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Electrical dipoles  on a thin long line © Frits F.M. de Mul
Electrical dipoles  on a thin long line Question: Calculate  E- field in arbitrary points around the line Available: Thin line, infinitely long, homogeneously filled with dipoles, each with dipole moment  p  [Cm]
Electrical dipoles  on a thin long line ,[object Object],[object Object],[object Object],[object Object]
Analysis and Symmetry (1) all points at  equal  r   are equivalent, even if at  different  z  or   1.  Cylinder : infinitely long and thin 2.   Distribution of dipoles: n  dipoles / meter; homogeneous; all directions uniform  each: dipole moment  p   [Cm] 3.  Coordinate axes:  X,Y,Z Z-axis = symm. axis Y X Z e z 4.  Cylinder symmetry: e r 
Analysis and Symmetry (2) Assume: field components of dipole field at some distance in point P àre known. p Dipole :  r P e r e  E r E 
Analysis and Symmetry (3) Possible directions of the dipoles: X Z e z e r  Y X Y X Z e z e r  Y Y X Z e z e r  Z
Approach to solution Use symmetry: dp  =  n  p  dz n =  dipole density (per meter) Z Y P y P Choose : Point  P  at distance  y P   from axis. r + r - Dipole elements at  + z  and - z  will contribute symmetrically dp   in  dz  at + z dp   in  dz  at - z +z -z O
Calculations (1): Dipoles // X-axis P dp   in  dz  at + z dp   in  dz  at - z Z +z -z y P r + r - O Y X Case 1: all dipoles // +X-axis dE    d   ==>  E // -X-axis E result    =  90 0   ==> e  + =e  
Calculations (2a): Dipoles // Y-axis resulting vector : ==>  E // Y-axis Find directions of the field contributions  dE dp   in  dz  at - z P dp   in  dz  at + z Z +z -z O Y X     Case 2: all dipoles // +Y-axis e   + e r+ e   - e r- dE r      cos   dE      sin   dipole  + z   - z   >0 d  r  >0 d  r  dE   >0 d   >0 E result
Calculations (2b): Dipoles // Y-axis Y-components: P Z +z -z O Y X dE   d       e   - e   + e r- e r+ d  r  d  r  E result r    dp = n.p.dz  ;       dE r+   and  dE r-   : each: dE  +   and  dE  -   : each:
Calculations (3a): Dipoles // Z-axis resulting vector : ==>  E // Z-axis P dp   in  dz  at + z dp   in  dz  at - z Z +z -z O Y X     Case 3: all dipoles // +Z-axis e   + e r+ e   - e r- dE r      cos   dE      sin   dipool  +  -  <0 d  r  >0 d  r  dE   >0 d   >0 dE result
Calculations (3b) Z-components: P Z +z -z O Y X dE   d       e   - e   + e r- e r+ d  r  d  r  dE result r r      dp = n p dz        dE r+   and  dE r-   : each: dE  +   and  dE  -   : each:
Conclusions  dipole direction:   dE  -  contributions:     and  r  are measured from dipole to P ) the end X Z e z e r  Y X Y Z

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E field line of dipoles

  • 1. Electrical dipoles on a thin long line © Frits F.M. de Mul
  • 2. Electrical dipoles on a thin long line Question: Calculate E- field in arbitrary points around the line Available: Thin line, infinitely long, homogeneously filled with dipoles, each with dipole moment p [Cm]
  • 3.
  • 4. Analysis and Symmetry (1) all points at equal r are equivalent, even if at different z or  1. Cylinder : infinitely long and thin 2. Distribution of dipoles: n dipoles / meter; homogeneous; all directions uniform each: dipole moment p [Cm] 3. Coordinate axes: X,Y,Z Z-axis = symm. axis Y X Z e z 4. Cylinder symmetry: e r 
  • 5. Analysis and Symmetry (2) Assume: field components of dipole field at some distance in point P àre known. p Dipole :  r P e r e  E r E 
  • 6. Analysis and Symmetry (3) Possible directions of the dipoles: X Z e z e r  Y X Y X Z e z e r  Y Y X Z e z e r  Z
  • 7. Approach to solution Use symmetry: dp = n p dz n = dipole density (per meter) Z Y P y P Choose : Point P at distance y P from axis. r + r - Dipole elements at + z and - z will contribute symmetrically dp in dz at + z dp in dz at - z +z -z O
  • 8. Calculations (1): Dipoles // X-axis P dp in dz at + z dp in dz at - z Z +z -z y P r + r - O Y X Case 1: all dipoles // +X-axis dE    d   ==> E // -X-axis E result  = 90 0 ==> e  + =e  
  • 9. Calculations (2a): Dipoles // Y-axis resulting vector : ==> E // Y-axis Find directions of the field contributions dE dp in dz at - z P dp in dz at + z Z +z -z O Y X     Case 2: all dipoles // +Y-axis e  + e r+ e  - e r- dE r  cos  dE    sin  dipole + z - z >0 d  r  >0 d  r  dE   >0 d   >0 E result
  • 10. Calculations (2b): Dipoles // Y-axis Y-components: P Z +z -z O Y X dE   d       e  - e  + e r- e r+ d  r  d  r  E result r    dp = n.p.dz ;      dE r+ and dE r- : each: dE  + and dE  - : each:
  • 11. Calculations (3a): Dipoles // Z-axis resulting vector : ==> E // Z-axis P dp in dz at + z dp in dz at - z Z +z -z O Y X     Case 3: all dipoles // +Z-axis e  + e r+ e  - e r- dE r  cos  dE    sin  dipool + - <0 d  r  >0 d  r  dE   >0 d   >0 dE result
  • 12. Calculations (3b) Z-components: P Z +z -z O Y X dE   d       e  - e  + e r- e r+ d  r  d  r  dE result r r      dp = n p dz       dE r+ and dE r- : each: dE  + and dE  - : each:
  • 13. Conclusions dipole direction: dE - contributions:   and r are measured from dipole to P ) the end X Z e z e r  Y X Y Z