# How to prove that path of a charged particle in a uniform magnetic field is circular or helical? [duplicate]

I know that Lorentz force acts perpendicularly with the motion of the charged particle and curves the path into a circular one ,but how to show it mathematically?

Let's suppose we have a uniform magnetic field in the $$z$$ direction, $$\vec{B}=B_0\hat{z}$$ $$\\$$
Now let's give our particle a charge $$q$$, and an initial velocity vector $$\vec{v} = \begin{pmatrix} v_x \\ v_y \\ v_z \\ \end{pmatrix}$$
And we have the Lorentz force, $$\vec{F} = q\vec{v}\times\vec{B}$$
To make things easier, we're going to view things as seen from a frame where we are moving at $$\vec{v}_{frame} = v_z\hat{z}$$, so the new velocity of our particle is $$\vec{v}\space' = \begin{pmatrix} v_x \\ v_y \\ 0 \\ \end{pmatrix}$$ What makes this useful is that is it perpendicular to the magnetic field, so our cross product just becomes the product of the magnitudes of $$\vec{B}$$ and $$\vec{v}\space'$$, $$F = qv'B$$ where $$x = \lvert\vec{x}\rvert$$ We know, due to the properties of the cross product, that $$\vec{F}$$ is perpendicular to the velocity vector $$\vec{v}\space'$$, and the magnetic field vector $$\vec{B}$$. This means the force constant, is in the x-y place, and always perpendicular to velocity in that plane, which is sufficient for circular motion. The radius of this motion, you can determine using the formula for centripetal force, $$\vec{F} = \frac{mv^2}{r}$$ Just make sure you use the right velocity!  So now we know that the motion is circular in the reference where we're travelling at $$v_z\hat{z}$$, all we have to do is transform back, and now our circle isn't staying in the same place, it's moving, so we have helical motion!
A quick disclaimer: if $$v_z$$ is near the speed of light, this derivation with the frame transform is a bit shake, since relativity comes into play, but in most cases this is fine.