A Circulating Charged Particle — Question 4

Question 4

A finite magnetic slab may turn a particle back before it reaches the far boundary

A positive particle enters x=0 along +x into a field $-B\hat z$ occupying $0<x<d$. Define $r=mv/(qB)$ and $\omega=qB/m$. The field is zero outside. Ignore edge fringing.

1. For d<r, derive the first exit angle, transverse displacement and transit time.

2. Evaluate angle and displacement for d=.600r. Explain what happens if d>r.

Question 4 - Solution

A finite magnetic slab may turn a particle back before it reaches the far boundary

(a) Method and calculation

$$x=r\sin\phi,\quad y=r(1-\cos\phi),\quad\phi=\omega t.$$$$\phi_{\mathrm{exit}}=\sin^{-1}(d/r),\quad y_{\mathrm{exit}}=r-\sqrt{r^2-d^2},\quad t_{\mathrm{exit}}=\phi_{\mathrm{exit}}/\omega.$$

(b) Interpretation and further calculation

$$d=.600r:\quad\phi_{\mathrm{exit}}=36.8699^\circ,\quad y_{\mathrm{exit}}=.200r.$$$$d>r:\quad x_{\max}=r<d;\ \text{the particle returns to }x=0\text{ at }t=\pi/\omega.$$

Physical check

The field rotates the velocity without increasing the speed; penetration depth is limited by orbit geometry.