Newton's third law in magnetic fields Say I have a charged particle moving through a magnetic field perpendicular to it. It will experience a force, but according to Newton third law

Every force has an equal and opposite reaction.

So what is the opposite reaction/force  of this magnetic force. 
Which body experiences this force?
 A: @Jon Custer is right if a magnet producing the magnetic field is present.
But there is more to learn of this question: As Hertz famously discovered, there are so called electro-magnetic waves.
These waves are made up of alternating electric and magnetic fields, that are unrelated to any physical object in the classical newtonian sense. This is different to the magnetic field of the magnet.
Since Newton's Third Law is very much equivalent to conservation of momentum I will concentrate on this formulation of Newton's theory.

*

*The downfall of classical conservation of momentum: These fields can of course exhibit force on a charged particle with
non-zero mass, very much like the magnetic field of the magnet.
Therefore the fields are changing the momentum of the particle. This
is the downfall of the classical concept of conservation of
momentum, since there is no other particle that can assert for the overall change of momentum of the entire system. By classical I mean that momentum is just \begin{equation}
\mathbf{p}=m\mathbf{v} \end{equation} and therefore only associated
with mass. This is the Newtonian view on monentum.

*Why momentum is still conserved in a broader sense: Experiments have
shown that the fields themselves or the electromagnetic wave for
that purpose carry momentum themselves. So the change in momentum of
the carged particle is compensated by the change of momentum of the
electromagnetic wave. To fully understand this concept you shouldy
study Maxwell's theory.

Remark: I edited large parts of this answer, as it didn't meet my quality standard anymore and caused misunderstandings in the comment section.
A: The answer by user224659 is correct but I want to focus on the second part and I hope bring some added clarity.
The magnetic force is an interaction between a charged particle and an electromagnetic field. The rate of change of the momentum of the charged particle is
$$
\frac{d{\bf p}}{dt} = q {\bf v} \times {\bf B}
$$
and the rate of change of momentum in the local electromagnetic field (at the location of the particle) is
$$
\frac{d{\bf p}_{\rm em}}{dt} = -q {\bf v} \times {\bf B}.
$$
There you have an example of an "action and reaction" pair, to use the terminology of Newton's third law, but writing it in terms of rate of change of momentum might make it easier to see what is happening to the electromagnetic field.
Notice that when the charge is accelerating, the electromagnetic field is also changing: its momentum is changing and therefore it is not completely static. This comes about because there is both the applied magnetic field (which is owing to other things, not the charge under discussion) and also the electric field caused by the charge we are thinking about. The combination of these leads to the changing momentum in the electromagnetic field. (Keep in mind that
in order to carry a non-zero momentum, the electromagnetic field does not necessarily have to be in a wavelike motion.)
When we apply the above facts in practice, we don't treat point particles but rather local concentrations of charge, with a finite amount of charge per unit volume, and then $q$ refers to the total charge of such a body. The formulae above apply when the radius of the charged body is small compared to other relevant distances in the physical situation, but not so small as to yield unphysical predictions for the field very near the charge.
A: A charge may, besides potential energy, experience potential momentum given by $q\vec A$. In the presence of currents therefore the kinetic momentum $mv$ is not conserved, but $mv+q\vec A$ is. The rate of change of this total momentum is equal and opposite for two particles that magnetically interact.
Note that this statement assumes that radiation effects are negligible, which is reasonable for the quasi static case.
