# Homework Help: Magnetic Field Problem - Tricky Wording.

1. Nov 28, 2007

### PFStudent

1. The problem statement, all variables and given/known data

21. An electron follows a helical path in a uniform magnetic field of magnitude $0.300{{.}}T$. The pitch of the path is $6.00{{.}}{\mu}m$, and the magnitude of the magnetic force on the electron is ${2.00{\times}{{10}^{-15}}}{{.}}N$. What is the electron's speed?

2. Relevant equations

$${\vec{F}_{E}} = {q{\vec{E}}}$$

$${\Delta{V}_{p}} = {{-}{\int_{r_{0}}^{r_{1}}}{\vec{E}_{p1}(r)}{\cdot}{d{\vec{r}}}}$$

$${\vec{E}} = {{-}{\nabla}{V(r)}}$$

$${\vec{E}} = {{-}{\frac{\partial}{\partial{r}}}{\left[{V(r)}\right]}{\hat{r}}}$$

$${\vec{F}_{B}} = {q{\vec{v}}{\times}{\vec{B}}}$$

3. The attempt at a solution

This problem isn't so hard to solve, as intrepreting the information the problem is giving.

Like, when the problem says,

What exactly do they mean by "pitch of the path," and why are they giving it as a measure of distance (ie. ${m}$)?

In addition, they mention that the electron, "follows a helical path," by helical path--they're basically saying an upward spiral path, is that correct?

Also, since a helical path has some curvature to it, does the centripetal force apply to this problem? (even though a helical path is not the same as going around in a cricle)

And if the centripetal force does apply, why?

Any help is appreciated.

Thanks,

-PFStudent

Last edited: Nov 28, 2007
2. Nov 29, 2007

### PFStudent

Hey,

Yea..., so does anyone have any ideas on this?

Thanks,

-PFStudent

Last edited: Nov 29, 2007
3. Nov 30, 2007

### PFStudent

Hey,

Yea, still a little stuck on this,..does anyone know how the wording in this problem is supposed to be interpreted?

Thanks,

-PFStudent

4. Nov 30, 2007

### Staff: Mentor

The pitch of a helix is the distance along the axis for one complete turn. If it had a pitch of zero, it would just be a circle.

It's a spiral.

Sure it will have a centripetal force, since it will have an acceleration perpendicular to the helix axis. But you don't need to worry about the details to solve this problem, since you're given the force and the magnetic field strength.