Deriving the Magnetic Field from a Magnetic Dipole

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SUMMARY

The discussion focuses on deriving the magnetic field from a magnetic dipole, specifically the equation ##B=\frac { 1 }{ 4\pi \epsilon r^{ 3 } } \left[ 3\hat { r } \left( \hat { r } \cdot \vec { m } \right) -\vec { m } \right]##. Participants address the challenge of using the BAC CAB rule versus the determinant method to simplify the expression ##\vec { \nabla } \times \left( \vec { m } \times \vec { r } \right)##, with one user, Chris Maness, noting a discrepancy in results. The discussion highlights the importance of understanding vector calculus and electromagnetic principles in solving such problems.

PREREQUISITES
  • Understanding of vector calculus, specifically the BAC CAB rule
  • Familiarity with electromagnetic theory, particularly magnetic dipoles
  • Knowledge of the curl and divergence operations in vector fields
  • Proficiency in using mathematical notation and derivations in physics
NEXT STEPS
  • Study the derivation of the magnetic field from a magnetic dipole in detail
  • Learn about the application of the BAC CAB rule in vector calculus
  • Explore the implications of the product rule for derivatives in vector calculus
  • Review the concepts of curl and divergence in electromagnetic fields
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying electromagnetism and vector calculus, as well as educators seeking to clarify concepts related to magnetic dipoles and field derivations.

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Homework Statement


Show that: ##B=\frac { 1 }{ 4\pi \epsilon r^{ 3 } } \left[ 3\hat { r } \left( \hat { r } \cdot \vec { m } \right) -\vec { m } \right] ##

Homework Equations



B=delxA, m=a*I

The Attempt at a Solution


[/B]
I follow my professors derivation. However, she expands the term: ##\vec { \nabla } \times \left( \vec { m } \times \vec { r } \right) =2 \vec m## by just using the determinate. I figured I would go with the less messy "BAC CAB" like she uses on another term. However, I can't get ##2 \vec m## by using BAC CAB. I get ##3 \vec m## instead. I have tried both ways using spherical and cartesian. Does not ##\nabla \cdot \vec m=0##? If I can show that ##\vec { r } \left(\vec \nabla \cdot \vec m \right)=-\vec m##. I would be home free, but I think it should equal zero.

Any E&M or vector calc. gurus, your help would be appreciated.

Thanks,
Chris Maness
 
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Dang, ok. That was a pretty basic oversight. Thanks

Chris
 

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