# Motion of charged particle in magnitc field given by potential of magnetic dipole

1. Jan 2, 2010

### Mumba

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 $$\Phi$$= 0)

(i) of a constant magnetic dipole $$m_{d}$$

$$A=\frac{\mu_{0}}{4 pi}\frac{m_{d} \times r}{r^{3}}$$

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 dont really understand what to do with the magnetic dipole (as a vector)? Whats the story whith that scalar potential?

Mumba
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 2, 2010

### diazona

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.
$\vec{m}_d$ 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.

3. Jan 3, 2010

### Mumba

I mean, using cyl. coordinates, my degrees of freedom would be just R and $$\Theta$$.
So i can get the components of r=(Rcos$$\Theta$$, Rsin$$\Theta$$, z), where the z-axis is pointing upwards and $$\Theta$$ the angle between the x-axis (pointing towards you) and R.
What can i do with $$m_{d}$$? Should be a vector so i cant really just set it on the origin?! Can i give it an arbitrary direction, say $$m_{d}=(m_{d},0,0)$$?