- #1

- 208

- 2

## Homework Statement

Given the solar wind has velocity ##450## km s##^{−1}##, i.e. it is slow solar wind, and if the heliopause is ##150## AU from the Sun, how many turns are wound into each ﬁeld line between the Sun and the heliopause? Assume the magnetic field is ##6##nT, and the rotational period of the Sun is ∼27 days.

## Homework Equations

## The Attempt at a Solution

Not sure where the orbital period of the sun comes into it. And I assume the ##450## km s##^{−1}## are the velocity of solar wind particles radially, but I'm not 100% on that.

What I thought was if it takes ##\frac{150AU}{450 \times 10^3 ms^{-1}} = 5 \times 10^7s## to get to the heliopause, then in that time since a particle in the magnetic field will move in a helix with frequency ##\frac{eB}{m}##, and with a frequency and a time I could work out how many times a particle would loop the field line.

Problem is I haven't been give a mass and I feel like the number of turns in the field line shouldn't depend on the mass, it should be a property of the field. Even if that's not true and it should depend on particle type, I haven't been given a particle type to go look up the mass of. And this answer doesn't involve the rotational period of the sun. So I'm stuck! Thanks for any help! :)

Last edited: