A particle, of mass m and with electric charge Q, falls vertically with a constant velocity v0. Under these conditions, the air resistance force can be considered as $R_{ar} = k \cdot v$, where $k$ is a constant and $v$ is the velocity. The particle then penetrates a region where a uniform and constant magnetic field B acts, perpendicular to the plane of the paper and entering it, as shown in the figure. The speed of the particle is then changed, acquiring, after a certain period of time, a new value vl, constant. (remember that the intensity of the magnetic force is $|F_M| = |q| |v| |B|$, in SI units for v perpendicular to B)
Sketch the force vectors (Weight, $R_ {ar}$ and $F_M$) that act on the particle, in the presence of the B field, in the situation where the speed becomes the speed $v_L$. Represent, by a dashed line, the direction and direction of $v_L$
Attemp: I had only been able to express the value of the particle velocity vl, in the region where the B field operates as a function of m, g, k, B and Q.
The tangential velocity has a constant modulus, so we have an MRU: the tangential acceleration is zero.
Because it is a uniform circular motion, there will be a constant centripetal acceleration: $ac = \frac{V^2}{R}$
With this there is a constant centripetal force ($F_c = m \cdot \frac{V^2}{R}$) with direction pointing towards the center of the path. And this does not influence the value of V
$$v = \frac{mg}{\sqrt{k^2+q^2B^2}}$$
The formula is correct, but the justification is wrong. Let's suppose that you have $\vec v_L$ making an angle $\theta$ with respect to $\vec g$ The air resistance will be opposite of $\vec v_L$, the magnetic force is perpendicular to $\vec v_L$.
If the particle would move on a circular trajectory with a constant velocity, the magnetic force would be always perpendicular to the trajectory and constant in magnitude, the air resistance will be always tangent (and constant in magnitude), and the gravity will be constant, but always pointing down. Therefore the sum of them will never be constant.
The other option you have is that at some point the particle will move in a straight line. With the requirement that the sum of the forces is zero, you get a constant velocity. If we write the sum of the forces along $v_L$ and perpendicular to that, we get: $$mg\cos\theta=kv_L\\mg\sin\theta=qv_LB$$ Squaring the equations and adding them together: $$m^2g^2=(k^2+q^2B^2)v_L^2$$ In the initial phase, without the magnetic field, the air resistance completely cancels gravity, so $$kv_0=mg$$ Putting it all together: $$v_L=\sqrt{\frac{m^2g^2}{\frac{m^2g^2}{v_0^2}+q^2B^2}}=\frac{v_0}{\sqrt{1+\left(\frac{qv_0B}{mg}\right)^2}}$$