# How does the motion of a charged particle in a magnetic field with the right-hand rule obey mirror symmetry?

In The New Cosmic Onion, I read the following:

If mirror symmetry were an exact property of Nature, it would be impossible to tell whether a film of an experiment has been made directly or by filming the view in a mirror in which the experiment has been reflected. This is equivalent to saying that Nature does not distinguish between left and right in an absolute way. This is the case for phenomena controlled by gravity, the electromagnetic, or strong forces. As these control most observed phenomena, it had been assumed that left-right symmetry is an inherent property of all subatomic processes. But in 1956 mirror symmetry was discovered to be broken in weak interactions.

How does the following situation with electromagnetism follow mirror symmetry? A positively charged particle travels upwards in the plane of the screen, and a magnetic field points to the right in the plane of the screen. By the right-hand rule, we see that the particle experiences a force into the screen. However, when we reflect this situation left-to-right across a plane perpendicular to the screen, the right-hand rule then tells us that the particle experiences a force out of the plane of the screen.

So what gives?

When in doubt, it is typically helpful to analyze the reflection properties of magnetic fields in terms of a plausible source for them. In your case, you can think of the magnetic field as being generated by a ring of current in a plane that's parallel to the reflection plane (i.e. containing the directions 'up' and 'into the screen'). This current is not affected by the reflection $$-$$ so, therefore, neither is the magnetic field.
• @Emilio Pisanty Under parity (${\bf r}\mapsto -{\bf r}$) the field ${\bf B}$ does not change sign, but the OP is talking about a mirror in which $(x,y,z)\mapsto (-x,y,z)$ and not the usual parity operation. The mirror image of a LH screw is a RH screw. – mike stone Jun 25 at 21:02
• Coordinates are useless if you don't define them. Assuming the plane of the screen is the $x,y$ plane, with $y$ the vertical, then yes, the OP's mirror does $(x,y,z)\mapsto(-x,y,z)$, but $B_x$ is not affected. Your solenoid has currents confined to the $y,z$ plane, so they are not touched by the reflection. – Emilio Pisanty Jun 25 at 21:06