Changing magnetic field (magnetic mirror)

[Kara386]

Homework Statement

I think this is a 'magnetic mirror' question - the field lines converge on the z axis.

For a particle moving into an area of increasing field strength, where the field lines converge. Assume the $B$ field is rotationally symmetric about the $z$ axis, with $B = B_z \hat{z}$ being the main field, $B_r$ pointing inwards and $B_φ = 0$ (cylindrical coordinates). Use the fact that this field has $div B = 0$ to find $B_r$, and show that $\frac{v_p^2}{B}$ is a conserved quantity by considering the Lorentz force.

Homework Equations

The Attempt at a Solution

I'm a little rusty on cylindrical co-ords, but I think the field can be written as
$B = B_r \hat{r} + B_z \hat{z}$
In cylindrical co-ordinates the divergence of the field is (using the fact it's divergence free):
$\frac{1}{r}\frac{\partial{(r B_r)}}{\partial r} + \frac{\partial B_z}{\partial z} = 0$

With a converging field clearly $B_z = B(z)$, and using the product rule on the first term:

$\frac{\partial B_r}{\partial r} + B_r + \frac{\partial B_z}{\partial z} = 0$

Which gives

$B_r = -\frac{\partial B_r}{\partial r} - \frac{\partial B_z}{\partial z}$

The equation of motion for this system is

$m\frac{dv}{dt} = q(v \times B)$
Which can be calculated from the determinant

$m\frac{dv}{dt} = q \left| \begin{array}{ccc} \hat{r} & \hat{\theta} & \hat{z} \\ v_r & v_{\theta} & v_z \\ B_r & 0 & B_z\end{array} \right|$

$= qv_{\theta} B_z \hat{r} - \hat{\theta} q(vB_z - v_zB_r) - \hat{z} q v_{\theta} B_r$

Can't see how to show from there that the quantity $\frac{v_p^2}{B}$ is conserved. Not even sure that determinant is right! Thanks for any help! :)

 

[Fred Wright]

First, your expression for $B_r$ is incorrect. You dropped a factor of r. Since the change in the field with respect to r is much greater than the change in the field with respect to z you can show that $B_r$ is approximately proportional to r.
Secondly, look up the expressions for velocity in cylindrical coordinates and plug them into your determinant.
Third, for $\frac{{v_p}^2}{B}$ to be a conserved quantity it must be a constant of motion. That is, $$ \frac{d({\frac{{v_p}^2}{B}})}{dt}=0$$

 

[Kara386]

First, your expression for $B_r$ is incorrect. You dropped a factor of r. Since the change in the field with respect to r is much greater than the change in the field with respect to z you can show that $B_r$ is approximately proportional to r.
Secondly, look up the expressions for velocity in cylindrical coordinates and plug them into your determinant.
Third, for $\frac{{v_p}^2}{B}$ to be a conserved quantity it must be a constant of motion. That is, $$ \frac{d({\frac{{v_p}^2}{B}})}{dt}=0$$

Yes, I made lots of mistakes, I got there in the end though. Thanks anyway for your response! :)