Q.
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 $d < \frac{p}{Bq}$. If 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) $\sin\theta = \frac{pd}{Bq}$
Explanation
(a) From figure it is clear that
$\sin\theta = \frac{d}{r}$ also $r = \frac{p}{qB}$
$\therefore\sin\theta = \frac{Bqd}{p}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.