question_answer

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) \[\sin \theta =\frac{pd}{BQ}\]

Correct Answer: A

Solution :

[a] : A to D is part of circle with centre C and radius CD = r.                \[mv=p=BQr\]or\[r=\frac{p}{BQ}\] \[\sin \theta =\frac{ED}{CD}=\frac{d}{r}=\frac{BQd}{p}\]