Lorentz force - particle in an odd magnetic field

[Toby_phys]

Homework Statement

Particles of mass $m$ and charge $q$ are initially traveling in a beam along the $z$ direction with speed $v$ when they enter a long magnetic quadrupole lens, where there is no E-field and the magnetic flux density is $B = Ay\hat{i} + Ax\hat{j}$, and where A is a positive constant. Neglecting edge effects of the magnet, write down the equations of motion for one of the particles. Solve these equations under the assumption that the particle’s path always makes a small angle with the $z$-direction, and that the particle had $x = x_0$ and $y = y_0$ before entering the magnetic field.

Homework Equations

We have the Lorentz force in the absence of any electric fields:

$$
\mathbf{F}=q\mathbf{v\times B}
$$

We have Newtons 2nd law:

$$\mathbf{F}=m\mathbf{a}$$

The Attempt at a Solution

Equating the Lorentz force with Newtons second law:

$$
\begin{align}
m\mathbf{a} &= \begin{bmatrix}
V_x \\
V_y \\
V_z
\end{bmatrix}
\times
\begin{bmatrix}
Ay \\
Ax \\
0
\end{bmatrix}&=\begin{bmatrix}
-AxV_z \\
AyV_z \\
AxV_x-AyV_y
\end{bmatrix}
\end{align}
$$

Which gets us 3 differential equations:
$$ m \ddot{x}=-Ax\dot{z}$$
$$m \ddot{y}=Ay\dot{z} $$
$$m \ddot{z}=Ax\dot{x}-Ay\dot{y} $$I couldn't see how to progress from here. Any help would be appreciated. thank you.

 

[kuruman]

I think the key to solving this is here

... under the assumption that the particle’s path always makes a small angle with the z-direction,

In the first two equations you can approximate $\dot{z} \approx v_0$. In this approximation you have two decoupled differential equations which you can easily solve for $x(t)$ and $y(t)$. Put these back in the third equation which will take the form $m\ddot{z}=f(t)$.

Edited for language clarifications.