How is "time inversion symmetry" violated in Hall effect?

Question

I recently read this

"It (the Hall effect) describes the transverse deflection of moving conduction electrons in a Hall bar when time-reversal symmetry is broken."

in an article. As far as I know, the Hall effect is the formation of a voltage difference across an electrical conductor, when a magnetic field is applied transverse to the current direction. This is generated due to the deflection of the electron by the Lorentz force.

$$\mathbf{F} = q \left( \mathbf{E} + \mathbf{V} \times \mathbf{B} \right)$$

When time is reversed,

$$ \mathbf{F}' = q \left( \mathbf{E} + \mathbf{(-V)} \times \mathbf{(-B)} \right) $$ $$ \mathbf{F}' = \mathbf{F} $$

So basically, even when the current, velocity of charge carriers ($\mathbf{V}$), and magnetic field $(\mathbf{B})$ are reversed. The direction of the force $\mathbf{F}$ will remain the same and the Hall voltage will remain in the same direction.

for example, if the current is in the $+X$ direction, (electron velocity along the $-X$ direction) and the magnetic field is in the $+Z$ direction, then the force will be along the $+Y$ direction. If we reverse time, electron velocity will be along $+X$ and the field will be along $-Z$, and the force will remain the same along $+Y$.

Where exactly do we break TR here?

Answer 1

It is true that Maxwell's equations exhibit TRS. However, when thinking about the Hall effect, you have to consider the external magnetic field as EXTERNAL (fixed) and thus does not get affected by the time reversal operation, and thus the presence of an external magnetic field itself breaks time reversal (do your same calculation except keep B fixed, then the forces are different). So, if time reversal symmetry is present, the external field must be zero and thus there will be no Hall effect. Another way to see this is that if you flip the magnetic field, the Hall voltage switches sign.