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(B) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path with radius $r = \frac{mv}{qB}$.T

Vedclass · Accepted Answer

$q(b-a)B/m$

Vedclass · Answer

(B) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it follows a circular path with radius $r = \frac{mv}{qB}$.The magnetic field exists in the region $a \leq x \leq b$. The width of this region is $d = b - a$.For the particle to emerge from the region $x > b$,the radius of the circular path $r$ must be greater than the width of the magnetic field region $d$.If $r \leq (b - a)$,the particle will be deflected back into the region $x  b$.Therefore,the condition to just enter the region $x > b$ is $r = b - a$.Substituting the expression for $r$,we get $\frac{mv}{qB} = b - a$.Solving for $v$,we find the minimum velocity $v = \frac{q(b - a)B}{m}$.

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