# It seems like a silly question but doesnt the magnetic lorentz force increase with increasing velocity

We know the equation $$F = q \left( v \times B \right)$$.

So force Lorentz is directly proportional to the component of velocity perpendicular to the magnetic field vector. So an increase in velocity must lead to an increase in magnetic Lorentz force.

We also know that in a uniform magnetic field, a charged particle moving with a velocity $$v$$ executes uniform circular motion with a radius $$r = \frac{m v}{q B}$$ where $$q$$ is charge $$B$$ is the magnetic field, $$m$$ is the mass of the particle and $$v$$ is the velocity.

So by that relation, an increase in velocity must lead to an increase in radius. But the same increase in velocity increases Lorentz magnetic force which leads to a decrease in radius. So does the radius stay constant? But then how would instruments like a cyclotron work?

"But the same increase in velocity increases lorentz magnetic force which leads to a decrease in radius."

It doesn't lead to a decrease in radius because the centripetal force needed to keep the particle moving in a circle is of magnitude $$mv^2/r$$, so although doubling $$v$$ doubles the magnetic Lorentz force, it quadruples the force that would be needed to keep the charge moving in a circle of the same radius. The circle radius has to double.

The correct equation that you quote, $$r=mv/qB$$, emerges simply by equating the centripetal force needed ($$mv^2/r$$) to the magnetic Lorentz force ($$Bqv$$)!

So by that relation an increase in velocity must lead to a increase in radius.

Yes.

But the same increase in velocity increases lorentz magnetic force which leads to a decrease in radius.

No, it is growing too slow for that.

Yes, the correct Abraham-Lorentz force law, ignoring the electric field part, is $$\vec F=q\,\vec v\times\vec B$$ And considering perpendicular case for simplicity, it is $$F=qvB$$

If you want to have circular motion, you must have centripetal acceleration. i.e. $$F=\frac{mv^2}r=qvB$$ That is, if you want the radius to even just stay constant, the force should be proportional to the square of the velocity. But it is only proportional to the velocity, not the square, and so it will not be able to do that.

The only way to fix this, is to have the radius be linearly growing with velocity. Indeed, $$\frac{mv}r=qB\qquad\implies\qquad r=\frac{mv}{qB}$$

For a charged particle in a uniform magnetic field, the radius of motion is completely determined by its velocity, which does not change. Left to its own devices, this charged particle will follow a circular (or helical) path indefinitely.

A cyclotron uses (oscillating) electric fields to accelerate the charged particles; the magnetic fields are effectively just there to keep the particle confined long enough to get up to speed. Effectively, the magnetic field exists in two hollow D-shaped cavities (called, rather uncreatively, "dees"), which can be held at different potentials. By alternating the voltage of the dees, we can make it so that a particle always travels from a dee at higher potential to one at lower potential, gaining energy in the transition. See here for a bit more detailed description.