# Limits of the Path Length of B-field induced gyroradius

1. Sep 2, 2014

### thefireman

Hi,
I am trying to derive the path length of a charged particle in a B-field. I am assuming the particle will travel a distance L along the applied field. Using the following equations for the path length of a helix and gyroradius:
Helix defined as
$[a*cos(t),a*cos(t),b*t]$for t on [0,T] has a path length of
$P=\sqrt{(a^2+b^2)}$
$r=\frac{v_{perp}m^*}{qB}$
and assuming that bt must equal L at T=1 (parametrize helix from 0 to 1), I get the following expression:

$P=\sqrt{(\frac{v_{perp}m^*}{qB})^2+L^2}$

However, the bounds do not make sense. At zero field, It should simply travel in a straight line, i.e. P=L.
I am not sure about the "infinite" field, since it is oscillating more rapidly but with an ever decreasing radius. I could argue that the radius being 0 means only the vertical distance is traveled, or that it just rotates infinity at 0 radius?
In either case, I cannot get the BC for B=0, as this causes the expression to be infinite.

Can I simply not apply the gyroradius in a zero field condition, or did I derive this incorrectly? What is the limiting case of infinite field? Actually quite an interesting problem!

2. Sep 7, 2014