A particle with charge $q$, moving with a momentum $p$, enters a uniform magnetic field normally. The magnetic field has magnitude $B$ and is confined to a region of width $d$, where $d < \frac{p}{{Bq}}$, The particle is deflected by an angle $\theta $ in crossing the field
Diffcult
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(a) From figure it is clear that
$\sin \theta = \frac{d}{r}$ also $r = \frac{p}{{qB}}$
$\therefore $$\sin \theta = \frac{{Bqd}}{p}$
$\sin \theta = \frac{d}{r}$ also $r = \frac{p}{{qB}}$
$\therefore $$\sin \theta = \frac{{Bqd}}{p}$
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