Would small polarons in a ferrihydrite nanoparticle moving at velocity v in the geomagnetic field B experience a Lorentz force related to v cross B?

Question

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1. What the question is really asking

A ferrihydrite nanoparticle contains small polarons – essentially electrons that are “dressed” by a local lattice distortion and hop from one Fe site to another.
Suppose the whole particle drifts east‑or‑west with a speed v through the Earth’s magnetic field B (≈ 5 × 10⁻⁵ T).

Will those electrons feel a Lorentz force q (v × B) that makes them move (or drift) inside the particle?

In other words, does the motion of a neutral solid through the geomagnetic field produce a magnetic force on its internal charge carriers that is large enough to influence their transport?

2. Detailed step‑by‑step answer

2.1 The Lorentz force on a moving charge

For a particle of charge (q) moving with velocity (\mathbf{v}) in a magnetic field (\mathbf{B}),

[ \boxed{\mathbf{F}_L = q\,\mathbf{v}\times\mathbf{B}} \qquad (1) ]

Only the component of velocity perpendicular to (\mathbf{B}) matters; the magnitude is

[ F_L = |q|\,v_{\perp} B . ]

2.2 What velocity appears in (1)?

The velocity that matters is the velocity of the charge relative to the laboratory (inertial) frame.

Inside a solid we can split the charge’s motion into

[ \mathbf{v}{\text{lab}} = \mathbf{v}{\text{particle}} + \mathbf{v}_{\text{drift}} . ]

(\mathbf{v}_{\text{particle}}) – the translational velocity of the whole nanoparticle (the “east‑west” motion).
(\mathbf{v}_{\text{drift}}) – the velocity of the charge relative to the lattice, i.e. the hopping or conduction velocity of the polaron.

If the polarons are not conducting (no external electric field), (\mathbf{v}{\text{drift}}\approx 0).
Thus the only velocity that enters the Lorentz force is the bulk velocity (\mathbf{v}{\text{particle}}).

2.3 Charge neutrality of the particle

A ferrihydrite particle is electrically neutral overall.
Each electron that hops leaves behind a hole (a positive charge) on the Fe site from which it moved. The pair (electron + hole) is neutral, so the net charge of the particle is zero.

Consequences:

The total magnetic force on the particle as a whole is the sum of the forces on all its constituent charges.
For every electron feeling a force (-e\mathbf{v}\times\mathbf{B}) there is a corresponding positive hole feeling (+e\mathbf{v}\times\mathbf{B}). The forces cancel exactly if the electron and hole move together with the same bulk velocity.

Hence, no net mechanical force acts on the particle.

2.4 Force on an individual polaron

Even though the net force on the particle vanishes, we can still ask whether a single electron experiences a noticeable magnetic force that could bias its hopping direction.

Take typical numbers:

Quantity	Symbol	Typical value
Elementary charge	(e)	(1.60\times10^{-19}\,\text{C})
Particle speed (east‑west)	(v)	(10^{-2}\,\text{m s}^{-1}) (≈ cm s⁻¹, a generous Brownian drift)
Geomagnetic field	(B)	(5\times10^{-5}\,\text{T})
Perpendicular component (worst case)	(v_{\perp}=v)	–

[ F_L = e\,vB = (1.60\times10^{-19}\,\text{C})(10^{-2}\,\text{m s}^{-1})(5\times10^{-5}\,\text{T})
\boxed{F_L \approx 8\times10^{-26}\,\text{N}} \qquad (2) ]

2.5 Compare with other forces acting on a polaron

Thermal force: The thermal energy at room temperature is (k_{!B}T \approx 4\times10^{-21}\,\text{J}). Over an atomic distance (a\sim 3\times10^{-10}\,\text{m}) the corresponding “force scale” is

[ F_{\text{th}} \sim \frac{k_{!B}T}{a} \approx \frac{4\times10^{-21}}{3\times10^{-10}} \approx 1.3\times10^{-11}\,\text{N}. ]

Electric field that drives hopping: Typical hopping fields in ferrihydrite are (E \sim 10^{5}\,\text{V m}^{-1}). The electric force on an electron is

[ F_E = eE \approx 1.6\times10^{-19}\times10^{5} = 1.6\times10^{-14}\,\text{N}. ]

Both thermal and electric forces are 12–14 orders of magnitude larger than the magnetic force (2).

