Particle motion in a magnetic field¶
Before treating a plasma as a fluid, it helps to see how a single charged particle moves. The equation of motion is the Lorentz force law:
Gyration¶
In a uniform magnetic field with \(\mathbf{E}=0\), a particle circles the field line at the gyrofrequency (cyclotron frequency)
with a gyroradius (Larmor radius)
Motion along \(\mathbf{B}\) is unaffected, so the full trajectory is a helix. Ions and electrons gyrate in opposite directions, and because \(m_i \gg m_e\), ion gyroradii are much larger than electron gyroradii.
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Drifts: when something else pushes¶
Question: in a uniform magnetic field a particle just circles. What makes it wander across the field?
Split the motion into a fast circle (the gyration) and the slow motion of the circle's centre, the guiding centre. Anything that makes the gyroradius larger on one side of the orbit than the other stops the circle from closing, so each orbit ends a little to one side of where it began: the guiding centre drifts.
The general rule: a force drift¶
Suppose a constant force \(\mathbf{F}\) acts perpendicular to \(\mathbf{B}\). Move into a frame travelling at velocity \(\mathbf{v}_F\). If, in that frame, the magnetic force \(q\,\mathbf{v}_F\times\mathbf{B}\) exactly cancels \(\mathbf{F}\), the particle just gyrates there. That requires
(check: \(q\,\mathbf{v}_F\times\mathbf{B} = (\mathbf{F}\times\mathbf{B})\times\mathbf{B}/B^2 = -\mathbf{F}\) for \(\mathbf{F}\perp\mathbf{B}\)). The drift is perpendicular to both the force and the field. Every drift below is this formula with a different force. The key question for each one is whether \(\mathbf{F}/q\) depends on the charge: if it does, ions and electrons drift in opposite directions and the drift carries an electric current.
See the drifts¶
The orbits below come from solving \(m\,d\mathbf{v}/dt = q(\mathbf{E}+\mathbf{v}\times\mathbf{B}) + \mathbf{F}\) numerically. Use the buttons to switch between drifts, and drag to rotate.
Drag to rotate · scroll to zoom · double-click to reset
Units are normalised (ion gyrofrequency = 1). The electron mass is set to 1/4 of the ion mass so that both orbits are visible (the real ratio is 1/1836), and both species have the same kinetic energy.
E×B drift¶
The force is \(q\mathbf{E}\), so
Why: on the half of the orbit where the particle moves along \(\mathbf{E}\) it gains speed and its gyroradius grows; on the other half it slows and the radius shrinks. The charge cancels out of \(\mathbf{F}/q\), so ions and electrons drift together, at the same speed, and no current flows. The whole plasma moves.
Why this matters for reconnection
The E×B drift is the single-particle version of the statement that the plasma and the magnetic field move together. In reconnection this breaks down inside a thin layer, first for ions (on the scale of the ion inertial length) and then for electrons.
Gravity drift¶
The force is \(m\mathbf{g}\), so \(\mathbf{v}_g = m\,\mathbf{g}\times\mathbf{B}/(qB^2)\). Now \(\mathbf{F}/q\) contains the charge and the mass: ions and electrons drift in opposite directions, and the ions much faster. The result is a current. Gravity drifts are tiny in most of space, but they drive currents in the equatorial ionosphere.
Grad-B drift¶
If \(|\mathbf{B}|\) is stronger on one side, the gyroradius \(r_L = mv_\perp/(|q|B)\) is smaller there. The orbit is tighter on the strong-field side and wider on the weak-field side, so it does not close:
It depends on the sign of \(q\), so ions and electrons drift oppositely: a current. Around Earth, where \(B\) grows towards the planet, this drift carries ions westward and electrons eastward, forming the ring current that weakens the surface field during geomagnetic storms.
Curvature drift¶
A particle sliding along a curved field line at speed \(v_\parallel\) feels a centrifugal force \(m v_\parallel^2/R_c\) outward from the centre of curvature. Using that force in the general rule:
where \(\mathbf{R}_c\) points outward from the centre of curvature to the particle, and \(\boldsymbol{\kappa} = (\hat{\mathbf{b}}\cdot\nabla)\hat{\mathbf{b}} = -\mathbf{R}_c/R_c^2\) is the magnetic field curvature vector pointing inward toward the centre of curvature. Again ions and electrons drift oppositely. In a current-free field, curved field lines always come with a stronger field on the inside of the bend (see Bending flux tubes), so the curvature and grad-B drifts add up:
Magnetic mirror¶
Not a sideways drift, but a push along the field. A gyrating particle behaves like a small current loop with magnetic moment \(\mu = mv_\perp^2/(2B)\), and \(\mu\) stays constant as long as the field changes slowly over one orbit. If the particle moves into a stronger field, \(v_\perp^2\) must grow in proportion to \(B\); its total energy is fixed, so \(v_\parallel\) must shrink. If \(B\) grows enough, \(v_\parallel\) reaches zero and the particle is reflected. A particle whose velocity makes an angle \(\alpha\) (its pitch angle) with \(\mathbf{B}\) where the field is weakest, \(B_{\min}\), turns back where
Particles bounce between two such mirror points: this traps the Van Allen radiation belts in Earth's field and the energetic particles in solar coronal loops.
