Question

A particle with charge $q$ having momentum $p$ enters a uniform magnetic field normally. The magnetic filed 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.

• A. $\sin \theta =\frac{pd}{Bq}$
• B. $\sin \theta =\frac{Bq}{qd}$
• C. $\sin \theta =\frac{p}{Bdq}$
• D. $\sin \theta =\frac{Bqd}{p}$

Answer 1

The correct option is D $\sin \theta =\frac{Bqd}{p}$

The condition can be shown as
where, $r$ is the radius of circular path.
From figure it is clear that
$\sin \theta =\frac{d}{r}$
also, $r=\frac{p}{qB}$
$\therefore \sin \theta =\frac{d}{p/qB}=\frac{Bqd}{p}$.