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(B) The trajectory of a charged particle in a region of a uniform magnetic field perpendicular to its velocity is a circle.The radius $R$ of this circ

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Ions with $q < \frac{2mV}{B^2\omega^2}$ will hit the screen

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(B) The trajectory of a charged particle in a region of a uniform magnetic field perpendicular to its velocity is a circle.The radius $R$ of this circular path is given by $R = \frac{mv}{Bq}$.The kinetic energy gained by the ion accelerated through a potential difference $V$ is $\frac{1}{2}mv^2 = qV$,which implies $v = \sqrt{\frac{2qV}{m}}$.Substituting $v$ into the expression for $R$,we get $R = \frac{m}{Bq} \sqrt{\frac{2qV}{m}} = \frac{\sqrt{2mqV}}{Bq} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.For the ion to hit the screen,the radius of its path must be greater than the width $\omega$ of the magnetic field region,i.e.,$R > \omega$.Therefore,$\frac{1}{B} \sqrt{\frac{2mV}{q}} > \omega$.Squaring both sides,we get $\frac{2mV}{B^2q} > \omega^2$.Rearranging for $q$,we find $q < \frac{2mV}{B^2\omega^2}$.

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