Polarization drift¶
If \(\mathbf{E}\) changes in time, so does the E×B drift, and the particle must be accelerated to keep up. Its inertia makes it lag, giving a drift along \(\mathbf{E}\):
It is proportional to the mass, so ions dominate, and it carries a current (the polarization current). It is what lets a plasma respond to changing electric fields, for example in Alfvén waves.
Summary¶
| Drift | Force | Velocity | Ions and electrons | Current? | Where it matters |
|---|---|---|---|---|---|
| E×B | \(q\mathbf{E}\) | \(\mathbf{E}\times\mathbf{B}/B^2\) | Same direction, same speed | No | Plasma convection; reconnection inflow |
| Gravity | \(m\mathbf{g}\) | \(m\,\mathbf{g}\times\mathbf{B}/(qB^2)\) | Opposite | Yes | Equatorial ionosphere |
| Grad-B | \(-\mu\nabla B\) | \(\dfrac{mv_\perp^2}{2qB}\dfrac{\mathbf{B}\times\nabla B}{B^2}\) | Opposite | Yes | Ring current |
| Curvature | \(mv_\parallel^2/R_c\) outward | \(\dfrac{mv_\parallel^2}{qB}\dfrac{\mathbf{R}_c\times\mathbf{B}}{R_c^2B}\) | Opposite | Yes | Ring current, magnetotail |
| Polarization | inertia, \(-m\,d\mathbf{v}_E/dt\) | \(\dfrac{m}{qB^2}\dfrac{d\mathbf{E}_\perp}{dt}\) | Opposite, ions faster | Yes | Waves, time-varying fields |
| Mirror (along B) | \(-\mu\nabla_\parallel B\) | Reflects at \(B = B_{\min}/\sin^2\alpha\) | Both trapped | (bounce motion) | Radiation belts, coronal loops |
Answer
A particle drifts across the field whenever a force pushes it perpendicular to \(\mathbf{B}\); the drift is \(\mathbf{F}\times\mathbf{B}/(qB^2)\). An electric field moves all particles together (no current). Every other force here (gravity, field gradients, field-line curvature, inertia) moves ions and electrons in opposite directions and so creates currents, which is how a plasma carries current across a magnetic field.
Practice problems¶
Try each problem before opening its solution. Click an option to check a concept question;
type a number and press Check for a numerical one (forms like 6.2e7 or 6.2x10^7 both work).
Concept questions¶
Problem 1
In crossed electric and magnetic fields, how do the E×B drifts of an ion and an electron compare?
- Same direction and same speed
- Opposite directions, same speed
- Same direction, but ions are faster
- Same direction, but electrons are faster
\(\mathbf{v}_E = \mathbf{E}\times\mathbf{B}/B^2\) contains neither the charge nor the mass. The whole plasma drifts together, so no current flows.
Problem 2
Ions and electrons have the same temperature. Roughly how does the proton gyroradius compare with the electron gyroradius?
- About 43 times larger (\(\sqrt{m_p/m_e}\))
- About 1836 times larger
- The same
- About 43 times smaller
\(r_L = mv_\perp/(|q|B)\) and at equal temperature \(v_\perp\propto m^{-1/2}\), so \(r_L\propto\sqrt{m}\): \(\sqrt{1836}\approx 43\).
Problem 3
A charged particle moves in a uniform magnetic field with no electric field. What happens to its speed?
- It stays constant
- It increases
- It decreases
- It oscillates
The magnetic force \(q\mathbf{v}\times\mathbf{B}\) is always perpendicular to \(\mathbf{v}\), so it changes the direction of motion but does no work.
Show that…¶
Problem 4: magnetic forces do no work
Show that the magnetic force does no work on a charged particle.
Hint
Work per unit time is \(\mathbf{F}\cdot\mathbf{v}\).
