Derivation of ideal magnetic dipole field strength

Click For Summary
SUMMARY

The discussion focuses on the derivation of the ideal magnetic dipole field strength as presented in Griffiths' "Introduction to Quantum Mechanics Electrodynamics." The vector potential A is expressed as A = (μ₀/4π)(m sin(θ)/r²)−φ, indicating that the magnetic moment m is aligned along the z-axis. The transition from the vector potential A to its φ component is clarified through the cross product of the magnetic moment and the unit vector in spherical coordinates. The discussion also addresses the definition of φ in spherical coordinates when θ = 0 and r = 0.

PREREQUISITES
  • Understanding of vector calculus and cross products
  • Familiarity with spherical coordinate systems
  • Knowledge of magnetic dipole moments and their representations
  • Basic principles of electromagnetism as outlined in Griffiths' texts
NEXT STEPS
  • Study Griffiths' "Introduction to Electrodynamics" for a deeper understanding of magnetic dipoles
  • Learn about the mathematical derivation of vector potentials in electromagnetism
  • Explore the implications of magnetic dipole fields in quantum mechanics
  • Investigate the properties of spherical coordinates and their applications in physics
USEFUL FOR

Students and professionals in physics, particularly those focusing on electromagnetism and quantum mechanics, as well as educators looking to clarify concepts related to magnetic dipole fields.

RedDeer44
Messages
3
Reaction score
1
For reference, this is from Griffiths, introduction to quantum mechanics electrodynamics, p253-255

When deriving the ideal magnetic dipole field strength, if we put the moment m at origin and make it parallel to the z-axis,
the book went from the vector potential A

$$
A= \frac{\mu_0}{4\pi}\frac{\vec{m}\times \hat{r}}{r^2}
$$
to
$$
A = \frac{\mu_0}{4\pi}\frac{m\sin{\theta}}{r^2}\hat{\phi}
$$

Can someone explain how the single ##\phi## component come about? This to me seems to indicate ##r## has non-zero ##\theta## component and zero ##\phi## component. But I thought ##r## is any point?

Also, for a point in spherical coordinates, is ##\phi## value defined when ##\theta = 0##? Or when ##r=0##?
 
Last edited:
Physics news on Phys.org
You have ##\vec{m}=(0,0,m)## and thus
$$\vec{m} \times \hat{r} = \begin{pmatrix} 0\\0\\m \end{pmatrix} \times \begin{pmatrix} \cos \varphi \sin \vartheta \\ \sin \varphi \sin \vartheta \\ \cos \vartheta \end{pmatrix} =m \sin \vartheta \begin{pmatrix} -\sin \varphi \\ \cos \varphi \\ 0 \end{pmatrix} = m \sin \vartheta \hat{\varphi}.$$
 
  • Informative
  • Like
Likes   Reactions: Dale and RedDeer44

Similar threads

  • · Replies 5 ·
Replies
5
Views
890
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 1 ·
Replies
1
Views
627
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K