Electron's Path Through Solenoid's Magnetic Field

[mbrmbrg]

Homework Statement

An electron is shot into one end of a solenoid. As it enters the uniform magnetic field within the solenoid, its speed is 800 m/s and its velocity vector makes an angle of 30° with the central axis of the solenoid. The solenoid carries 4.2 A and has 8000 turns along its length. How many revolutions does the electron make along its helical path within the solenoid by the time it emerges from the solenoid's opposite end? (In a real solenoid, where the field is not uniform at the two ends, the number of revolutions would be slightly less than the answer here.)

Homework Equations

$$B=\mu_0in$$ (For a solenoid)
$$qv\times B=\frac{mv^2}{r}$$
# revolutions=length of solenoid/pitch of path

The Attempt at a Solution

I've combined and simplified my first two equations. However, I still have an undefined n (I've been given total # loops rather than loops/unit length) and r floating around. I'm hoping that my third equation will help me out, but I don't know how to calculate pitch; all I know is that pitch is determined by the parallel component of the velocity.

What's the formula for pitch?

My work leading up to my current equation:
$$qvB\sin{30^o}=\frac{mv_{\perp}^2}{r}$$
$$qv\mu_0in\sin{30^o}=\frac{mv^2\sin^2{30^o}}{r}$$
$$e\mu_0in=\frac{mv\sin{30^o}}{r}$$

 

[mbrmbrg]

Some further thoughts (over dinner, no less!)

solenoid's length = (v_parallel)(t)
n = N/length = 8000/(v cos30)(t)

total revolutions = (frequency)(time) = (2pi r)(t)/(v cos30)


So I now I need time rather than length...

edit: (frequency)(time) = [(v*cos30)/(2*pi*r)]*t

 

[mbrmbrg]

Another edit to last post: that should be sin30 rather than cos 30.

But no matter! My professor told me how to combine my two eqations so the radii magically cancel (no time necessary, solve one of my earlier equations for L).

JOY! JUBILATION! Why is there no emoticon that sings and dances for the sheer joy of life?