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(B) The particle moves in a straight line until it reaches $x = a$. Upon entering the magnetic field region $(a \le x \le b)$,the magnetic force acts

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$q(b - a)B/m$

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(B) The particle moves in a straight line until it reaches $x = a$. Upon entering the magnetic field region $(a \le x \le b)$,the magnetic force acts as a centripetal force,causing the particle to move in a circular path of radius $r = \frac{mv}{qB}$.For the particle to just reach the region $x > b$,the radius of the circular path must be at least equal to the width of the magnetic field region,which is $(b - a)$.Therefore,the condition is $r \ge (b - a)$.Substituting the expression for $r$,we get $\frac{mv}{qB} \ge (b - a)$.Solving for $v$,we find $v \ge \frac{q(b - a)B}{m}$.Thus,the minimum velocity required is $v_{\min} = \frac{q(b - a)B}{m}$.

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