2.6 Effect on hopping probability

Polaron hopping rates are governed by an Arrhenius law

[ \Gamma = \Gamma_0 \exp!\left(-\frac{E_a - q\mathbf{E}\cdot\mathbf{a}}{k_{!B}T}\right), ]

where (E_a) is the activation energy and (\mathbf{a}) the hop vector.
A magnetic field can only enter through the Lorentz force, which would modify the trajectory of a moving charge but does not change the activation barrier because the magnetic force does no work ((\mathbf{F}_L\cdot\mathbf{v}=0)).

Hence, even if the tiny magnetic force were present, it would not bias the hop direction; it would merely curve the path, and the curvature radius is astronomically large (see below).

2.7 Radius of curvature for an electron in the geomagnetic field

If an electron were moving at (v=10^{-2}\,\text{m s}^{-1}) in a uniform (\mathbf{B}),

[ r = \frac{m_e v}{|q| B}. ]

Using the effective mass of a small polaron (≈ 10 (m_e) in ferrihydrite) :

[ r \approx \frac{10\,m_e\,v}{eB} = \frac{10(9.11\times10^{-31}\,\text{kg})(10^{-2}\,\text{m s}^{-1})}{1.6\times10^{-19}\,\text{C}(5\times10^{-5}\,\text{T})} \approx 1.1\times10^{5}\,\text{m}. ]

A curvature radius of ~100 km means the magnetic field does not noticeably bend the electron’s microscopic hop (≈ Å).

2.8 Putting it all together

Question	Answer
Does a Lorentz force act on the polarons?	Yes, in principle each electron experiences (\mathbf{F}=q\mathbf{v}_{\text{particle}}\times\mathbf{B}).
Is the force large enough to cause measurable motion inside the particle?	No. The magnitude is ~(10^{-25}) N, 12–14 orders smaller than thermal/electric forces, and the resulting curvature radius is ~100 km, far larger than the nanoparticle (tens of nanometers).
Does the particle as a whole feel a magnetic force?	No, because it is electrically neutral; forces on electrons cancel those on the accompanying holes.

Final answer:
A ferrihydrite nanoparticle moving through the Earth’s magnetic field does not experience any appreciable Lorentz‑force‑induced drift of its internal small polarons. The magnetic force on an individual electron is astronomically tiny compared with thermal and electric forces, and the particle’s overall neutrality makes the net magnetic force zero.

3. Common Mistakes

Mistake	Why it’s wrong	How to avoid it
Assuming the particle’s net charge is non‑zero	Ferrihydrite is electrically neutral; each hopping electron leaves a compensating positive hole.	Explicitly write down charge balance before applying (\mathbf{F}=q\mathbf{v}\times\mathbf{B}).
Using the polaron hopping velocity instead of the bulk velocity	The Lorentz force depends on the lab‑frame velocity. If no external field drives hopping, (\mathbf{v}_{\text{drift}}\approx0).	Separate (\mathbf{v}{\text{particle}}) and (\mathbf{v}{\text{drift}}) and insert the appropriate one into Eq. (1).
Neglecting the magnetic field’s inability to do work	A magnetic force can only change direction, not kinetic energy, so it cannot change hopping rates.	Remember that (\mathbf{F}_L\cdot\mathbf{v}=0).
Over‑estimating the particle’s speed	Typical drift speeds of nanoparticles are ≤ cm s⁻¹; using km s⁻¹ inflates the force by many orders.	Use realistic Brownian or sedimentation velocities (≈ 10⁻³–10⁻² m s⁻¹).
Comparing the Lorentz force to the activation energy directly	Force and energy have different units; you must compare a force to a force scale (e.g., thermal force (k_BT/a)).	Convert energies to forces using a characteristic length (≈ Å).
Ignoring the effective mass of a polaron	The curvature radius depends on mass; using the free‑electron mass underestimates (r).	Use the known polaron effective mass (≈ 5–20 (m_e) for iron oxides).

By checking each of these points you can reliably conclude that magnetic forces are negligible for small‑polaron transport in moving ferrihydrite nanoparticles.

Original question: Would small polarons in a ferrihydrite nanoparticle moving at velocity v in the geomagnetic field B experience a Lorentz force related to v cross B? on Physics Stack Exchange, licensed CC BY-SA.