# Angular momentum of particle in dipole magnetic field

Basically I'm just trying to find the expression for the angular momentum of a particle of mass $m$ and charge $q$ in a dipole magnetic field. In cylindrical coordinates, $\vec{v}=v_{\rho}\hat{\rho}+v_{\phi}\hat{\phi}+v_{z}\hat{z} = v_{\rho}\hat{\rho}+ \rho \dot{\phi}\hat{\phi}+v_{z}\hat{z}$ with $\rho^2 = x^2 + y^2$. The Lagrangian is: $$\mathcal{L}=\frac{1}{2}m\vec{v}\cdot\vec{v}+q\vec{v}\cdot\vec{A}$$ Where $\vec{A}$ is the vector potential

$$\vec{A} = \frac{M}{r^3}\hat{M}\times\vec{r}=\frac{M}{r^3}\hat{z}\times(x\hat{x}+y\hat{y}+z\hat{z})=\frac{M}{r^3}(-y\hat{x}+x\hat{y})=\frac{M\rho}{r^3}(-\frac{y}{\rho}\hat{x}+\frac{x}{\rho}\hat{y}) = \frac{M\rho}{r^3}\hat{\phi}$$ and $r^2 = \rho^2 + z^2$. Therefore, the Lagrangian is: $$\mathcal{L}=\frac{1}{2}m\rho^2\dot{\phi}^2+q\rho\dot{\phi}A_{\phi}+f(v_{\rho},v_{z})$$ So the generalized angular momentum is: $$p_{\phi}=\frac{\partial\mathcal{L}}{\partial \dot{\phi}}=m\rho^2\dot{\phi}+q\rho A_{\phi} = m\rho v_{\phi}+\frac{qM\rho^2}{r^3}$$ Does this seem right?

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