Why would charged particles travel to the poles of a planet given the magnetic field is like a bar magnet's? I've been told that the magnetic field of a planet (say, Earth) causes charged particles traveling towards it to go toward the poles. Why? 
$$
\mathbf{F_m} =q(\mathbf{v} \times \mathbf{B})
$$
The above would say that given a magnetic field that points towards the poles, the charged particles experience a force in a direction perpendicular to the direction of the poles. And I also just can't imagine when the magnetic field lines we're usually shown would lead a particle to the poles. Would anyone mind explaining this?
 A: You're right, the charged particles don't experience a force directed toward the poles. So, a particle that is initially traveling exactly perpendicular to the magnetic field would just spin tight circles around the field lines. But there are virtually no particles that impact our field lines exactly perpendicularly. Nearly all of them have some component of velocity that is parallel to the field lines. Since there is no force in that direction, the particle keeps moving parallel to the field line, making tight circles around the line as it does so. We often say that fast-moving charged particles are "pinned" to the field lines because of this. All field lines hit the ground near the poles, so the particles, following the field lines, hit the poles.
A: The magnetic field for a dipole moment $\vec m$
$$
{\vec B}(\vec m,\vec r)~=~\frac{\mu_0}{4\pi}\frac{3(\hat m\cdot\hat r)\hat r~-~\vec m}{r^3}.
$$
The magnetic moment we orient along the $z$ axis and $\vec r$ is the radial vector to a point in space. For a charged mass at that point in space moving with velocity $\vec v$ we have of course the Lorentz force $\vec F~=~q\vec v\times\vec B$. The motion is going to be a bit complicated, but we can look at the relevant part of the magnetic field which is $3(\hat m\cdot\hat r)\hat m~-~\vec m$ Let us assume the particle has a trajectory perpendicular to the magnetic moment. We then clearly have a component $\vec m\times\vec v$ to the force. The magnitude of this force is then going to be determined by the $\vec m\cdot\hat r$. So if the particle is on the equatorial plane of the dipole the force is smaller than if it is coming in at a plane tangent to the poles. Also note there is a difference in sign.
While the motion is a bit complicated, there are from the above some qualitative features we can see. A charged particle incident on the Earth's magnetic field will have a Lorentz force inducing circular motion around the magnetic field lines. However, the sign of the force can change. What can happen is that the charged particle will get close to the pole and the trajectory will reverse and wind around the magnetic field until it reaches the other pole. The charged particle is in a sort of magnetic bottle. Also as the particle approaches the poles the field density increases which increases the frequency of its motion. This circular acceleration can produce Brehmsstralung radiation that is picked up on low frequency radio tuners as whistles.
This is also why the van Allen belts are these torus shaped regions around the Earth. Jupiter has much larger ones. The charged particles in these belts are trapped in a magnetic bottle. With a solar Coronal Mass Ejection (CME) that dumps lots of charged particles into the Earth's magnetic field, some of these particles wind around close enough to the poles to cause auroral lights 
