Mechanics, charge in vector potential

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SUMMARY

A particle of charge q in a vector potential defined as \(\vec{A} = (B/2)(-y\vec{i} + x\vec{j})\) exhibits circular motion with an angular frequency \(\omega_0 = \frac{qB}{\mu c}\), where \(\mu\) is the mass of the particle. The Hamiltonian is formulated as \(H = \frac{1}{2\mu} \left( \vec{p} - \frac{q}{c} \vec{A} \right)^2\). The equations of motion derived from the Hamiltonian lead to coupled first-order equations that can be decoupled to demonstrate harmonic motion, confirming the circular orbits and allowing extraction of the angular frequency.

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Homework Statement


A particle of charge q in a vector potential

[tex]\vec{A} = (B/2) (-y\vec{i} + x\vec{j} )[/tex]

show that a classical particle in this potential will move in circles at an angular frequency [tex]\omega_0 = qB/ \mu c[/tex], [tex]\mu[/tex] is the mass of the particle.

Homework Equations


The Attempt at a Solution



I write a Hamiltionian as

[tex]H = \frac{1}{2\mu} \left( \vec{p} - \frac{q}{c} \vec{A} \right)^2[/tex]

and then get

[tex]\mu \dot{x} = p_x + \frac{qB}{2c}y[/tex]
[tex]\mu \dot{y} = p_y - \frac{qB}{2c}x[/tex]

and then I'm lost. I'm not sure if this is right, and if it is: how do I see how this gives circular orbits and what the angular frequency is?
 
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Write down the Lagrangian for the particle. What happens to [tex]\vec{A} \cdot \vec{r}[/tex]? So, what are the cyclic coordinates and what's conserved?

Once you figure out what the constants of motion are, you can solve the last two equations you wrote.
 
If you have to use the Hamiltonian formulation, don't forget the other set of equations
p'= -dH/dq.
The 4 1st order equations here correspond to the 2 2nd order equations of the Lagrangian formulation. Either way, they are coupled, and you can decouple them in a simple fashion. Circular motion corresponds to harmonic motion in each dimension, so if you can show the equations are of the form
x'' +w^2 x+constant=0
then that's it (cause you should know the solutions of such an equation already);
then you can pull off the angular frequency (speed) w. The constant appears from integration but you can verify yourself that it's zero.
 

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