Question

$A$ 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. Then

• A. $sin\,\theta =\frac{BQD}{P}$
• B. $sin\,\theta =\frac{P}{BQd}$
• C. $sin\,\theta =\frac{BP}{Qd}$
• D. $sin\,\theta =\frac{Pd}{BQ}$

Answer 1

Answer: (A)

$A$ to $D$ is part of circle with centre $C$ and radius $CD = r$
$mv=p=BQr$ or $r=\frac{P}{BQ}$
$sin\, \theta = \frac{ED}{CD}=\frac{d}{r}$
$=\frac{BQd}{P}$