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(B) When a charged particle enters a perpendicular magnetic field,it follows a circular path of radius $r$ given by the balance between the magnetic f

Vedclass · Accepted Answer

$q(b - a)B/m$

Vedclass · Answer

(B) When a charged particle enters a perpendicular magnetic field,it follows a circular path of radius $r$ given by the balance between the magnetic force and the centripetal force:$qvB = \frac{mv^2}{r}$Rearranging for velocity,we get:$v = \frac{qBr}{m}$For the particle to just enter the region $x > b$,the radius of its circular path must be at least equal to the width of the magnetic field region,which is $(b - a)$.Thus,the minimum radius required is $r_{\min} = b - a$.Substituting this into the velocity equation:$v_{\min} = \frac{qB(b - a)}{m}$Therefore,the correct option is $B$.

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