Potential energy of a charge in a magnetic quadrupole field

Question

I have a charged particle of charge $q$ that moves with velocity $\vec{V}$ from a position $\vec{r}$, inside a magnetic quadrupole field of the form: $$\vec{B}=B_{0}(x,y, -2z)$$ The Lorentz force acts upon this particle: $$\vec{f}_{Lorentz}=q\vec{V}\times\vec{B}=qB_0[(-2V_{y}z-V_zy)\hat{x}+(V_zx+2V_xz)\hat{y}+(V_xy-V_yx)\hat{z}]$$ I know the trajectory and velocity of the particle as a function of time $(\hat{r}(t)$, $\hat{V}(t))$, but I can't manage to find the potential of this force. I suppose that I should use the integral $V_{Lorentz}=-\int{\vec{F}_{Lorentz}\dot{}d\vec{r}}$, But i'm not sure how or if maybe I could get a closed expression for this potential.

Thanks!

Answer 1

The force is not necessarily conservative:

$$\nabla\times\left(V\times B\right)=V\overbrace{\nabla\cdot B}^{=0}-B\nabla\cdot V + B\cdot\nabla V - V\cdot\nabla B \neq 0$$

$\vec{V}$ must make the above zero: otherwise the force may not be written with a scalar potential.

Answer 2

The Lorentz force is at right angles to the velocity and does no work. That would suggest that there is no change in the potential energy.