Lagrangian, particle/magnetic need solution check

[AbigailM]

Preparing for classical prelim, just wondering if this solution is correct.

Homework Statement

A particle with mass m and charge q moves in a uniform magnetic field $\boldsymbol{B}=B\boldsymbol{\hat{z}}$. Write a Lagrangian describing the motion of the particle in the xy plane that gives the correct Lorentz-force equation of motion,
$m\mathbf{a}=q\mathbf{v}\times\mathbf{B}$

Homework Equations

L=T - U

$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)=\frac{\partial L}{\partial q}$

$\boldsymbol{B}=\boldsymbol{\nabla \times A}$

$\boldsymbol{E}=-\boldsymbol{\nabla}\phi +\frac{\partial \boldsymbol{A}}{\partial t}$

The Attempt at a Solution

I'm going to start with the complete lorentz force and remove the electric potential from the lagrangian later.

$\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B})$

$\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times (\boldsymbol{\nabla \times A}))$

$\mathbf{F}=q(\mathbf{E}+\boldsymbol{\nabla} (\mathbf{v.A})-\boldsymbol{A}(\boldsymbol{\nabla . v}))$

$\mathbf{F}=q(-\boldsymbol{\nabla}\phi+\frac{\partial\boldsymbol{A}}{\partial t}+\boldsymbol{\nabla}(\boldsymbol{v.A})-\boldsymbol{A}(\boldsymbol{\nabla v}))$

$U=q(\phi -\boldsymbol{v.A})=q\phi-q\boldsymbol{v.A}$

We are only interested in the magnetic field so we'll ignore $q\phi$.

$U=q\boldsymbol{v.A}$

$L=\frac{1}{2}mv^{2} + q\boldsymbol{v.A} \hspace{2 mm}\mathbf{:Answer}$

As always thanks for the help!

 

[gabbagabbahey]

$\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times (\boldsymbol{\nabla \times A}))$

$\mathbf{F}=q(\mathbf{E}+\boldsymbol{\nabla} (\mathbf{v.A})-\boldsymbol{A}(\boldsymbol{\nabla . v}))$

This is correct, but you should explain how you got it (i.e that you used a specific vector product rule and that 2 of the terms were zero and why) on your exam if you want full marks.

$\mathbf{F}=q(-\boldsymbol{\nabla}\phi+\frac{\partial\boldsymbol{A}}{\partial t}+\boldsymbol{\nabla}(\boldsymbol{v.A})-\boldsymbol{A}(\boldsymbol{\nabla v}))$

$U=q(\phi -\boldsymbol{v.A})=q\phi-q\boldsymbol{v.A}$

Your choice of generalized potential is not immediately obvious when looking at this. How did you choose $U$ from this when there are 2 other terms present that depend on your generalized coordinates, momenta and time? There is a step missing in between which will make the choice of $U$ much easier to argue.

$L=\frac{1}{2}mv^{2} + q\boldsymbol{v.A} \hspace{2 mm}\mathbf{:Answer}$

As always thanks for the help!

Did you actually show that this choice of Lagrangian gives the correct force law?

I only see an incomplete motivation for choosing such a Lagrangian, and no calculations to show that it is indeed a correct choice.