Proof
\(\mathbf{F}\cdot\mathbf{v} = q(\mathbf{v}\times\mathbf{B})\cdot\mathbf{v} = 0\), because \(\mathbf{v}\times\mathbf{B}\) is perpendicular to \(\mathbf{v}\). So \(d(\tfrac12mv^2)/dt = 0\): the speed is constant.
Problem 5: the E×B frame
For \(\mathbf{E}\perp\mathbf{B}\), show that in a frame moving at \(\mathbf{v}_E = \mathbf{E}\times\mathbf{B}/B^2\) the electric field vanishes, \(\mathbf{E}' = \mathbf{E} + \mathbf{v}_E\times\mathbf{B} = 0\).
Hint
Use \((\mathbf{a}\times\mathbf{b})\times\mathbf{c} = \mathbf{b}(\mathbf{a}\cdot\mathbf{c}) - \mathbf{a}(\mathbf{b}\cdot\mathbf{c})\).
Proof
\(\mathbf{v}_E\times\mathbf{B} = \dfrac{(\mathbf{E}\times\mathbf{B})\times\mathbf{B}}{B^2} = \dfrac{\mathbf{B}(\mathbf{E}\cdot\mathbf{B}) - \mathbf{E}B^2}{B^2} = -\mathbf{E}\), since \(\mathbf{E}\cdot\mathbf{B}=0\). So \(\mathbf{E}' = 0\): in the drifting frame the particle just gyrates.
Numbers¶
Problem 6
What is the proton gyrofrequency \(f = \Omega_p/2\pi\) in the solar wind, where \(B = 5\) nT?
Solution
\(\Omega_p = eB/m_p = 0.479\) rad/s, so \(f = 0.076\) Hz: one gyration every 13 seconds.
Problem 7
A solar-wind proton has \(v_\perp = 50\) km/s in \(B = 5\) nT. What is its gyroradius, in km?
Solution
\(r_L = v_\perp/\Omega_p = 5\times10^4/0.479\approx 1.04\times10^5\) m \(= 104\) km.
Problem 8
What is the E×B drift speed for \(E = 1\) mV/m and \(B = 5\) nT, in km/s?
Solution
\(v_E = E/B = 10^{-3}/(5\times10^{-9}) = 2\times10^5\) m/s \(= 200\) km/s.
Problem 9
What is the electron gyrofrequency \(f_{ce}\) in the same 5 nT field?
Solution
\(f_{ce} = eB/(2\pi m_e)\approx 28\) Hz/nT \(\times\,5\) nT \(= 140\) Hz, 1836 times the proton value.
More on drifts¶
Problem 10
Which of these drifts does not produce an electric current?
- Grad-B drift
- Curvature drift
- E×B drift
- Gravity drift
In \(\mathbf{v}_F = \mathbf{F}\times\mathbf{B}/(qB^2)\) with \(\mathbf{F} = q\mathbf{E}\), the charge cancels. Ions and electrons drift together, so there is no net current.
Problem 11
Why does a particle drift in a magnetic field whose strength varies across it?
- The particle speeds up in the stronger field
- Its gyroradius is smaller on the strong-field side, so the orbit does not close
- The field lines are curved
- An electric field appears
\(r_L\propto 1/B\): the orbit is tighter where \(B\) is strong and wider where it is weak. Its speed is unchanged (a magnetic field does no work).
Problem 12
A 10 keV proton (all its energy perpendicular to \(\mathbf{B}\)) sits where \(B = 100\) nT, and \(B\) changes over a length \(L_B = |B/\nabla B| = 6400\) km (one Earth radius). What is its grad-B drift speed, in km/s?
Solution
\(v_{\nabla B} = \dfrac{mv_\perp^2}{2qBL_B} = \dfrac{W_\perp}{qBL_B} = \dfrac{10^4\ \text{V}}{10^{-7}\ \text{T}\times6.4\times10^6\ \text{m}} = 1.56\times10^4\) m/s \(\approx 15.6\) km/s. A storm-time ring-current proton circles Earth in a few hours.
Problem 13
A magnetic bottle has a mirror ratio \(B_{\max}/B_{\min} = 4\). Particles with pitch angles smaller than some angle at \(B_{\min}\) escape through the ends (the loss cone). What is that angle, in degrees?
Solution
A particle is reflected before \(B_{\max}\) if \(B_{\min}/\sin^2\alpha < B_{\max}\), i.e. \(\sin^2\alpha > 1/4\). So the loss cone is \(\sin\alpha < 1/2\), \(\alpha < 30°\).