If a charged particle moves in magnetic field it experiences Lorentz force given as : $$\vec F = q \vec v \times \vec B$$
But I couldn't find a place where a reaction force as per Newton's third law could act.
Does it act upon source creating magnetic field or something else?
Force is defined as: $$\vec{F}=\frac{d}{dt}\vec{p}$$ So the Newton's third law implies the conservation of linear momentum. In general, for a zero net force one can write: $$\sum_{k}F_{k}=\frac{d}{dt}\sum_{k}p_{k}=0$$ So $\sum_{k}p_{k} =$ constant.
Just as the field has energy density, it also has a momentum density. Let's call the momentum density $\mathcal{P}$. Knowing that the Poynting vector is defined as the energy transfer per unit area per unit time of an electromagnetic field, we can define the momentum density of the electromagnetic field as: $$\mathcal{P}=\frac{1}{c^2}S$$ where $c$ is the speed of light and $S$ is the Poynting vector. So there is an exchange of momentum between the charged particle and the electromagnetic field.
First you have to know that a particle with some velocity $\vec v$ and some charge $q$ in presence of a magnetic field $\vec B_0$ perform a uniform circular motion. And the magnetic field does not make work at all. This means that the speed $v$ of the particle has no change. (Because of the energy and work theorem)
Because of the circular motion, the charged particle generates a magnetic field $\vec B _q$ that is exactly equal in magnitude and in oposite direction to the field $\vec B_0$.
This is a little tricky way to explain the Newton's third law, but if you know the source of the magnetic field $\vec B_0$, you can easy use the common formula $\vec F_{12}=\vec F_{21}$