case k2 0 2 sin + Ia r ( ) = A r 0 ( ) 3/2 4 + 2 2 2 sin r a ra ( ) ( , ) r a r A r 1 1 r ( ) = = ( , ) A r r B A sin r sin r ( ) ( ) ( 5/2 ) + + + 2 2 2 2 cos r 2 2 sin sin 2 sin a r ar r ar 2 Ia = 0 ( ) 4 + + 2 2 2 sin r a ra 4 a sin 2 + ar 2 where 0 k + 2 2 sin r ra For r ( ) ( , ) r r A r 1 1 r ( ) = = ( , ) A r r B A s in r sin r ( ) 2 Ia r ( ) + 2 r = cos s i n 0 3 4 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 17

  18. Magnetostatics example: current loop -- continued ( ) ( + ) 2 x + y sin a cos ra 2 ( k ) 2 ( ) k K k E k ( ) r = A 0 4 Ia ( ) / 1 2 2 4 2 2 2 sin r ra 4 a sin + ar where : k + 2 2 2 sin r ( ) r spherical for that Note = A ( ) r = B A = x y polar coordinate : s sin cos ( ) r ( ) r A ( ) 2 4 2 ( k ) 2 ( ) Ia k K k E k ( ) r A = 0 where ( ) r / 1 2 2 4 + + 2 2 2 sin rA r a ra ( ) ( ) ( ) r A ( ) r sin 1 1 r ( ) r = r B sin r For : r ( ) r 2 I a ( ) r + r B 0 2 cos sin 3 4 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 18

  19. Other examples of current density sources: expression mechanical Quantum current for density r particle a for of mass M charge and e and of probabilit amplitude y : ( ) ( ) ) r e = * * J r r r r ( ) ( ) ( ) ( ) ( 2 Mi atom) an in as (such potential spherical a in electron an For Y R : ( ) r ( ) r ( ) r ( ) r = nlm nl lm l l ( ) r ( ) r * Y Y 1 e ( ) r ( ) r ( ) r ( ) r 2 * lm lm = J R Y Y l l nl lm lm 2 sin Mi r l l m e ( ) r 2 = l nlm sin M r l z sin r r = + = x y Note that : sin cos m e ( ) r ( ) ( r ) r 2 = J z l nlm 2 2 sin M r l 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 19

  20. Magnetic vector potential for this case: ( ) r J ' ( ) r = 3 A 0 ' d r r r 4 ' m e ( ) r ( ) ( ' r ) 2 = J z r ' ' l nlm 2 2 ' sin ' M r l ( ) r 2 ( ) ' ' z r r ' e m ( ) r nlm = 3 A 0 ' l l d r 2 2 r 4 ' sin ' M r = example For electron : in the 211 state of H : nlm l 2 1 ' r ( ) r 2 = / ' r 2 a ' sin ' e 211 3 64 a a ( ) r 2 3 z r e r r r ( ) r = + + + / r a A 0 1 1 e 3 2 3 8 2 8 M a a a 02/06/2017 PHY 712 Spring 2017 -- Lecture 11 20

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