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In regards to the right hand rule, given Earth's electric and magnetic field, in which "direction" would a particle go?

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I don't really see the point of this question. You're familiar with the Lorentz force law $\mathbf{F} = q(\mathbf{E} + \mathbf{v}\times\mathbf{B})$, right? What are you asking about that you can't get from that equation? (Just to be clear, I'm not trying to be insulting or anything, I really just don't understand what you're trying to find out) –  David Z Apr 2 '11 at 1:55
Not familiar enough to answer the equation. Might you do an example equation with the applicable units of measurement (real numbers) for each variable? –  earls Apr 2 '11 at 4:08
if you're not familiar with the Lorentz force law, I'm not sure I can do justice to it in a comment. Are you actually asking how to determine the force on a particle given the fields $\mathbf{E}$ and $\mathbf{B}$ at the particle's location? In that case, I'd suggest that you edit your question to just ask that (don't make it involve the Earth's fields) and it will be a much better question. Also, have a look at this question and this Wikipedia article and see if they help clarify anything for you. –  David Z Apr 2 '11 at 4:16

1 Answer 1

The field lines of earth's magnetic field go from south pole to north pole. So they are parallel to the surface at the equator, and perpendicular at the (magnetic) poles (going vertically into the ground).

Now, what happens to a moving charge in a magnetic field? A particle moving parallel to the field lines will not feel any force, however when it's moving perpendicularly to the field, it will be deflected according to the right hand rule. Note that the direction of $\vec v$ will change, but not the magnitude.

(The source of this behavior comes from the cross product in the formulation of Lorentz' force: $F_L = q \, (\vec v \times \vec B)$.)

Say your thumb points into the direction of the field line, your index finger in the direction of the particle movement. Then it will be deflected in the direction of your middle finger. Now turn your hand such that your index finger points into the new direction. You will see that the particle will describe a circle around the field line.

An incoming particle, say an electron from the ion wind of the sun, will in general have both velocity components perpendicular and parallel to the field. It will be moved away while describing the circle, and thus fly on a corkscrew-like path.

By the way, this is also the source of the northern lights (aurora borealis):

  • Near the equator, the component of $\vec v$ pointing to the surface is perpendicular to the field lines, so that component is "trapped" in the circular motion. The other component parallel to the field lines leads the particle to the poles.

  • Near the poles, the component of $\vec v$ pointing to the surface is parallel to the field lines, so the particles can descend to the ground while describing circles.

Because of the circular motion, the particles a) need much longer to reach the ground, and b) loose energy due to syncroton radiation. In this way, earth's magnetic field protects us from certain types of cosmic radiation!

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though you have only examined the effect of the magnetic field. There exists an electric field too en.wikipedia.org/wiki/Natural_electric_field_of_the_Earth –  anna v May 5 '11 at 14:22
That's interesting. I wasn't aware of a sizeable electric field (besides what you get in thunderstorms). –  jdm May 5 '11 at 14:32
jdm, great answer, thank you very much. But having learning of Earth's electric field, how does it factor into Lorentz's equation concerning the velocity of the particle? Consider three particles: -1, 0, +1 charge. Please note if the north pole, south pole, and equator have a significant difference as before... Thanks! –  earls May 21 '11 at 14:32

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