A particle with charge q,moving with a moment | vedclass

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Question

(A) When a charged particle enters a uniform magnetic field perpendicularly,it follows a circular path of radius $r = \frac{mv}{qB} = \frac{p}{qB}$.Fr

Vedclass · Accepted Answer

$\sin 	heta = \frac{Bqd}{p}$

Vedclass · Answer

(A) When a charged particle enters a uniform magnetic field perpendicularly,it follows a circular path of radius $r = \frac{mv}{qB} = \frac{p}{qB}$.From the geometry of the path,the particle travels a horizontal distance $d$ within the magnetic field region.In the right-angled triangle formed by the radius $r$,the horizontal distance $d$,and the deflection angle $	heta$,we have $\sin 	heta = \frac{d}{r}$.Substituting the value of $r$,we get $\sin 	heta = \frac{d}{p / (qB)} = \frac{Bqd}{p}$.Thus,the correct option is $A$.

$A$ 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.

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