How do I find the radius of motion of this particle? [closed]

Question

A charged particle $Q$ centers a region of uniform magnetic field $\vec B=B\hat x$ at the origin, moving in the positive $z$-direction. It follows a circular arc of less than $\frac{\pi}{2}$ and is observed later at $\vec r= a \hat{x}+ b\hat{y}$. Find the momentum of this particle.

I derived $p=qBr$ from the centripetal force and Lorentz force equations.

I'm having trouble identifying $r$. I was thinking of using a triangle and therefore

$$\tan\frac{\pi}{2}= \frac{\sqrt{a^2+b^2}}{r}$$

but that doesn't seem correct as it would end up being undefined.

I was also thinking of taking the limit as the angle approaches pi/2 but then r would approach 0.

Besides these, I'm fresh out of ideas

Any help or hints are greatly appreciated!

Answer 1

There are two phases to the particle's motion:

1. Circular motion inside the magnetic field.
2. Straight-line motion towards the detector at $(a,b)$.

The first step in most physics problems is to draw a picture:

The particle (with mass $m$, velocity $v$, momentum $p$, and charge $Q$) enters the magnetic field from below at the origin $(0,0)$. The magnetic field bends the particle's velocity in an arc of radius $r$ and angular distance $\theta$. The particle exits the magnetic field at the point labeled $\vec{r}_{\textrm{exit}}$. The particle then takes a straight-line path (since it is no longer in the magnetic field) to the point labeled $(a,b)$. All this just restates the problem.

Ultimately, you want to find an equation for $p$ in terms of quantities you are given ($Q$, $B$, $a$, $b$). The link between these quantities is the exit point: from the origin to the exit point involves the magnetic field equations, and from the exit point to the detection point involves the rotated velocity and the exit point.