Why do charged particles follow magnetic field lines?

Question

This may seem like a dumb question, but I can't think of the answer. The context I am curious about is the solar wind. Specifically particles flowing out of coronal holes and following the magnetic field lines arching out into space. Why do both the positive and negative particles follow these field lines and flow into space?

Answer 1

Background

In the absence of an electric field, a charged particle experiences a force that is perpendicular to the magnetic field and its velocity relative to that field, called the Lorentz force. This is given by: $$ \mathbf{F}_{s} = q_{s} \ \mathbf{v}_{s} \times \mathbf{B} \tag{1} $$ where $q_{s}$ is the charge of species $s$, $\mathbf{v}_{s}$ is the velocity of species $s$ with respect to the magnetic field, $\mathbf{B}$. We can approximate the plasma as a fluid, i.e., magnetohydrodynamics, which will help simplify some things.

In the fluid limit, we can show that Ohm's law, in a frame of reference moving relative to the fluid at velocity $\mathbf{U}$ (or equivalently, one can say we are at rest and the fluid moves at $\mathbf{U}$), is given by: $$ \mathbf{j} = \sigma \left( \mathbf{E} + \mathbf{U} \times \mathbf{B} \right) \tag{2} $$ where $\mathbf{j}$ is the current density and $\sigma$ is the electrical conductivity. Note that this is in the non-relativistic limit, where we can approximate by using gallilean transformations (e.g., $\mathbf{E}' = \mathbf{E} + \mathbf{U} \times \mathbf{B}$) instead of Lorentz transformations. We can then use Ampère's law combined with Ohm's law (e.g., Equation 2 above) to show: $$ \nabla \times \mathbf{B} = \mu_{o} \ \sigma \left( \mathbf{E} + \mathbf{U} \times \mathbf{B} \right) \tag{3} $$ where $\mu_{o}$ is the permeability of free space. We can take the curl of Equation 3 and combine with Faraday's law to show: $$ \partial_{t} \mathbf{B} = \nabla \times \left( \mathbf{U} \times \mathbf{B} \right) + \frac{1}{\mu_{o} \ \sigma} \nabla^{2} \mathbf{B} \tag{4} $$

In the limit as $\sigma \rightarrow \infty$, one can further show that: $$ \frac{d \Phi_{B}}{dt} = \int_{S} \ dA \ \left[ \partial_{t} \mathbf{B} - \nabla \times \left( \mathbf{U} \times \mathbf{B} \right) \right] \cdot \hat{\mathbf{n}} = 0 \tag{5} $$ where $\Phi_{B}$ is the magnetic flux and $dA$ is an arbitrary surface with unit normal vector, $\hat{\mathbf{n}}$. Equation 5 is known as the frozen-in condition because it states that the magnetic fields are tied to the fluid.

Answer

Why do both the positive and negative particles follow these field lines and flow into space?

The answer is two fold relating to the frozen-in condition being partially satisfied and the Lorentz force (Equation 1 above). Equation 1 shows that if a particle tries to move orthogonal to a magnetic field, the field will turn it back towards the field resulting in roughly circular motion (when considering only magnetic fields and only the perpendicular components of the velocity). There are intuitive reasons for why this should be so, as I previously stated in this answer.

If, however, a particle's velocity ($\mathbf{v}_{s}$) is parallel to the magnetic field, it will experience no force in the absence of electric and gravitational fields. Thus, it is very easy for particles to move along the magnetic field but difficult to move across it.

The frozen-in condition also shows that the field and particles are tied to each other, such that if one changes the other changes to compensate (assuming the changes are slow enough).

Answer 2

Start off with the Lorentz force of a magnetic field on a moving charged particle $$\vec{F} = Q (\vec{v} \times \vec{B})$$

Notice that this force is always directed perpendicularly to the charged particle velocity and to the magnetic field.

So, consider a particle moving in the direction of the field lines - it must feel no force (since $\vec{v} \times \vec{B} = 0$) and therefore no acceleration one way or another. It is free to move along the field lines.

Now consider a particle moving at right angles to the field lines. It experiences a force of magnitude $QvB$ at right angles to its velocity and at right angles to the magnetic field. The force leads to a centripetal acceleration. The particle trajectory is curved into a circular motion around the field lines. The speed of the particle is unchanged and the radius of the circular motion will decreases as the magnetic field gets stronger. $$ r = \frac{mv}{|Q|B}$$

Thus for small velocities/large fields, the particle will be confined to the field lines on a very tight radius, but free to move along them. Thus the particle will describe a helical motion.

This is what we mean by particles being "tied" to the field lines.

Changing the sign of the charge does not alter the "gyroradius", but simply changes the direction in which the particles will circle. The force is zero along the field line direction for particles charged with either sign.

Answer 3

In simple terms:

As lemon commented: They don't follow along the field lines per se, they spiral around the field lines. The charge affects whether they spiral in a clockwise or counter-clockwise direction.

To be a little more precise, as long as a particle has an initial speed along the field direction, the resulting trajectory is a spiral. When a spiral is long enough, when viewed from far enough, it looks like a line, so it looks like the particles follow the field lines. And where you can't distinguish the spiral from a line, you can also not distinguish whether the spiral is "clockwise" or "counterclockwise".

This is the case with the solar wind, because most particles do have a speed along the field direction, and those that don't soon will (because they get hit by others).

However, when a particle has no initial speed in the field direction, the trajectory becomes a circle.

The reason why the trajectory is a spiral lies in the right-hand rule as applied to electromagnetic fields.