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

Question

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?

Answer 1

I've put down an answer that should help you but doesn't dive deeply into the mathematics, if you're feeling confident I would suggest trying to solve it using equations of motion and coupled differential equations, but if not, I think what's below should lead you in the right direction.

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.