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(C) When a charged particle enters a uniform magnetic field,it follows a circular path with radius $R = \frac{mv}{qB}$.The magnetic field exists in th

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$\frac{qB(b-a)}{m}$

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(C) When a charged particle enters a uniform magnetic field,it follows a circular path with radius $R = \frac{mv}{qB}$.The magnetic field exists in the region between $x = a$ and $x = b$. The width of this magnetic field region is $d = b - a$.For the particle to cross the magnetic field region and enter the region $x > b$,the radius of its circular path must be greater than the width of the magnetic field region.Therefore,$R > (b - a)$.Substituting the expression for $R$,we get $\frac{mv}{qB} > (b - a)$.Solving for $v$,we find $v > \frac{qB(b - a)}{m}$.

In the figure shown,a charge $q$ of mass $m$ moving with a velocity $v$ along the $x$-axis enters a region of uniform magnetic field $B$ directed into the page. If the particle is able to enter the region $x > b$,then the velocity $v$ must be greater than:

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