Conversion factor when converting from SI to natural units

[Dex_]

Homework Statement

I'm trying to answer this question from my particle physics lab, I can obtain the equation quite easily by using the definition of the force as rate of change of momentum and equating it to the Lorentz force, but I cannot get the factor of 0.3 from unit conversions. I've also attached the experimental diagram of the problem.

"Show that, for a dipole magnetic field in y direction and an initial momentum in
z direction, the change in the x component of the momentum in units of GeV/c is given
by

$$\Delta p_{x} = 0.3 q \int B_{y} dz$$

with $q$ the particle charge in units of e. You may assume that $p_{z} \gg p_x$."

p.s - sorry if my post is formatted wrong this is my first homework post, pleass point out my mistakes in the replies.

Relevant Equations

$c = 3 \times 10^{8} m \, s^{-1}$
$GeV = 1.602 \times 10^{-10} J$

For the right hand side I expressed each of the quantities in SI units,


$$C \cdot \frac{kg}{A \cdot s^{2}} \cdot m$$


Then substitute newtons in as $N = kg \cdot ms^{-2}$ and we get,


$$C \cdot \frac{N}{A \cdot m} \cdot m$$


Using the definition of work as $J = N \cdot m$ and Amperes as $C \cdot s^{-1}$ we get,


$$ C \cdot \frac{J}{C \cdot s^{-1} \cdot m^{2}} \cdot m $$


Then after cancelling we are left with,


$$\frac{J}{m \, s^{-1}}$$


To convert from Joules to GeV we divide by $1.602 \times 10^{-10}$ and to get it into units of c we multiply by $3 \times 10^{8}$.


However the factor I am left with is,


$1.875 \times 10^{18}$


I can actually see how we could get the 0.3 if we divide $3 \times 10^{8}$ by $10^{9}$, but I don't see how to get there. I also don't see how the units of charge in terms of $e$ helps if charge is being cancelled out anyways.

 

[TSny]

Note that $q$ is the charge in units of $e$.