Find Integrals of Motion for Particle in Magnetic Dipole Field

Click For Summary
SUMMARY

The discussion focuses on finding the first integrals of motion for a particle with mass m and charge q in a magnetic field represented by the vector potential due to a constant magnetic dipole, defined as A = (μ₀ / 4π) (mₑ × r / r³). Participants suggest using cylindrical coordinates for calculations and applying Lagrangian mechanics to derive the equations of motion. The scalar potential is confirmed to be zero, clarifying the conditions of the problem. The magnetic dipole vector mₑ is treated as a constant throughout the calculations.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with vector potentials in electromagnetism
  • Knowledge of cylindrical coordinate systems
  • Basic concepts of motion in magnetic fields
NEXT STEPS
  • Study the derivation of equations of motion using Lagrangian mechanics
  • Learn about vector potentials and their applications in electromagnetism
  • Explore the implications of scalar and vector potentials in magnetic fields
  • Investigate the behavior of charged particles in magnetic dipole fields
USEFUL FOR

This discussion is beneficial for physicists, particularly those specializing in classical mechanics and electromagnetism, as well as students seeking to understand the dynamics of charged particles in magnetic fields.

Mumba
Messages
26
Reaction score
0
Find the fi rst integrals of motion for a particle of mass m and charge q in a magnetic field given by the vector potential (scalar potential [tex]\Phi[/tex]= 0)

(i) of a constant magnetic dipole [tex]m_{d}[/tex]

[tex]A=\frac{\mu_{0}}{4 pi}\frac{m_{d} \times r}{r^{3}}[/tex]

Hint: Cylindrical coordinates are useful.


I think what i should do is to compute A for cylindrical coordinate system and then use Lagrangian mechanics to get a equation of motion? Is this correct? (we have the charge q given, so we can use the kinetic engergy?)

I tried to compute A but i don't really understand what to do with the magnetic dipole (as a vector)? Whats the story whith that scalar potential?

Thanks for your help,
Mumba
 
Physics news on Phys.org
Mumba said:
I think what i should do is to compute A for cylindrical coordinate system and then use Lagrangian mechanics to get a equation of motion? Is this correct? (we have the charge q given, so we can use the kinetic engergy?)
That would probably be my first thought as well. But I wouldn't write A in cylindrical coordinates just yet. Try to go as far as you can without expressing it in any particular coordinate system.
Mumba said:
I tried to compute A but i don't really understand what to do with the magnetic dipole (as a vector)? Whats the story whith that scalar potential?
[itex]\vec{m}_d[/itex] is just a given vector. You don't really do anything with it, except carry it through the calculation. And I think they just tell you that the scalar potential is zero to clarify the conditions of the problem... I can't imagine why you would think that the scalar potential would be nonzero, unless you were specifically told so.
 
Hi Thanks for the answer.

But i can't go very far without a coordinate system, can I?
I mean, using cyl. coordinates, my degrees of freedom would be just R and [tex]\Theta[/tex].

So i can get the components of r=(Rcos[tex]\Theta[/tex], Rsin[tex]\Theta[/tex], z), where the z-axis is pointing upwards and [tex]\Theta[/tex] the angle between the x-axis (pointing towards you) and R.

What can i do with [tex]m_{d}[/tex]? Should be a vector so i can't really just set it on the origin?! Can i give it an arbitrary direction, say [tex]m_{d}=(m_{d},0,0)[/tex]?

Thanks,
Mumba
 

Similar threads

Replies
7
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
1
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
3K
Replies
2
Views
3K
Replies
3
Views
2K
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K